Chapter 7 Permutations And Combinations

The given word is ROSE.

The letters in the word ROSE are R, O, S, E. There are 4 distinct letters.

We need to form 4-letter words using these letters.

The repetition of letters is not allowed.

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Since the order of the letters matters and repetition is not allowed, this is a problem of permutations.

We need to find the number of permutations of 4 distinct letters taken all at a time ($n=4$, $r=4$).

The number of permutations of $n$ distinct objects taken $r$ at a time is given by the formula: $P(n, r) = \frac{n!}{(n-r)!}$

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In this case, $n=4$ (number of distinct letters) and $r=4$ (number of letters in the word to be formed).

Number of 4-letter words = $P(4, 4)$

$P(4, 4) = \frac{4!}{(4-4)!} = \frac{4!}{0!}$

We know that $0! = 1$.

So, $P(4, 4) = \frac{4!}{1} = 4!$

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Now, calculate the factorial of 4:

$4! = 4 \times 3 \times 2 \times 1$

$4! = 12 \times 2 \times 1$

$4! = 24 \times 1$

$4! = 24$

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Therefore, the number of 4-letter words that can be formed from the letters of the word ROSE without repetition is 24.

We have 4 flags of different colours.

A signal is generated by using 2 flags, placed one below the other.

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This problem involves selecting 2 flags out of 4 and arranging them in a specific order (one below the other). The order matters (the top flag is different from the bottom flag).

This is a problem of permutations of $n$ distinct objects taken $r$ at a time.

Here, $n$ is the total number of distinct flags, so $n=4$.

$r$ is the number of flags used in a signal, so $r=2$.

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The number of permutations of $n$ distinct objects taken $r$ at a time is given by the formula: $P(n, r) = \frac{n!}{(n-r)!}$

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We need to calculate $P(4, 2)$:

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Now, calculate the factorials:

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Substitute these values back into the formula:

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Alternatively, we can think of this using the Fundamental Principle of Counting.

For the top position, there are 4 choices of flags.

After choosing a flag for the top position, there are 3 remaining flags for the bottom position.

Total number of signals = (Choices for top) $\times$ (Choices for bottom)

Total number of signals = $4 \times 3 = 12$

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Therefore, the number of different signals that can be generated is 12.

We need to form 2-digit even numbers using the digits 1, 2, 3, 4, 5.

A 2-digit number has two places: the tens place and the units place.

For the number to be even, the digit in the units place must be an even digit from the given set of digits.

The even digits in the set {1, 2, 3, 4, 5} are 2 and 4.

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Let the 2-digit number be represented as $\text{_}\text{_}$, where the first blank is the tens place and the second blank is the units place.

Consider the units place:

The digit in the units place must be even. The available even digits are 2 and 4.

Number of choices for the units place = 2 (either 2 or 4).

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Consider the tens place:

The digits can be repeated, so any of the given digits {1, 2, 3, 4, 5} can be used for the tens place.

Number of choices for the tens place = 5 (1, 2, 3, 4, or 5).

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Using the Fundamental Principle of Counting, the total number of 2-digit even numbers is the product of the number of choices for each place.

Total number of even numbers = (Choices for tens place) $\times$ (Choices for units place)

Total number of even numbers = $5 \times 2 = 10$

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The 10 possible numbers are: 12, 14, 22, 24, 32, 34, 42, 44, 52, 54.

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Therefore, the number of 2-digit even numbers that can be formed from the digits 1, 2, 3, 4, 5 with repetition allowed is 10.

Given:

• Number of different flags available = 5
• A signal requires arranging at least 2 flags in order on a vertical staff.
• Repetition of flags is not allowed (since the flags are distinct and arranged one below the other).

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To Find:

The total number of different signals that can be generated.

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Solution:

The condition "at least 2 flags" means a signal can be formed using 2 flags, or 3 flags, or 4 flags, or 5 flags from the available 5 distinct flags.

Since the order of flags in a signal matters and repetition is not allowed, this is a problem of finding the number of permutations of 5 distinct objects taken $r$ at a time, where $r$ can be 2, 3, 4, or 5.

The number of permutations of $n$ distinct objects taken $r$ at a time is given by the formula:

Here, $n=5$ (total number of flags).

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Case 1: Using 2 flags ($r=2$)

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Case 2: Using 3 flags ($r=3$)

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Case 3: Using 4 flags ($r=4$)

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Case 4: Using 5 flags ($r=5$)

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The total number of different signals is the sum of the numbers of signals from each case, because these cases are mutually exclusive (a signal cannot use exactly 2 flags and exactly 3 flags at the same time).

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Alternatively, using the Fundamental Principle of Counting:

For a signal using 2 flags:

Choices for the top flag: 5

Choices for the bottom flag: 4 (since it must be different)

Number of signals with 2 flags = $5 \times 4 = 20$

For a signal using 3 flags:

Choices for 1st flag: 5

Choices for 2nd flag: 4

Choices for 3rd flag: 3

Number of signals with 3 flags = $5 \times 4 \times 3 = 60$

For a signal using 4 flags:

Choices for 1st flag: 5

Choices for 2nd flag: 4

Choices for 3rd flag: 3

Choices for 4th flag: 2

Number of signals with 4 flags = $5 \times 4 \times 3 \times 2 = 120$

For a signal using 5 flags:

Choices for 1st flag: 5

Choices for 2nd flag: 4

Choices for 3rd flag: 3

Choices for 4th flag: 2

Choices for 5th flag: 1

Number of signals with 5 flags = $5 \times 4 \times 3 \times 2 \times 1 = 120$

Total signals = $20 + 60 + 120 + 120 = 320$.

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Therefore, the number of different signals that can be generated by arranging at least 2 flags from five different flags is 320.

The given digits are 1, 2, 3, 4, and 5. There are a total of 5 distinct digits available.

We need to form 3-digit numbers.

A 3-digit number has three places: the hundreds place, the tens place, and the units place.

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(i) Repetition of the digits is allowed:

For the hundreds place, there are 5 choices (any of the digits 1, 2, 3, 4, 5).

For the tens place, since repetition is allowed, there are still 5 choices (any of the digits 1, 2, 3, 4, 5).

For the units place, since repetition is allowed, there are still 5 choices (any of the digits 1, 2, 3, 4, 5).

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Using the Fundamental Principle of Counting, the total number of 3-digit numbers is the product of the number of choices for each place.

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Thus, if repetition of digits is allowed, 125 different 3-digit numbers can be formed.

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(ii) Repetition of the digits is not allowed:

For the hundreds place, there are 5 choices (any of the digits 1, 2, 3, 4, 5).

For the tens place, since repetition is not allowed, one digit has already been used for the hundreds place. So, there are $5 - 1 = 4$ choices remaining for the tens place.

For the units place, since repetition is not allowed, two distinct digits have already been used for the hundreds and tens places. So, there are $5 - 2 = 3$ choices remaining for the units place.

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Using the Fundamental Principle of Counting, the total number of 3-digit numbers is the product of the number of choices for each place.

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Alternatively, this is a permutation problem of selecting 3 digits out of 5 distinct digits and arranging them in order. The number of permutations of $n$ distinct objects taken $r$ at a time is given by $P(n, r) = \frac{n!}{(n-r)!}$.

Here, $n=5$ (total number of digits) and $r=3$ (number of digits in the number).

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Thus, if repetition of digits is not allowed, 60 different 3-digit numbers can be formed.

The given digits are 1, 2, 3, 4, 5, and 6. There are a total of 6 distinct digits available.

We need to form 3-digit even numbers.

A 3-digit number has three places: the hundreds place, the tens place, and the units place.

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For a number to be even, the digit in the units place must be an even digit from the given set.

The even digits in the set {1, 2, 3, 4, 5, 6} are 2, 4, and 6.

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Let the 3-digit number be represented as $\text{_}\text{_}\text{_}$, where the blanks represent the hundreds, tens, and units places, respectively.

Consider the units place:

The digit in the units place must be one of the even digits {2, 4, 6}.

Number of choices for the units place = 3.

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Consider the tens place:

Since repetition of digits is allowed, any of the 6 given digits {1, 2, 3, 4, 5, 6} can be used for the tens place.

Number of choices for the tens place = 6.

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Consider the hundreds place:

Since repetition of digits is allowed, any of the 6 given digits {1, 2, 3, 4, 5, 6} can be used for the hundreds place.

Number of choices for the hundreds place = 6.

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Using the Fundamental Principle of Counting, the total number of 3-digit even numbers is the product of the number of choices for each place.

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Thus, if repetition of digits is allowed, 108 different 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6.

Given:

• The first 10 letters of the English alphabet are A, B, C, D, E, F, G, H, I, J.
• Total number of distinct letters available ($n$) = 10.
• We need to form a 4-letter code. This means we select and arrange 4 letters ($r = 4$).
• No letter can be repeated.

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To Find:

The number of different 4-letter codes that can be formed.

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Solution:

Since we are forming a code, the order of the letters matters (e.g., 'ABCD' is different from 'ACBD'). Also, repetition is not allowed.

This is a problem of finding the number of permutations of 10 distinct objects taken 4 at a time.

The number of permutations of $n$ distinct objects taken $r$ at a time is given by the formula:

Here, $n=10$ and $r=4$.

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Using the formula (i):

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Expand the factorials:

Cancel out the common terms:

Calculate the product:

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Alternatively, using the Fundamental Principle of Counting:

We need to fill 4 positions for the 4-letter code.

• For the first position, there are 10 choices (any of the 10 letters).
• For the second position, since no repetition is allowed, there are 9 choices remaining.
• For the third position, there are 8 choices remaining.
• For the fourth position, there are 7 choices remaining.

Total number of codes = (Choices for 1st) $\times$ (Choices for 2nd) $\times$ (Choices for 3rd) $\times$ (Choices for 4th)

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Therefore, the number of different 4-letter codes that can be formed using the first 10 letters of the English alphabet without repetition is 5040.

Given information:

• The digits available are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. There are 10 distinct digits in total.
• We need to construct a 5-digit telephone number.
• Each number must start with 67.
• No digit can appear more than once (repetition is not allowed within the 5 digits of the number).

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The telephone number has 5 digits, let's represent the positions as follows:

Position 1 Position 2 Position 3 Position 4 Position 5

We are given that the number starts with 67. This means:

• The digit in Position 1 is fixed as 6.
• The digit in Position 2 is fixed as 7.

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Since no digit appears more than once, the digits 6 and 7 cannot be used in the remaining positions (Position 3, Position 4, and Position 5).

The original set of digits is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.

The digits 6 and 7 have been used. The remaining available digits are {0, 1, 2, 3, 4, 5, 8, 9}.

Number of remaining available digits = $10 - 2 = 8$ digits.

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We need to fill the remaining 3 positions (Position 3, Position 4, and Position 5) using these 8 available digits, without repetition.

Since the order of the digits matters in a telephone number and repetition is not allowed among the available digits, this is a problem of finding the number of permutations.

We need to find the number of permutations of 8 distinct objects (the remaining digits) taken 3 at a time (for the 3 remaining positions).

The number of permutations of $n$ distinct objects taken $r$ at a time is given by the formula: $\mathbf{P(n, r) = \frac{n!}{(n-r)!}}$

Here, $n=8$ (number of available digits) and $r=3$ (number of positions to fill).

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Using the permutation formula:

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Expand the factorials and simplify:

Calculate the product:

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Alternatively, using the Fundamental Principle of Counting for the remaining 3 positions:

• For the 3rd position, there are 8 choices (any of the remaining 8 digits).
• For the 4th position, since one digit has been used for the 3rd position, there are $8 - 1 = 7$ choices remaining.
• For the 5th position, since two distinct digits have been used for the 3rd and 4th positions, there are $8 - 2 = 6$ choices remaining.

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Since the first two digits (6 and 7) are fixed, there is only 1 way to choose and place them in the first two positions.

The total number of 5-digit telephone numbers is the product of the number of ways to fill each position.

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Therefore, the number of 5-digit telephone numbers that can be constructed using the digits 0 to 9, starting with 67 and with no digit appearing more than once, is 336.

Given that a coin is tossed 3 times.

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For a single toss of a coin, there are two possible outcomes:

• Head (H)
• Tail (T)

So, the number of possible outcomes for a single toss is 2.

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Since the outcome of each toss is independent of the others, we can use the Fundamental Principle of Counting to find the total number of possible outcomes for 3 tosses.

The Fundamental Principle of Counting states that if an event can occur in $m$ ways and another independent event can occur in $n$ ways, then the two events can occur in $m \times n$ ways.

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Applying this principle to the coin tosses:

• Number of outcomes for the 1st toss = 2 (H or T)
• Number of outcomes for the 2nd toss = 2 (H or T)
• Number of outcomes for the 3rd toss = 2 (H or T)

Total number of possible outcomes = (Outcomes for 1st toss) $\times$ (Outcomes for 2nd toss) $\times$ (Outcomes for 3rd toss)

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The possible outcomes can be listed as follows:

• HHH
• HHT
• HTH
• THH
• HTT
• THT
• TTH
• TTT

There are 8 distinct outcomes.

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Therefore, the number of possible outcomes when a coin is tossed 3 times is 8.

Given:

• Number of different flags available = 5.
• Each signal requires the use of 2 flags arranged one below the other.

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To Find:

The number of different signals that can be generated.

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Solution:

We need to select 2 flags out of 5 distinct flags and arrange them in a specific order (top and bottom position). Since the order matters and repetition is not allowed (as the flags are distinct and used in two different positions for the signal), this is a problem of permutations.

We need to find the number of permutations of $n$ distinct objects taken $r$ at a time.

Here, $n$ is the total number of distinct flags, so $n=5$.

$r$ is the number of flags used in a signal, so $r=2$.

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Using the Fundamental Principle of Counting:

A signal consists of two positions: the top flag and the bottom flag.

• For the top position, there are 5 choices (any of the 5 available flags).
• For the bottom position, since one flag has been used for the top position and repetition is not allowed, there are $5 - 1 = 4$ choices remaining.

By the Fundamental Principle of Counting, the total number of different signals is the product of the number of choices for each position:

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Alternatively, using the Permutation Formula:

The number of permutations of $n$ distinct objects taken $r$ at a time is given by the formula:

Here, $n=5$ and $r=2$.

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We need to calculate $P(5, 2)$:

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Expand the factorials:

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Substitute these values back into the formula:

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Both methods give the same result.

Therefore, the number of different signals that can be generated using 2 flags out of 5 different flags is 20.

We are asked to evaluate the given factorial expressions.

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The factorial of a non-negative integer $n$, denoted by $n!$, is the product of all positive integers less than or equal to $n$. Mathematically, it is defined as:

Also, by definition, $0! = 1$.

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(i) Evaluate 5!

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(ii) Evaluate 7!

We can also write $7!$ in terms of $5!$ as $7! = 7 \times 6 \times (5 \times 4 \times 3 \times 2 \times 1) = 7 \times 6 \times 5!$

Using the value of $5!$ from part (i):

Let's perform the multiplication:

$\begin{array}{cc}& & 1 & 2 & 0 \\ \times & & & 4 & 2 \\ \hline && 2 & 4 & 0 \\ & 4 & 8 & 0 & \times \\ \hline 5 & 0 & 4 & 0 \\ \hline \end{array}$

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(iii) Evaluate 7! – 5!

Using the values calculated in parts (i) and (ii):

Let's perform the subtraction:

$\begin{array}{cc} & 5 & 0 & 4 & 0 \\ - & & 1 & 2 & 0 \\ \hline & 4 & 9 & 2 & 0 \\ \hline \end{array}$

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Alternatively, we can factor out $5!$:

Using the value of $5! = 120$:

Let's perform the multiplication:

$\begin{array}{cc}& & 1 & 2 & 0 \\ \times & & & 4 & 1 \\ \hline && 1 & 2 & 0 \\ & 4 & 8 & 0 & \times \\ \hline 4 & 9 & 2 & 0 \\ \hline \end{array}$

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The results are:

• $5! = \mathbf{120}$
• $7! = \mathbf{5040}$
• $7! - 5! = \mathbf{4920}$

We are asked to compute the values of the given factorial expressions.

The factorial of a non-negative integer $n$, denoted by $n!$, is defined as $n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1$, with $0! = 1$.

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(i) Compute $\frac{7!}{5!}$

We can write $7!$ in terms of $5!$ as follows:

Now substitute this into the expression:

Cancel out the common term $5!$ from the numerator and the denominator:

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(ii) Compute $\frac{12!}{(10!) (2!)}$

We can write $12!$ in terms of $10!$ as follows:

Also, calculate $2!$:

Now substitute (b) and (c) into the expression:

Cancel out the common term $10!$:

Simplify the expression:

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The results are:

• $\frac{7!}{5!} = \mathbf{42}$
• $\frac{12!}{(10!) (2!)} = \mathbf{66}$

We are asked to evaluate the expression $\frac{n!}{r!(n-r)!}$ for $n=5$ and $r=2$.

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Substitute the given values of $n$ and $r$ into the expression:

Simplify the term in the parenthesis:

So the expression becomes:

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Now, calculate the factorial values:

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Substitute these values back into the expression:

Perform the division:

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Alternatively, we can simplify the factorial expression before calculating the full factorials:

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The value of the expression $\frac{n!}{r!(n-r)!}$ when $n=5$ and $r=2$ is 10.

We are given the equation:

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We can express the larger factorials in terms of the smallest factorial present, which is $8!$.

Recall that $n! = n \times (n-1)!$.

So, we have:

Substitute (ii) into (iii):

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Now substitute the expressions for $9!$ and $10!$ from (ii) and (iv) into the original equation (i):

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To eliminate the factorials from the denominators, multiply both sides of equation (v) by $90 \times 8!$:

Distribute on the left side and simplify:

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Alternatively, find a common denominator on the left side of equation (i):

So, the equation becomes:

We know that $10! = 10 \times 9! = 10 \times 9 \times 8!$.

Substitute this into the right side of (vi):

Multiply both sides by $10 \times 9 \times 8!$:

Simplify:

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The value of $x$ is 100.

We are asked to evaluate the given factorial expressions.

The factorial of a non-negative integer $n$, denoted by $n!$, is the product of all positive integers less than or equal to $n$. Mathematically, it is defined as $n! = n \times (n-1) \times (n-2) \times \cdots \times 3 \times 2 \times 1$, with $0! = 1$.

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(i) Evaluate 8!

We can group the terms for easier calculation:

We know that $5! = 120$. So,

Let's perform the multiplication:

$\begin{array}{cc}& & 3 & 3 & 6 \\ \times & & 1 & 2 & 0 \\ \hline && 0 & 0 & 0 \\ & 6 & 7 & 2 & \times \\ 3 & 3 & 6 & \times & \times \\ \hline 4 & 0 & 3 & 2 & 0 \\ \hline \end{array}$

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(ii) Evaluate 4! – 3!

First, calculate the individual factorials:

Now, perform the subtraction:

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Alternatively, we can factor out the smaller factorial:

Using the value $3! = 6$:

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The results are:

• $8! = \mathbf{40320}$
• $4! - 3! = \mathbf{18}$

We need to check if the equation $3! + 4! = 7!$ is true.

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First, evaluate the factorials on the left side of the equation:

Now, calculate the sum of the factorials on the left side:

So, the left side of the equation is $30$.

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Next, evaluate the factorial on the right side of the equation:

We know that $4! = 24$, so we can write:

Let's perform the multiplication:

$\begin{array}{cc}& & 2 & 1 & 0 \\ \times & & & 2 & 4 \\ \hline & & 8 & 4 & 0 \\ 4 & 2 & 0 & \times & \times \\ \hline 5 & 0 & 4 & 0 \\ \hline \end{array}$

So, the right side of the equation is $5040$.

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Now, compare the values of the left side and the right side:

Left side = $30$

Right side = $5040$

Since $30 \neq 5040$, the equation $3! + 4! = 7!$ is false.

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The answer is No, $3! + 4!$ is not equal to $7!$.

We are asked to compute the value of the expression $\frac{8!}{6! \times 2!}$.

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We can express $8!$ in terms of $6!$ since $6!$ is in the denominator.

Recall that $n! = n \times (n-1) \times \cdots \times (n-k+1) \times (n-k)!$

Also, calculate the value of $2!$:

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Now substitute the expressions from (i) and (ii) into the given expression:

Cancel out the common term $6!$ from the numerator and the denominator:

Simplify the expression:

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Alternatively, calculate the full factorial values first:

Substitute these values into the expression:

Perform the division:

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Both methods yield the same result.

The value of $\frac{8!}{6! \times 2!}$ is 28.

We are given the equation:

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We can express the larger factorials in terms of the smallest factorial present, which is $6!$.

Recall that $n! = n \times (n-1)!$.

So, we have:

Substitute (ii) into (iii):

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Now substitute the expressions for $7!$ and $8!$ from (ii) and (iv) into the original equation (i):

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To eliminate the factorials from the denominators, multiply both sides of equation (v) by $56 \times 6!$:

Distribute on the left side and simplify:

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Alternatively, find a common denominator on the left side of equation (i):

So, the equation becomes:

We know that $8! = 8 \times 7! = 8 \times 7 \times 6!$.

Substitute this into the right side of (vi):

Multiply both sides by $8 \times 7 \times 6!$:

Simplify:

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The value of $x$ is 64.

We are asked to evaluate the expression $\frac{n!}{(n-r)!}$ for the given values of $n$ and $r$.

Recall that the expression $\frac{n!}{(n-r)!}$ represents the number of permutations of $n$ distinct objects taken $r$ at a time, denoted by $P(n, r)$.

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(i) Evaluate $\frac{n!}{(n-r)!}$ when $n = 6$ and $r = 2$.

Substitute the given values of $n$ and $r$ into the expression:

Simplify the term in the parenthesis:

So the expression becomes:

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Expand $6!$ in terms of $4!$:

Substitute this into the expression:

Cancel out the common term $4!$:

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Alternatively, calculate the full factorial values:

Substitute these values into the expression:

Perform the division:

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The value of $\frac{n!}{(n-r)!}$ when $n=6$ and $r=2$ is 30.

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(ii) Evaluate $\frac{n!}{(n-r)!}$ when $n = 9$ and $r = 5$.

Substitute the given values of $n$ and $r$ into the expression:

Simplify the term in the parenthesis:

So the expression becomes:

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Expand $9!$ in terms of $4!$:

Substitute this into the expression:

Cancel out the common term $4!$:

Calculate the product:

$\begin{array}{cc}& & & 7 & 2 \\ \times & & 2 & 1 & 0 \\ \hline && & 0 & 0 \\ & & 7 & 2 & \times \\ 1 & 4 & 4 & \times & \times \\ \hline 1 & 5 & 1 & 2 & 0 \\ \hline \end{array}$

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The value of $\frac{n!}{(n-r)!}$ when $n=9$ and $r=5$ is 15120.

The given word is ALLAHABAD.

First, let's count the total number of letters in the word and the frequency of each distinct letter.

• Total number of letters = 9
• The distinct letters are A, L, H, B, D.
• Frequency of A = 4 (A appears 4 times)
• Frequency of L = 2 (L appears 2 times)
• Frequency of H = 1 (H appears 1 time)
• Frequency of B = 1 (B appears 1 time)
• Frequency of D = 1 (D appears 1 time)

The sum of the frequencies is $4 + 2 + 1 + 1 + 1 = 9$, which matches the total number of letters.

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We need to find the number of permutations of these 9 letters, where some letters are repeated.

The formula for the number of permutations of $n$ objects, where $p_1$ objects are of one kind, $p_2$ objects are of a second kind, ..., $p_k$ objects are of a $k$th kind, and the rest are distinct, is given by:

In this problem:

• $n = 9$ (total number of letters)
• $p_1 = 4$ (frequency of A)
• $p_2 = 2$ (frequency of L)
• The frequencies of H, B, D are 1, so $1! = 1$, and they do not affect the denominator beyond a factor of 1.

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Using the formula (i):

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Expand the factorials:

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Substitute these values into the expression:

Alternatively, write $9!$ in terms of $4!$:

Cancel out the common term $4!$ and substitute the value of $2!$:

Simplify:

Calculate the product:

$\begin{array}{cc}& & 2 & 1 & 0 \\ \times & & & 3 & 6 \\ \hline & 1 & 2 & 6 & 0 \\ & 6 & 3 & 0 & \times \\ \hline 7 & 5 & 6 & 0 \\ \hline \end{array}$

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Therefore, the number of different permutations of the letters of the word ALLAHABAD is 7560.

Given information:

• The digits available are 1, 2, 3, 4, 5, 6, 7, 8, 9.
• Total number of distinct digits available ($n$) = 9.
• We need to form a 4-digit number. This means we select and arrange 4 digits ($r = 4$).
• Repetition of digits is not allowed.

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To Find:

The number of different 4-digit numbers that can be formed.

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Solution:

We need to arrange 4 distinct digits out of 9 distinct digits to form a 4-digit number. The order of the digits matters (e.g., 1234 is a different number from 4321).

This is a problem of finding the number of permutations of 9 distinct objects taken 4 at a time.

Using the Fundamental Principle of Counting:

A 4-digit number has four places: the thousands place, the hundreds place, the tens place, and the units place.

• For the thousands place, there are 9 choices (any digit from 1 to 9).
• For the hundreds place, since repetition is not allowed, one digit has been used. So, there are $9 - 1 = 8$ choices remaining.
• For the tens place, two distinct digits have been used. So, there are $9 - 2 = 7$ choices remaining.
• For the units place, three distinct digits have been used. So, there are $9 - 3 = 6$ choices remaining.

By the Fundamental Principle of Counting, the total number of different 4-digit numbers is the product of the number of choices for each place:

Calculate the product:

$\begin{array}{cc}& & 7 & 2 \\ \times & & 4 & 2 \\ \hline & 1 & 4 & 4 \\ 2 & 8 & 8 & \times \\ \hline 3 & 0 & 2 & 4 \\ \hline \end{array}$

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Alternatively, using the Permutation Formula:

The number of permutations of $n$ distinct objects taken $r$ at a time is given by the formula:

Here, $n=9$ and $r=4$.

---

Using the formula (i):

---

Expand $9!$ in terms of $5!$ and simplify:

---

Both methods yield the same result.

Therefore, the number of 4-digit numbers that can be formed using the digits 1 to 9 without repetition is 3024.

Given information:

• The digits available are 0, 1, 2, 3, 4, 5.
• Total number of distinct digits available ($n$) = 6.
• We need to form numbers lying between 100 and 1000.
• Repetition of digits is not allowed.

---

Numbers lying between 100 and 1000 are 3-digit numbers.

A 3-digit number has three places: the hundreds place, the tens place, and the units place.

Let the 3-digit number be represented by $\text{_}\text{_}\text{_}$.

---

The digits available are {0, 1, 2, 3, 4, 5}.

Consider the hundreds place (first digit):

For a number to be a 3-digit number, the hundreds digit cannot be 0.

So, the digit in the hundreds place can be any digit from {1, 2, 3, 4, 5}.

Number of choices for the hundreds place = 5.

---

Consider the tens place (second digit):

We have 6 digits available in total. One digit has been used for the hundreds place.

Repetition is not allowed, so we cannot use the digit already used in the hundreds place.

The remaining number of available digits is $6 - 1 = 5$. These digits include 0 (if 0 was not used in the hundreds place) or exclude 0 (if 0 was used in the hundreds place, which is not possible for the first digit).

So, there are 5 choices for the tens place (any of the remaining 5 digits).

---

Consider the units place (third digit):

Two distinct digits have already been used (one for the hundreds place and one for the tens place).

Repetition is not allowed.

The remaining number of available digits is $6 - 2 = 4$.

Number of choices for the units place = 4.

---

Using the Fundamental Principle of Counting, the total number of 3-digit numbers that can be formed is the product of the number of choices for each place.

---

Alternatively, using Permutations (considering the constraint of the first digit):

We are forming 3-digit numbers from 6 distinct digits {0, 1, 2, 3, 4, 5} without repetition.

This is equivalent to finding the number of permutations of 6 objects taken 3 at a time, $P(6, 3)$, but we must exclude the cases where 0 is in the hundreds place.

Total number of permutations of 6 distinct digits taken 3 at a time is:

Now consider the cases where 0 is in the hundreds place. If 0 is in the hundreds place, we need to arrange the remaining 5 digits {1, 2, 3, 4, 5} in the tens and units places (2 positions) without repetition.

Number of such arrangements = Permutations of 5 objects taken 2 at a time.

These 20 arrangements are 3-digit numbers starting with 0 (e.g., 012, 034), which are essentially 2-digit numbers and are not between 100 and 1000.

The number of 3-digit numbers between 100 and 1000 (which are actual 3-digit numbers) is the total number of permutations of 6 digits taken 3 at a time, minus the number of permutations where 0 is in the first position.

---

Both methods confirm the result.

Therefore, the number of numbers lying between 100 and 1000 that can be formed using the digits 0, 1, 2, 3, 4, 5 without repetition is 100.

The formula for the number of permutations of $n$ distinct objects taken $r$ at a time is given by:

$^nP_r = \frac{n!}{(n-r)!}$, where $n \geq r$ and $n$ is a non-negative integer.

---

(i) Find $n$ such that $^nP_5 = 42\; ^nP_3$, where $n>4$.

Given the equation:

$^nP_5 = 42\; ^nP_3$

Apply the permutation formula:

$\frac{n!}{(n-5)!} = 42 \times \frac{n!}{(n-3)!}$

Since $n>4$, $n \geq 5$, so $n!$ is non-zero. Divide both sides by $n!$:

$\frac{1}{(n-5)!} = 42 \times \frac{1}{(n-3)!}$

Rearrange the equation by multiplying both sides by $(n-3)!$:

$\frac{(n-3)!}{(n-5)!} = 42$

Expand $(n-3)!$ until $(n-5)!$ is a factor:

$(n-3)! = (n-3)(n-4)(n-5)!$

Substitute this into the equation:

$\frac{(n-3)(n-4)(n-5)!}{(n-5)!} = 42$

Since $n>4$, $n-5 \geq 0$, so $(n-5)!$ is non-zero. Cancel out $(n-5)!$ from the numerator and denominator:

$(n-3)(n-4) = 42$

Expand the left side:

$n^2 - 4n - 3n + 12 = 42$

$n^2 - 7n + 12 = 42$

Move all terms to one side to form a quadratic equation:

$n^2 - 7n + 12 - 42 = 0$

$n^2 - 7n - 30 = 0$

Factor the quadratic equation:

We look for two numbers that multiply to -30 and add up to -7. These numbers are -10 and +3.

$(n - 10)(n + 3) = 0$

This gives two possible values for $n$:

$n - 10 = 0 \implies n = 10$

$n + 3 = 0 \implies n = -3$

We are given the condition that $n > 4$.

Check the possible values against the condition:

• If $n = 10$, is $10 > 4$? Yes.
• If $n = -3$, is $-3 > 4$? No.

Therefore, the only valid value for $n$ is 10.

The value of $n$ is 10.

---

(ii) Find $n$ such that $\frac{^nP_4}{^{n-1}P_4} = \frac{5}{3}$, where $n>4$.

Given the equation:

$\frac{^nP_4}{^{n-1}P_4} = \frac{5}{3}$

Apply the permutation formula. Note the constraint $n>4$ means $n \geq 5$. For $^nP_4$, we need $n \geq 4$. For $^{n-1}P_4$, we need $n-1 \geq 4$, which means $n \geq 5$. The condition $n>4$ already ensures this.

$\frac{\frac{n!}{(n-4)!}}{\frac{(n-1)!}{((n-1)-4)!}} = \frac{5}{3}$

Simplify the denominator in the lower part of the fraction:

$\frac{\frac{n!}{(n-4)!}}{\frac{(n-1)!}{(n-5)!}} = \frac{5}{3}$

Rewrite the left side by multiplying the numerator by the reciprocal of the denominator:

$\frac{n!}{(n-4)!} \times \frac{(n-5)!}{(n-1)!} = \frac{5}{3}$

Expand $n!$ and $(n-4)!$ to reveal common factors with $(n-1)!$ and $(n-5)!$:

$n! = n \times (n-1)!$

$(n-4)! = (n-4) \times (n-5)!$

Substitute these expansions into the equation:

$\frac{n \times (n-1)!}{(n-4) \times (n-5)!} \times \frac{(n-5)!}{(n-1)!} = \frac{5}{3}$

Cancel out the common terms $(n-1)!$ and $(n-5)!$:

$\frac{n \times \cancel{(n-1)!}}{(n-4) \times \cancel{(n-5)!}} \times \frac{\cancel{(n-5)!}}{\cancel{(n-1)!}} = \frac{n}{n-4}$

So the equation simplifies to:

$\frac{n}{n-4} = \frac{5}{3}$

Cross-multiply:

$3 \times n = 5 \times (n-4)$

$3n = 5n - 20$

Rearrange the terms to solve for $n$:

$20 = 5n - 3n$

$20 = 2n$

Divide by 2:

$n = \frac{20}{2}$

$n = 10$

We are given the condition that $n > 4$.

Check the value against the condition: If $n = 10$, is $10 > 4$? Yes.

Therefore, the value for $n$ is 10.

The value of $n$ is 10.

We are given the equation:

---

The formula for the number of permutations of $n$ distinct objects taken $r$ at a time is $^nP_r = \frac{n!}{(n-r)!}$.

For the term $^4P_r$, we must have $4 \geq r$ and $r \geq 0$. So, $\mathbf{0 \leq r \leq 4}$.

For the term $^5P_{r-1}$, we must have $5 \geq r-1$ and $r-1 \geq 0$.

• $5 \geq r-1 \implies 6 \geq r$
• $r-1 \geq 0 \implies r \geq 1$

So, $\mathbf{1 \leq r \leq 6}$.

Combining the conditions $0 \leq r \leq 4$ and $1 \leq r \leq 6$, the valid range for the integer $r$ is $\mathbf{1 \leq r \leq 4}$.

---

Apply the permutation formula to the terms in equation (i):

---

Substitute (ii) and (iii) into equation (i):

---

We can expand $5!$ as $5 \times 4!$ and $(6-r)!$ as $(6-r)(5-r)(4-r)!$:

---

Substitute (v) and (vi) into equation (iv):

---

Since $1 \leq r \leq 4$, both $4!$ and $(4-r)!$ are non-zero. We can divide both sides of equation (vii) by $5 \times 4!$:

---

Multiply both sides of equation (viii) by $(4-r)!$:

---

Multiply both sides by $(6-r)(5-r)$:

---

Expand the left side of equation (ix):

Subtract 6 from both sides to form a quadratic equation:

---

Factor the quadratic equation (x). We look for two numbers that multiply to 24 and add up to -11. These numbers are -3 and -8.

Setting each factor to zero gives the possible values for $r$:

---

Now, check these possible values against the valid range for $r$, which is $1 \leq r \leq 4$:

• For $r = 3$: Does $1 \leq 3 \leq 4$? Yes.
• For $r = 8$: Does $1 \leq 8 \leq 4$? No, 8 is outside the range.

Thus, the only valid value for $r$ is 3.

---

The value of $r$ is 3.

The given word is DAUGHTER.

Let's analyze the letters in the word:

• Total number of letters = 8
• The letters are D, A, U, G, H, T, E, R.
• All letters are distinct.

The vowels in the word are A, U, E.

The consonants are D, G, H, T, R.

---

Total number of different 8-letter arrangements (without any restrictions):

Since all 8 letters are distinct, the total number of permutations of 8 distinct letters taken all at a time is $8!$.

---

(i) All vowels occur together:

Consider the three vowels (A, U, E) as a single block or unit. Let this block be VVV.

The remaining letters are the consonants: D, G, H, T, R. There are 5 consonants.

Now, we are arranging the block VVV and the 5 consonants. This gives us a total of $1 + 5 = 6$ entities to arrange: (VVV), D, G, H, T, R.

These 6 entities are distinct (since the block VVV is treated as one item, and the consonants are distinct).

The number of ways to arrange these 6 distinct entities is $6!$.

---

Within the vowel block VVV, the three vowels (A, U, E) can arrange themselves in any order. Since the vowels are distinct, the number of ways to arrange the 3 vowels within their block is $3!$.

---

To find the total number of arrangements where all vowels occur together, we multiply the number of arrangements of the blocks/consonants by the number of arrangements within the vowel block (by the Fundamental Principle of Counting).

---

Thus, the number of different 8-letter arrangements where all vowels occur together is 4320.

---

(ii) All vowels do not occur together:

The number of arrangements where all vowels do not occur together is the total number of possible arrangements minus the number of arrangements where all vowels occur together.

Using the calculated values:

$\begin{array}{cc} & 4 & 0 & 3 & 2 & 0 \\ - & & 4 & 3 & 2 & 0 \\ \hline & 3 & 6 & 0 & 0 & 0 \\ \hline \end{array}$

---

Thus, the number of different 8-letter arrangements from the letters of the word DAUGHTER where all vowels do not occur together is 36000.

Given information:

• Number of red discs = 4
• Number of yellow discs = 3
• Number of green discs = 2
• Discs of the same colour are indistinguishable.

---

Total number of discs = Number of red discs + Number of yellow discs + Number of green discs

---

We need to find the number of different arrangements of these 9 discs in a row.

This is a problem of finding the number of permutations of $n$ objects where some objects are identical.

The number of permutations of $n$ objects, where $p_1$ objects are of one kind, $p_2$ objects are of a second kind, ..., $p_k$ objects are of a $k$th kind, is given by the formula:

In this problem:

• $n = 9$ (total number of discs)
• $p_1 = 4$ (number of red discs)
• $p_2 = 3$ (number of yellow discs)
• $p_3 = 2$ (number of green discs)

---

Using the formula (ii):

---

Expand the factorials:

---

Substitute these values into the expression:

Perform the division:

---

Alternatively, expand $9!$ in terms of the largest factorial in the denominator ($4!$) and simplify:

Cancel out $4!$ and substitute the values of $3!$ and $2!$:

Simplify further:

---

Both methods give the same result.

Therefore, the number of different arrangements of 4 red, 3 yellow, and 2 green discs in a row is 1260.

The given word is INDEPENDENCE.

First, let's count the total number of letters and the frequency of each distinct letter:

• Total number of letters in the word ($n$) = 12
• The distinct letters are I, N, D, E, P, C.
• Frequency of I = 1
• Frequency of N = 3
• Frequency of D = 2
• Frequency of E = 4
• Frequency of P = 1
• Frequency of C = 1

The sum of the frequencies is $1+3+2+4+1+1 = 12$, which matches the total number of letters.

The vowels in the word are I, E, E, E, E (1 'I' and 4 'E's). Total number of vowels = 5.

The consonants in the word are N, N, N, D, D, P, C (3 'N's, 2 'D's, 1 'P', 1 'C'). Total number of consonants = 7.

---

Total number of arrangements of the letters of the word INDEPENDENCE:

This is the number of permutations of 12 objects where 4 are of one kind (E), 3 are of another kind (N), and 2 are of a third kind (D), and the rest are distinct.

Using the formula for permutations with repetitions $\frac{n!}{p_1! p_2! \cdots p_k!}$:

Calculate the factorials:

Substitute and compute:

---

(i) Arrangements that start with P:

Fix the letter 'P' in the first position. We need to arrange the remaining 11 letters: I, N, D, E, E, N, D, E, N, C, E.

Total remaining letters = 11.

The repetitions among the remaining 11 letters are: N (3 times), D (2 times), E (4 times), I (1 time), C (1 time).

Number of arrangements of these 11 letters = $\frac{11!}{3! 2! 4! 1! 1!} = \frac{11!}{3! 2! 4!}$.

Calculate the factorials and compute:

---

(ii) Arrangements where all the vowels always occur together:

The vowels are I, E, E, E, E. Treat these 5 vowels as a single block (let's call it 'VowelBlock').

The number of ways to arrange the 5 vowels (1 'I', 4 'E's) within the 'VowelBlock' is $\frac{5!}{4! 1!} = \frac{120}{24} = 5$.

The remaining letters are the consonants: N, N, N, D, D, P, C. There are 7 consonants.

Now, we are arranging 8 entities: the 'VowelBlock' and the 7 consonants (3 'N's, 2 'D's, 1 'P', 1 'C').

Number of arrangements of these 8 entities = $\frac{8!}{3! 2! 1! 1! 1!} = \frac{8!}{3! 2!}$.

Calculate the factorials and compute:

To get the total number of arrangements where all vowels occur together, we multiply the number of ways to arrange the blocks/consonants by the number of ways to arrange the vowels within their block:

---

(iii) Arrangements where the vowels never occur together:

The number of arrangements where the vowels never occur together is the total number of arrangements minus the number of arrangements where all vowels occur together.

Using the results from the total arrangements and part (ii):

---

(iv) Arrangements where the words begin with I and end in P:

Fix the letter 'I' in the first position and the letter 'P' in the last position.

The word is INDEPENDENCE (12 letters).

The letters 'I' and 'P' are used. We need to arrange the remaining $12 - 2 = 10$ letters in the middle positions.

The remaining 10 letters are: N, D, E, E, N, D, E, N, C, E.

The repetitions among these 10 letters are: N (3 times), D (2 times), E (4 times), C (1 time).

Number of arrangements of these 10 letters = $\frac{10!}{3! 2! 4! 1!} = \frac{10!}{3! 2! 4!}$.

Calculate the factorials and compute:

---

Summary of results:

• Total number of arrangements = 1663200
• (i) Arrangements starting with P = 138600
• (ii) Arrangements where all vowels always occur together = 16800
• (iii) Arrangements where the vowels never occur together = 1646400
• (iv) Arrangements where the words begin with I and end in P = 12600

Given information:

• The digits available are 1, 2, 3, 4, 5, 6, 7, 8, 9.
• Total number of distinct digits available ($n$) = 9.
• We need to form a 3-digit number. This means we select and arrange 3 digits ($r = 3$).
• No digit is repeated.

---

To Find:

The number of different 3-digit numbers that can be formed.

---

Solution:

We need to arrange 3 distinct digits out of 9 distinct digits to form a 3-digit number. The order of the digits matters.

This is a problem of finding the number of permutations of 9 distinct objects taken 3 at a time.

Using the Fundamental Principle of Counting:

A 3-digit number has three places: the hundreds place, the tens place, and the units place.

• For the hundreds place, there are 9 choices (any digit from 1 to 9).
• For the tens place, since repetition is not allowed, one digit has been used. So, there are $9 - 1 = 8$ choices remaining.
• For the units place, two distinct digits have been used. So, there are $9 - 2 = 7$ choices remaining.

By the Fundamental Principle of Counting, the total number of different 3-digit numbers is the product of the number of choices for each place:

Calculate the product:

$\begin{array}{cc}& & 7 & 2 \\ \times & & & 7 \\ \hline & 5 & 0 & 4 \\ \hline \end{array}$

---

Alternatively, using the Permutation Formula:

The number of permutations of $n$ distinct objects taken $r$ at a time is given by the formula:

Here, $n=9$ and $r=3$.

---

Using the formula (i):

---

Expand $9!$ in terms of $6!$ and simplify:

---

Both methods yield the same result.

Therefore, the number of 3-digit numbers that can be formed using the digits 1 to 9 without repetition is 504.

Given information:

• We need to form a 4-digit number.
• The digits available are the standard digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. There are 10 distinct digits in total.
• No digit is repeated.

---

To Find:

The number of different 4-digit numbers that can be formed with no digit repeated.

---

Solution:

A 4-digit number has four places: the thousands place, the hundreds place, the tens place, and the units place.

Let the 4-digit number be represented by $\text{_}\text{_}\text{_}\text{_}$.

---

Consider the thousands place (first digit):

For a number to be a 4-digit number, the thousands digit cannot be 0.

So, the digit in the thousands place can be any digit from {1, 2, 3, 4, 5, 6, 7, 8, 9}.

Number of choices for the thousands place = 9.

---

Consider the hundreds place (second digit):

We have 10 digits available in total {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.

One digit has been used for the thousands place.

Repetition is not allowed, so we cannot use the digit already used in the thousands place.

The remaining number of available digits is $10 - 1 = 9$. These 9 digits include 0 (since 0 was not used in the thousands place).

So, there are 9 choices for the hundreds place (any of the remaining 9 digits).

---

Consider the tens place (third digit):

Two distinct digits have already been used (one for the thousands place and one for the hundreds place).

Repetition is not allowed.

The remaining number of available digits is $10 - 2 = 8$.

Number of choices for the tens place = 8.

---

Consider the units place (fourth digit):

Three distinct digits have already been used.

Repetition is not allowed.

The remaining number of available digits is $10 - 3 = 7$.

Number of choices for the units place = 7.

---

Using the Fundamental Principle of Counting, the total number of 4-digit numbers with no digit repeated is the product of the number of choices for each place.

Calculate the product:

$\begin{array}{cc}& & 8 & 1 \\ \times & & 5 & 6 \\ \hline & 4 & 8 & 6 \\ 4 & 0 & 5 & \times \\ \hline 4 & 5 & 3 & 6 \\ \hline \end{array}$

---

Therefore, there are 4536 4-digit numbers with no digit repeated.

Given information:

• The digits available are 1, 2, 3, 4, 6, 7.
• Total number of distinct digits available = 6.
• We need to form a 3-digit even number.
• No digit is repeated.

---

To Find:

The number of different 3-digit even numbers that can be formed.

---

Solution:

A 3-digit number has three places: the hundreds place, the tens place, and the units place.

Let the 3-digit number be represented by $\text{_}\text{_}\text{_}$.

For the number to be even, the digit in the units place must be an even digit from the given set.

The even digits in the set {1, 2, 3, 4, 6, 7} are 2, 4, and 6.

---

It is usually easier to start filling the positions with the most restrictions. In this case, the units place has a restriction (must be even).

Consider the units place:

The digit in the units place must be one of {2, 4, 6}.

Number of choices for the units place = 3.

---

Consider the hundreds place:

We have 6 distinct digits available. One digit has been used for the units place.

Repetition is not allowed, so we cannot use the digit already in the units place.

The remaining number of available digits is $6 - 1 = 5$.

The hundreds place cannot be 0 (but 0 is not in the given set of digits anyway). So any of the remaining 5 digits can be used.

Number of choices for the hundreds place = 5.

---

Consider the tens place:

Two distinct digits have already been used (one for the units place and one for the hundreds place).

Repetition is not allowed.

The remaining number of available digits is $6 - 2 = 4$.

Number of choices for the tens place = 4.

---

Using the Fundamental Principle of Counting, the total number of 3-digit even numbers with no digit repeated is the product of the number of choices for each place.

---

Therefore, there are 60 3-digit even numbers that can be formed using the digits 1, 2, 3, 4, 6, 7 with no digit repeated.

Given information:

• The digits available are 1, 2, 3, 4, 5.
• Total number of distinct digits available ($n$) = 5.
• We need to form 4-digit numbers. This means we select and arrange 4 digits ($r = 4$).
• No digit is repeated.

---

Part 1: Total number of 4-digit numbers formed without repetition.

We need to arrange 4 distinct digits out of 5 distinct digits. The order of the digits matters.

This is a problem of finding the number of permutations of 5 distinct objects taken 4 at a time, denoted by $^nP_r = ^5P_4$.

Using the Permutation Formula:

---

Alternatively, using the Fundamental Principle of Counting:

A 4-digit number has four places: thousands, hundreds, tens, units.

• Thousands place: 5 choices (any of 1, 2, 3, 4, 5)
• Hundreds place: 4 choices (remaining digits)
• Tens place: 3 choices (remaining digits)
• Units place: 2 choices (remaining digits)

---

The total number of 4-digit numbers formed using the digits 1, 2, 3, 4, 5 without repetition is 120.

---

Part 2: Number of these 4-digit numbers that are even.

For a number to be even, the digit in the units place must be an even digit from the given set {1, 2, 3, 4, 5}.

The even digits are 2 and 4.

We will fill the positions using the Fundamental Principle of Counting, starting with the units place.

Let the 4-digit number be $\text{_}\text{_}\text{_}\text{_}$.

---

Consider the units place (fourth digit):

The digit must be even. The choices are 2 or 4.

Number of choices for the units place = 2.

---

Consider the thousands place (first digit):

One digit has been used for the units place. We have 5 total digits. Repetition is not allowed.

The number of remaining digits is $5 - 1 = 4$. Any of these 4 digits can be used for the thousands place (since 0 is not in the available digits, the thousands place will not be zero).

Number of choices for the thousands place = 4.

---

Consider the hundreds place (second digit):

Two distinct digits have been used (one for units, one for thousands).

The number of remaining digits is $5 - 2 = 3$.

Number of choices for the hundreds place = 3.

---

Consider the tens place (third digit):

Three distinct digits have been used.

The number of remaining digits is $5 - 3 = 2$.

Number of choices for the tens place = 2.

---

By the Fundamental Principle of Counting, the total number of 4-digit even numbers with no digit repeated is the product of the number of choices for each place:

---

Therefore, out of the 120 possible 4-digit numbers, 48 of them will be even.

Given information:

• Total number of persons in the committee ($n$) = 8.
• We need to choose a chairman and a vice chairman. This means we are selecting and arranging 2 persons ($r = 2$) in specific roles.
• One person cannot hold more than one position (repetition is not allowed, and the chosen persons must be distinct).

---

To Find:

The number of ways to choose a chairman and a vice chairman.

---

Solution:

We need to select 2 persons from 8 distinct persons and assign them to two distinct positions (Chairman and Vice Chairman). The order of selection and assignment matters (choosing person A as Chairman and person B as Vice Chairman is different from choosing person B as Chairman and person A as Vice Chairman).

This is a problem of finding the number of permutations of 8 distinct objects taken 2 at a time.

Using the Fundamental Principle of Counting:

• For the position of Chairman, there are 8 choices (any of the 8 persons).
• For the position of Vice Chairman, since the same person cannot hold both positions, one person has been chosen as Chairman. So, there are $8 - 1 = 7$ choices remaining for the Vice Chairman.

By the Fundamental Principle of Counting, the total number of ways to choose a chairman and a vice chairman is the product of the number of choices for each position:

---

Alternatively, using the Permutation Formula:

The number of permutations of $n$ distinct objects taken $r$ at a time is given by the formula:

Here, $n=8$ and $r=2$.

---

Using the formula (i):

---

Expand $8!$ in terms of $6!$ and simplify:

---

Both methods yield the same result.

Therefore, the number of ways to choose a chairman and a vice chairman from a committee of 8 persons, such that one person cannot hold more than one position, is 56.

We are given the ratio of two permutation terms:

This can be written as a fraction:

---

The formula for the number of permutations of $k$ distinct objects taken $m$ at a time is given by $^kP_m = \frac{k!}{(k-m)!}$.

For the term $^{n-1}P_3$, we have $k = n-1$ and $m = 3$. The condition for this permutation to be defined is $n-1 \geq 3$, which implies $\mathbf{n \geq 4}$.

For the term $^nP_4$, we have $k = n$ and $m = 4$. The condition for this permutation to be defined is $\mathbf{n \geq 4}$.

Thus, the possible integer values for $n$ must be greater than or equal to 4.

---

Apply the permutation formula to the terms in equation (i):

---

Substitute (ii) and (iii) into equation (i):

---

Simplify the left side of the equation. Dividing by a fraction is the same as multiplying by its reciprocal:

Cancel out the common term $(n-4)!$ (since $n \geq 4$, $(n-4)!$ is defined and non-zero):

---

Expand $n!$ in terms of $(n-1)!$:

Substitute (v) into the left side of equation (iv):

Cancel out the common term $(n-1)!$ (since $n \geq 4$, $(n-1)!$ is defined and non-zero):

---

From equation (vi), by cross-multiplication or by equating the denominators (since the numerators are equal), we get:

---

Check if this value of $n$ satisfies the condition $n \geq 4$. Since $9 \geq 4$, the value $n=9$ is valid.

---

The value of $n$ is 9.

The formula for the number of permutations of $n$ distinct objects taken $r$ at a time is given by:

For $^nP_r$ to be defined, we must have $n \geq r$ and $r \geq 0$.

---

For the terms in the given equations:

For $^5P_r$: $5 \geq r$ and $r \geq 0$. Thus, $0 \leq r \leq 5$.

For $^6P_{r-1}$: $6 \geq r-1$ and $r-1 \geq 0$. This implies $r \leq 7$ and $r \geq 1$. Thus, $1 \leq r \leq 7$.

For both terms to be defined simultaneously, $r$ must satisfy both conditions. The common range for $r$ is $1 \leq r \leq 5$. Note that $r$ must also be an integer.

---

(i) Find r if $5 \times ^5P_r = 2 \times ^6P_{r-1}$.

Given the equation:

Apply the permutation formula (A):

Simplify the denominator on the right side:

Expand $6!$ as $6 \times 5!$ and $(7-r)!$ as $(7-r)(6-r)(5-r)!$ (since $r \leq 5$, $5-r \geq 0$, so $(5-r)!$ is defined):

Substitute (3) and (4) into (2):

Divide both sides by $5!$ and $(5-r)!$ (which are non-zero since $1 \leq r \leq 5$):

Multiply both sides by $(7-r)(6-r)$:

Expand the left side of (5):

Rearrange into a quadratic equation:

We need to solve the quadratic equation (6). We can use the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=5$, $b=-65$, $c=198$.

The possible values for $r$ are $\frac{65 + \sqrt{265}}{10}$ and $\frac{65 - \sqrt{265}}{10}$. Since $\sqrt{265}$ is not an integer, these values are not integers.

However, $r$ must be an integer for $^5P_r$ and $^6P_{r-1}$ to be standard permutations. There might be a mistake in the problem statement as given in the example, or perhaps my initial reading of the question prompt Example 13 vs Question 7 was incorrect. Let me check the original example 13 I solved earlier. Example 13 was $5 \times ^4P_r = 6 \times ^5P_{r-1}$. Let me solve Question 7 (i) assuming it is $5 \times ^5P_r = 2 \times ^6P_{r-1}$ as written.

Given the equation $5 \times ^5P_r = 2 \times ^6P_{r-1}$.

From $5 \times \frac{5!}{(5-r)!} = 2 \times \frac{6!}{(7-r)!}$, we got $5 = \frac{12}{(7-r)(6-r)}$.

$(7-r)(6-r) = \frac{12}{5} = 2.4$

$42 - 13r + r^2 = 2.4$

$r^2 - 13r + 39.6 = 0$

This quadratic equation does not have integer solutions for $r$.

Let me re-read the question. It is Question 7. Example 13 was similar. It seems the question itself might be intended to match the style of Example 13, but the coefficients are different. I will proceed with the given equation $5 \times ^5P_r = 2 \times ^6P_{r-1}$. As the calculation resulted in non-integer $r$, it's possible there's an error in the question as presented or the expected answer type. However, based on standard problems of this type, $r$ is expected to be an integer.

Let me assume the equation in Question 7(i) was meant to yield an integer solution. Let's check if there's a typo, for example, if it was $5 \times ^5P_r = 3 \times ^6P_{r-1}$ or similar.

Let's stick strictly to the provided input: $5 \times ^5P_r = 2 \times ^6P_{r-1}$. My derivation led to $r^2 - 13r + 39.6 = 0$. This has no integer solutions. This implies there is no integer $r$ satisfying the equation within the valid range $1 \leq r \leq 5$. It's possible the question intends to show that not all such equations have integer solutions.

However, in a typical problem from this chapter, an integer solution is expected. Let's double-check the calculation.

$5 \times \frac{5!}{(5-r)!} = 2 \times \frac{6!}{(7-r)!}$ $5 \times \frac{5!}{(5-r)!} = 2 \times \frac{6 \times 5!}{(7-r)(6-r)(5-r)!}$ Divide by $5! / (5-r)!$ on both sides: $5 = 2 \times \frac{6}{(7-r)(6-r)}$ $5 = \frac{12}{(7-r)(6-r)}$ $5(7-r)(6-r) = 12$ $5(r^2 - 13r + 42) = 12$ $5r^2 - 65r + 210 = 12$ $5r^2 - 65r + 198 = 0$. The discriminant is $65^2 - 4(5)(198) = 4225 - 3960 = 265$. The roots are $r = \frac{65 \pm \sqrt{265}}{10}$. These are not integers. Given this result, and assuming the question is stated correctly and expects an integer answer, it's possible there's an error in the problem itself. However, I must provide a solution based on the input. I will state that there is no integer solution for $r$ in the valid range.

Based on the standard interpretation of these problems where $r$ must be an integer, there is no integer value of $r$ in the range $1 \leq r \leq 5$ that satisfies the equation $5 \times ^5P_r = 2 \times ^6P_{r-1}$.

---

(ii) Find r if $^5P_r = ^6P_{r-1}$.

Given the equation:

The valid range for integer $r$ is $1 \leq r \leq 5$.

Apply the permutation formula (A):

Simplify the denominator on the right side:

Expand $6!$ as $6 \times 5!$ and $(7-r)!$ as $(7-r)(6-r)(5-r)!$ (since $r \leq 5$, $5-r \geq 0$, so $(5-r)!$ is defined):

Substitute (9) and (10) into (8):

Divide both sides by $5!$ and $(5-r)!$ (which are non-zero since $1 \leq r \leq 5$):

Multiply both sides by $(7-r)(6-r)$:

Expand the left side of (11):

Rearrange into a quadratic equation:

Factor the quadratic equation (12). We look for two numbers that multiply to 36 and add up to -13. These numbers are -4 and -9.

Setting each factor to zero gives the possible values for $r$:

---

Now, check these possible integer values against the valid range for $r$, which is $1 \leq r \leq 5$:

• For $r = 4$: Is $1 \leq 4 \leq 5$? Yes.
• For $r = 9$: Is $1 \leq 9 \leq 5$? No, 9 is outside the range.

Thus, the only valid integer value for $r$ in part (ii) is 4.

The value of $r$ is 4.

The given word is EQUATION.

Let's analyze the letters in the word:

• Total number of letters = 8
• The letters are E, Q, U, A, T, I, O, N.
• All 8 letters are distinct.

---

We need to form words using all the letters of the word EQUATION, and each letter must be used exactly once.

This means we are arranging all 8 distinct letters in different possible orders.

This is a problem of finding the number of permutations of 8 distinct objects taken all at a time ($n=8$, $r=8$).

The number of permutations of $n$ distinct objects taken $n$ at a time is given by $P(n, n) = n!$

---

In this case, $n=8$.

---

Calculate the factorial of 8:

We know $4! = 24$ and $5! = 120$ and $6! = 720$ and $7! = 5040$.

$\begin{array}{cc}& 5 & 0 & 4 & 0 \\ \times & & & & 8 \\ \hline 4 & 0 & 3 & 2 & 0 \\ \hline \end{array}$

---

Alternatively, using the Fundamental Principle of Counting:

We have 8 distinct letters to fill 8 positions.

• For the 1st position, there are 8 choices.
• For the 2nd position, there are 7 choices remaining.
• For the 3rd position, there are 6 choices remaining.
• ... and so on, until the 8th position, for which there is 1 choice remaining.

---

Therefore, the number of words that can be formed using all the letters of the word EQUATION, using each letter exactly once, is 40320.

The given word is MONDAY.

The letters in the word MONDAY are M, O, N, D, A, Y. There are 6 distinct letters.

Repetition of letters is not allowed in forming the words.

---

(i) 4 letters are used at a time:

We need to form 4-letter words using 4 out of the 6 distinct letters available, without repetition.

This is a permutation problem of selecting 4 distinct objects from 6 distinct objects and arranging them in order.

Number of letters available ($n$) = 6.

Number of letters to be used ($r$) = 4.

The number of permutations of $n$ distinct objects taken $r$ at a time is given by $\mathbf{^nP_r = \frac{n!}{(n-r)!}}$.

---

Using the permutation formula:

Expand the factorials:

Cancel out the common term $2!$:

Calculate the product:

---

Alternatively, using the Fundamental Principle of Counting:

We need to fill 4 positions for the 4-letter word.

• For the 1st position, there are 6 choices.
• For the 2nd position, there are 5 choices remaining (no repetition).
• For the 3rd position, there are 4 choices remaining.
• For the 4th position, there are 3 choices remaining.

---

The number of 4-letter words formed using 4 letters from MONDAY without repetition is 360.

---

(ii) All letters are used at a time:

We need to form 6-letter words using all 6 distinct letters of MONDAY, without repetition.

This is a permutation of 6 distinct objects taken all at a time ($n=6$, $r=6$).

The number of permutations of $n$ distinct objects taken $n$ at a time is given by $\mathbf{^nP_n = n!}$.

---

Using the formula:

Calculate the factorial of 6:

---

The number of words formed using all letters of MONDAY without repetition is 720.

---

(iii) All letters are used but first letter is a vowel:

We need to form 6-letter words using all 6 distinct letters, with the constraint that the first letter is a vowel. No repetition is allowed.

The vowels in the word MONDAY are O and A. There are 2 vowels.

We have 6 positions to fill for the 6-letter word: $\text{_}\text{_}\text{_}\text{_}\text{_}\text{_}$

---

Consider the first position:

The first letter must be a vowel. The choices are O or A.

Number of choices for the first position = 2.

---

Consider the remaining 5 positions:

One letter (a vowel) has been used for the first position. We have 6 total letters.

The remaining number of letters is $6 - 1 = 5$. These 5 letters are all distinct and can be placed in the remaining 5 positions in any order.

The number of ways to arrange the remaining 5 distinct letters in the remaining 5 positions is $5!$.

---

Using the Fundamental Principle of Counting, the total number of words with the first letter as a vowel is the product of the number of choices for the first position and the number of ways to arrange the remaining letters.

---

The number of words formed using all letters of MONDAY where the first letter is a vowel is 240.

The given word is MISSISSIPPI.

First, let's count the total number of letters and the frequency of each distinct letter:

• Total number of letters in the word ($n$) = 11
• The distinct letters are M, I, S, P.
• Frequency of M = 1
• Frequency of I = 4
• Frequency of S = 4
• Frequency of P = 2

The sum of the frequencies is $1 + 4 + 4 + 2 = 11$, which matches the total number of letters.

---

Step 1: Calculate the total number of distinct permutations of the letters in MISSISSIPPI.

This is the number of permutations of 11 objects where 4 are of one kind (I), 4 are of another kind (S), and 2 are of a third kind (P), and the rest are distinct (M).

Using the formula for permutations with repetitions $\frac{n!}{p_1! p_2! \cdots p_k!}$:

Calculate the factorials:

Substitute and compute:

---

Step 2: Calculate the number of distinct permutations where the four I's always occur together.

Treat the four I's (IIII) as a single block or unit. Let this block be 'IIII'.

The remaining letters are M, S, S, S, S, P, P. There are 7 remaining letters.

Now, we are arranging 8 entities: the block 'IIII' and the 7 remaining letters (M, S, S, S, S, P, P).

The entities to arrange are {'IIII', M, S, S, S, S, P, P}. The repetitions among these 8 entities are: S (4 times), P (2 times), M (1 time), 'IIII' (1 time).

The number of distinct arrangements of these 8 entities is $\frac{8!}{1! 4! 2!}$.

Calculate the factorials:

Substitute and compute:

Note: Within the block of four I's (IIII), the letters are identical. The number of ways to arrange them within the block is $\frac{4!}{4!} = 1$, so multiplying by this does not change the count of distinct arrangements.

---

Step 3: Calculate the number of distinct permutations where the four I's do not come together.

The number of arrangements where the four I's do not come together is the total number of distinct permutations minus the number of distinct permutations where all four I's occur together.

Using the results from Step 1 and Step 2:

---

Therefore, the number of distinct permutations of the letters in MISSISSIPPI where the four I's do not come together is 68460.

The given word is PERMUTATIONS.

Let's analyze the letters in the word:

• Total number of letters = 12.
• The distinct letters are P, E, R, M, U, T, A, I, O, N, S.
• The letter T appears 2 times. All other letters (P, E, R, M, U, A, I, O, N, S) appear 1 time.

The vowels in the word are E, U, A, I, O. There are 5 distinct vowels.

The consonants are P, R, M, T, T, N, S. There are 7 consonants, with T repeated twice.

---

(i) Words start with P and end with S:

Fix the letter 'P' in the first position and the letter 'S' in the last position.

We have 12 positions for the letters. The first position is P and the twelfth position is S.

We need to arrange the remaining $12 - 2 = 10$ letters in the remaining $12 - 2 = 10$ positions (from position 2 to position 11).

The remaining letters are E, R, M, U, T, A, T, I, O, N.

Among these 10 letters, the letter T is repeated 2 times. All other letters (E, R, M, U, A, I, O, N) are distinct.

The number of ways to arrange these 10 letters with repetitions is given by the formula for permutations with repetition: $\frac{n!}{p_1! p_2! \cdots}$.

Here, $n=10$ (number of letters to arrange in the middle), and $p_1=2$ (frequency of T).

Calculate the factorials:

Compute the number of arrangements:

---

The number of arrangements where the words start with P and end with S is 1814400.

---

(ii) Vowels are all together:

The vowels are E, U, A, I, O. There are 5 distinct vowels.

The consonants are P, R, M, T, T, N, S. There are 7 consonants, with T repeated twice.

Treat the 5 distinct vowels as a single block or unit (EUAIO).

The entities we need to arrange are the vowel block and the 7 consonants. This gives us a total of $1 + 7 = 8$ entities.

The entities are: (EUAIO), P, R, M, T, T, N, S.

Among these 8 entities, the consonant T is repeated 2 times. The vowel block and the other consonants (P, R, M, N, S) are distinct.

The number of ways to arrange these 8 entities with T repeated twice is $\frac{8!}{2!}$.

Calculate the factorials:

Compute the number of arrangements of the block and consonants:

---

Within the vowel block (EUAIO), the 5 distinct vowels can be arranged among themselves. The number of ways to arrange 5 distinct vowels is $5!$.

---

To find the total number of arrangements where all vowels occur together, we multiply the number of arrangements of the block/consonants by the number of arrangements within the vowel block (by the Fundamental Principle of Counting).

$\begin{array}{cc}& & 2 & 0 & 1 & 6 & 0 \\ \times & & & & 1 & 2 & 0 \\ \hline && & & 0 & 0 & 0 \\ & 4 & 0 & 3 & 2 & 0 & \times \\ 2 & 0 & 1 & 6 & 0 & \times & \times \\ \hline 2 & 4 & 1 & 9 & 2 & 0 & 0 \\ \hline \end{array}$

---

The number of arrangements where the vowels are all together is 2419200.

---

(iii) There are always 4 letters between P and S:

The word has 12 letters. We need to place P and S such that there are exactly 4 letters between them.

Consider the positions of P and S. If P is at position $i$ and S is at position $j$, with $i < j$, then $j - i - 1 = 4$, which means $j - i = 5$. The distance between their indices is 5.

The possible pairs of positions (P position, S position) are:

• (1, 6)
• (2, 7)
• (3, 8)
• (4, 9)
• (5, 10)
• (6, 11)
• (7, 12)

There are 7 such pairs of positions where P comes before S.

Similarly, there are 7 such pairs of positions where S comes before P:

• (6, 1)
• (7, 2)
• (8, 3)
• (9, 4)
• (10, 5)
• (11, 6)
• (12, 7)

So, there are a total of $7 + 7 = 14$ possible pairs of positions for the letters P and S such that there are always 4 letters between them.

---

Consider one such placement, e.g., P at position 1 and S at position 6. The positions look like: P _ _ _ _ S _ _ _ _ _ _.

The remaining 10 letters (E, R, M, U, T, A, T, I, O, N) must be arranged in the remaining 10 empty positions.

Among these 10 letters, the letter T is repeated 2 times. The other 8 letters are distinct.

The number of ways to arrange these 10 letters in the 10 remaining positions is $\frac{10!}{2!} = \frac{3628800}{2} = 1814400$.

---

This number of arrangements of the remaining letters is the same for each of the 14 possible pairs of positions for P and S.

Using the Fundamental Principle of Counting, the total number of arrangements where there are always 4 letters between P and S is the product of the number of ways to place P and S (14 ways) and the number of ways to arrange the remaining 10 letters (1814400 ways).

$\begin{array}{cc}& & 1 & 8 & 1 & 4 & 4 & 0 & 0 \\ \times & & & & & & & 1 & 4 \\ \hline & 7 & 2 & 5 & 7 & 6 & 0 & 0 \\ 1 & 8 & 1 & 4 & 4 & 0 & 0 & \times \\ \hline 2 & 5 & 4 & 0 & 1 & 6 & 0 & 0 \\ \hline \end{array}$

---

The number of arrangements where there are always 4 letters between P and S is 25401600.

We are given the equation involving combinations:

---

We know the property of combinations that if $^nC_a = ^nC_b$, then either $a = b$ or $a + b = n$.

In our given equation (i), we have $a = 9$ and $b = 8$.

Clearly, $9 \neq 8$, so the case $a=b$ is not applicable here.

Therefore, we must have $a + b = n$.

Solving for $n$:

For the combinations to be defined, $n$ must be a non-negative integer and $n \geq 9$ and $n \geq 8$. Our calculated value $n=17$ satisfies these conditions.

---

Now we need to find the value of $^nC_{17}$.

Substitute the value of $n=17$ from (iii) into the expression $^nC_{17}$:

---

We know that for any non-negative integer $n$, $^nC_n = 1$.

Applying this property to $^{17}C_{17}$:

---

Alternatively, using the combination formula $^nC_r = \frac{n!}{r!(n-r)!}$ with $n=17$ and $r=17$:

Since $0! = 1$:

---

The value of $^nC_{17}$ is 1.

Given information:

• Group consists of 2 men and 3 women.
• Total number of persons in the group = $2 + 3 = 5$.
• We need to constitute a committee of 3 persons.

---

Part 1: Number of ways to constitute a committee of 3 persons from the group.

We need to select 3 persons from a total of 5 persons. The order in which the persons are selected does not matter (a committee of A, B, C is the same as a committee of B, A, C). This is a problem of combinations.

Number of objects available ($n$) = 5.

Number of objects to be selected ($r$) = 3.

The number of combinations of $n$ distinct objects taken $r$ at a time is given by $\mathbf{^nC_r = \frac{n!}{r!(n-r)!}}$.

---

Using the combination formula:

Expand the factorials:

---

Alternatively, using the property $^nC_r = ^nC_{n-r}$:

---

The number of ways to constitute a committee of 3 persons from the group is 10.

---

Part 2: Number of these committees that would consist of 1 man and 2 women.

To form a committee with 1 man and 2 women, we need to make two separate selections:

1. Select 1 man from the 2 available men.

2. Select 2 women from the 3 available women.

---

Number of ways to select 1 man from 2 men = $^2C_1$.

---

Number of ways to select 2 women from 3 women = $^3C_2$.

---

Since these two selections are independent events (selecting men does not affect the selection of women), we use the Fundamental Principle of Counting to find the total number of ways to form a committee with 1 man and 2 women.

---

The number of these committees that would consist of 1 man and 2 women is 6.

A standard pack of 52 playing cards has:

• Total number of cards = 52
• 4 suits: Spades (♠), Hearts (♥), Diamonds (♦), Clubs (♣)
• Each suit has 13 cards.
• Spades and Clubs are black suits (26 black cards).
• Hearts and Diamonds are red suits (26 red cards).
• Face cards are Jack (J), Queen (Q), King (K). There are 3 face cards in each suit, so $3 \times 4 = 12$ face cards in total.

We are choosing a committee of cards (order does not matter), so this involves combinations.

---

Total number of ways of choosing 4 cards from a pack of 52:

We need to choose 4 cards from 52 distinct cards. The number of ways is given by $^nC_r = \frac{n!}{r!(n-r)!}$ where $n=52$ and $r=4$.

Expand $52!$ in terms of $48!$ and calculate $4!$:

Simplify the expression:

Calculate the product:

$\begin{array}{cc}& & 1 & 2 & 2 & 5 \\ \times & & & 2 & 2 & 1 \\ \hline & & 1 & 2 & 2 & 5 \\ & 2 & 4 & 5 & 0 & \times \\ 2 & 4 & 5 & 0 & \times & \times \\ \hline 2 & 7 & 0 & 7 & 2 & 5 \\ \hline \end{array}$

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The total number of ways of choosing 4 cards from a pack of 52 playing cards is 270725.

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(i) Four cards are of the same suit:

To have 4 cards of the same suit, we must choose one of the four suits and then choose 4 cards from the 13 cards in that suit.

• Number of ways to choose one suit = $^4C_1 = 4$.
• Number of ways to choose 4 cards from the chosen suit = $^{13}C_4$.

Using the Fundamental Principle of Counting:

$\begin{array}{cc}& & 7 & 1 & 5 \\ \times & & & & 4 \\ \hline & 2 & 8 & 6 & 0 \\ \hline \end{array}$

---

The number of ways to choose 4 cards of the same suit is 2860.

---

(ii) Four cards belong to four different suits:

To have four cards belonging to four different suits, we must choose one card from each of the four suits.

• Number of ways to choose 1 card from Spades = $^{13}C_1 = 13$.
• Number of ways to choose 1 card from Hearts = $^{13}C_1 = 13$.
• Number of ways to choose 1 card from Diamonds = $^{13}C_1 = 13$.
• Number of ways to choose 1 card from Clubs = $^{13}C_1 = 13$.

Using the Fundamental Principle of Counting:

$\begin{array}{cc}& & 1 & 6 & 9 \\ \times & & 1 & 6 & 9 \\ \hline & 1 & 5 & 2 & 1 \\ 1 & 0 & 1 & 4 & \times \\ 1 & 6 & 9 & \times & \times \\ \hline 2 & 8 & 5 & 6 & 1 \\ \hline \end{array}$

---

The number of ways to choose 4 cards belonging to four different suits is 28561.

---

(iii) Are face cards:

There are 12 face cards in a standard deck.

We need to choose 4 cards, and all of them must be face cards.

This is a combination problem of choosing 4 cards from the 12 available face cards.

Expand $12!$ in terms of $8!$ and calculate $4!$:

Simplify the expression:

---

The number of ways to choose 4 face cards is 495.

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(iv) Two are red cards and two are black cards:

There are 26 red cards and 26 black cards in a standard deck.

We need to choose 2 red cards from the 26 red cards AND 2 black cards from the 26 black cards.

• Number of ways to choose 2 red cards = $^{26}C_2$.
• Number of ways to choose 2 black cards = $^{26}C_2$.

Using the Fundamental Principle of Counting:

$\begin{array}{cc}& & 3 & 2 & 5 \\ \times & & 3 & 2 & 5 \\ \hline & 1 & 6 & 2 & 5 \\ & 6 & 5 & 0 & \times \\ 9 & 7 & 5 & \times & \times \\ \hline 1 & 0 & 5 & 6 & 2 & 5 \\ \hline \end{array}$

---

The number of ways to choose two red cards and two black cards is 105625.

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(v) Cards are of the same colour:

The cards can be either all red or all black.

Case 1: All 4 cards are red.

There are 26 red cards. We need to choose 4 red cards from the 26 red cards.

$\begin{array}{cc}& & 6 & 5 & 0 \\ \times & & & 2 & 3 \\ \hline & 1 & 9 & 5 & 0 \\ 1 & 3 & 0 & 0 & \times \\ \hline 1 & 4 & 9 & 5 & 0 \\ \hline \end{array}$

---

Case 2: All 4 cards are black.

There are 26 black cards. We need to choose 4 black cards from the 26 black cards.

This is the same calculation as for 4 red cards.

---

Since these two cases are mutually exclusive (the 4 cards cannot be all red and all black at the same time), we add the number of ways for each case to find the total number of ways where the cards are of the same colour.

---

The number of ways to choose 4 cards of the same colour is 29900.

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Summary of results for choosing 4 cards from 52:

• Total ways = 270725
• (i) Four cards are of the same suit = 2860
• (ii) Four cards belong to four different suits = 28561
• (iii) Four cards are face cards = 495
• (iv) Two are red cards and two are black cards = 105625
• (v) Cards are of the same colour = 29900

We are given the equation involving combinations:

---

We use the property of combinations which states that if $^nC_a = ^nC_b$, then either $a = b$ or $a + b = n$.

In our given equation (i), we have $a = 8$ and $b = 2$.

Clearly, $8 \neq 2$, so the case $a=b$ is not applicable here.

Therefore, we must have $a + b = n$.

Solving for $n$:

For the combinations $^nC_8$ and $^nC_2$ to be defined, $n$ must be a non-negative integer and $n \geq 8$ and $n \geq 2$. Our calculated value $n=10$ satisfies these conditions ($10 \geq 8$ and $10 \geq 2$).

---

Now we need to find the value of $^nC_2$.

Substitute the value of $n=10$ from (iii) into the expression $^nC_2$:

---

We use the combination formula $^nC_r = \frac{n!}{r!(n-r)!}$ with $n=10$ and $r=2$:

Expand the factorials. We can expand $10!$ in terms of $8!$ and calculate $2!$:

Substitute these into the formula:

Cancel out the common term $8!$ and simplify:

---

The value of $^nC_2$ is 45.

The formula for the number of combinations of $k$ distinct objects taken $m$ at a time is given by $\mathbf{^kC_m = \frac{k!}{m!(k-m)!}}$.

For $^kC_m$ to be defined, $k$ must be a non-negative integer and $k \geq m$.

For the term $^{2n}C_3$, we require $2n \geq 3$ and $2n$ to be an integer. This implies $n \geq 1.5$ and $n$ is such that $2n$ is an integer. If $n$ is an integer, then $2n$ is an integer. So we need integer $n \geq 2$.

For the term $^nC_3$, we require $n \geq 3$ and $n$ to be an integer. This implies integer $n \geq 3$.

For both combinations to be defined, $n$ must be an integer and $\mathbf{n \geq 3}$.

---

(i) Determine n if ²ⁿC₃ : ⁿC₃ = 12 : 1.

Given the ratio:

Apply the combination formula to both terms:

Substitute these into equation (1):

Simplify the complex fraction:

Cancel out $3!$ from the numerator and denominator:

Expand the factorials to reveal the terms in the denominators:

Substitute these expansions into equation (2):

Cancel out $(2n-3)!$ and $(n-3)!$ terms (which are non-zero for $n \geq 3$):

Factor $2n-2 = 2(n-1)$ from the numerator:

Since $n \geq 3$, $n \neq 0$ and $n-1 \neq 0$. Cancel out $n$ and $(n-1)$ from the numerator and denominator in equation (3):

Divide both sides of equation (4) by 4:

Multiply both sides by $(n-2)$ (which is non-zero for $n \geq 3$):

Rearrange the terms to solve for $n$:

The value $n=5$ satisfies the condition $n \geq 3$.

---

The value of $n$ is 5.

---

(ii) Determine n if ²ⁿC₃ : ⁿC₃ = 11 : 1.

Given the ratio:

From the derivation in part (i), we know that $\frac{^{2n}C_3}{^nC_3} = \frac{4(2n-1)}{n-2}$ for $n \geq 3$.

Substitute this into equation (5):

Multiply both sides of equation (6) by $(n-2)$ (which is non-zero for $n \geq 3$):

Distribute on both sides:

Rearrange the terms to solve for $n$:

Divide both sides by 3:

The value $n=6$ satisfies the condition $n \geq 3$.

---

The value of $n$ is 6.

Given information:

• There are 21 points on a circle.
• A chord of a circle is formed by joining any two distinct points on the circle.

---

To Find:

The number of chords that can be drawn through these 21 points.

---

Solution:

To draw a chord, we need to select 2 points from the given 21 points on the circle.

The order in which we select the two points does not matter (selecting point A and then point B results in the same chord as selecting point B and then point A). This is a combination problem.

Number of objects available ($n$) = 21 (the points on the circle).

Number of objects to be selected ($r$) = 2 (to form a chord).

The number of combinations of $n$ distinct objects taken $r$ at a time is given by $\mathbf{^nC_r = \frac{n!}{r!(n-r)!}}$.

---

Using the combination formula:

---

Expand $21!$ in terms of $19!$ and calculate $2!$:

Substitute these into the formula:

Cancel out the common term $19!$ and simplify:

---

Therefore, 210 chords can be drawn through 21 points on a circle.

Given information:

• Number of boys available = 5.
• Number of girls available = 4.
• We need to select a team consisting of 3 boys and 3 girls.

---

To form the team, we need to perform two independent selections:

1. Select 3 boys from the 5 available boys.

2. Select 3 girls from the 4 available girls.

---

The selection of team members is a combination problem, as the order of selection does not matter.

Number of ways to select 3 boys from 5 boys is given by $^5C_3$.

Calculate the value of $^5C_3$:

---

Number of ways to select 3 girls from 4 girls is given by $^4C_3$.

Calculate the value of $^4C_3$:

---

By the Fundamental Principle of Counting, the total number of ways to select a team of 3 boys and 3 girls is the product of the number of ways to perform each selection step (i) and (ii).

---

Therefore, a team of 3 boys and 3 girls can be selected from 5 boys and 4 girls in 40 ways.

Given information:

• Number of red balls available = 6.
• Number of white balls available = 5.
• Number of blue balls available = 5.
• We need to select a total of 9 balls.
• Each selection must consist of 3 balls of each colour.

---

To make a selection consisting of 3 balls of each colour, we need to perform three independent selection steps:

1. Select 3 red balls from the 6 available red balls.

2. Select 3 white balls from the 5 available white balls.

3. Select 3 blue balls from the 5 available blue balls.

---

The selection of balls is a combination problem, as the order in which the balls are selected does not matter.

Number of ways to select 3 red balls from 6 red balls is given by $^6C_3$.

Calculate the value of $^6C_3$:

---

Number of ways to select 3 white balls from 5 white balls is given by $^5C_3$.

Calculate the value of $^5C_3$:

---

Number of ways to select 3 blue balls from 5 blue balls is given by $^5C_3$.

This is the same calculation as for selecting 3 white balls from 5.

---

By the Fundamental Principle of Counting, the total number of ways to select a team of 3 balls of each colour is the product of the number of ways to perform each selection step (i), (ii), and (iii).

---

Therefore, the number of ways of selecting 9 balls consisting of 3 balls of each colour is 2000.

Given information:

• A standard deck has 52 playing cards.
• The deck contains 4 Aces (one in each suit).
• The remaining cards are non-aces, so there are $52 - 4 = 48$ non-ace cards.
• We need to select a combination of 5 cards.
• Each combination must contain exactly one ace.

---

To form a 5-card combination with exactly one ace, we need to make two independent selections:

1. Select exactly one ace from the 4 available aces.

2. Select the remaining $5 - 1 = 4$ cards from the 48 available non-ace cards.

---

The selection of cards is a combination problem, as the order of selection does not matter.

Number of ways to select 1 ace from 4 aces is given by $^4C_1$.

---

Number of ways to select 4 non-ace cards from 48 non-ace cards is given by $^{48}C_4$.

Expand $48!$ in terms of $44!$ and calculate $4!$:

Simplify the expression:

Calculate the product:

---

By the Fundamental Principle of Counting, the total number of 5-card combinations with exactly one ace is the product of the number of ways to perform each selection step (i) and (ii).

$\begin{array}{cc}& 1 & 9 & 4 & 5 & 8 & 0 \\ \times & & & & & & 4 \\ \hline & 7 & 7 & 8 & 3 & 2 & 0 \\ \hline \end{array}$

---

Therefore, the number of 5 card combinations out of a deck of 52 cards with exactly one ace in each combination is 778320.

Given information:

• Total number of players available = 17.
• Number of players who can bowl = 5.
• Number of players who cannot bowl (non-bowlers) = $17 - 5 = 12$.
• We need to select a cricket team of 11 players.
• The team must include exactly 4 bowlers.

---

To form a team of 11 players with exactly 4 bowlers, we must make two independent selections:

1. Select exactly 4 bowlers from the 5 available bowlers.

2. Select the remaining players from the non-bowlers. The number of remaining players needed is $11 - 4 = 7$. These 7 players must be selected from the 12 available non-bowlers.

---

The selection of players for the team is a combination problem, as the order of selection does not matter.

Number of ways to select 4 bowlers from 5 available bowlers is given by $^5C_4$.

Calculate the value of $^5C_4$:

---

Number of ways to select 7 non-bowlers from 12 available non-bowlers is given by $^{12}C_7$.

Calculate the value of $^{12}C_7$:

Simplify the expression:

Using the property $^nC_r = ^nC_{n-r}$ might be quicker for the expansion:

Let me recheck my calculation of $12 \times 11 \times 10 \times 9 \times 8 / 120$.

$\frac{12 \times 11 \times 10 \times 9 \times 8}{120} = \frac{95040}{120} = 792$. My previous calculation of 66 was incorrect. $12 \times 11 \times 10 \times 9 \times 8 = 11880 \times 8 = 95040$. Yes, 792 is correct for $^{12}C_7$.

---

By the Fundamental Principle of Counting, the total number of ways to select a cricket team of 11 players with exactly 4 bowlers is the product of the number of ways to perform each selection step (i) and (ii).

Calculate the product:

$\begin{array}{cc}& & 7 & 9 & 2 \\ \times & & & & 5 \\ \hline & 3 & 9 & 6 & 0 \\ \hline \end{array}$

---

Therefore, one can select a cricket team of eleven from 17 players with exactly 4 bowlers in 3960 ways.

Given:

Number of black balls in the bag = 5

Number of red balls in the bag = 6

---

To Find:

The number of ways to select 2 black balls and 3 red balls from the bag.

---

Solution:

The number of ways to select 2 black balls from 5 black balls is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$, where $n$ is the total number of items and $k$ is the number of items to select.

Number of ways to select 2 black balls from 5 black balls = $\binom{5}{2}$

$\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!}$

$\binom{5}{2} = \frac{5 \times 4 \times \cancel{3 \times 2 \times 1}}{(2 \times 1) \times \cancel{(3 \times 2 \times 1)}}$

$\binom{5}{2} = \frac{5 \times \cancel{4}^2}{\cancel{2}_1 \times 1} = 5 \times 2 = 10$

The number of ways to select 3 red balls from 6 red balls is given by:

Number of ways to select 3 red balls from 6 red balls = $\binom{6}{3}$

$\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!}$

$\binom{6}{3} = \frac{6 \times 5 \times 4 \times \cancel{3 \times 2 \times 1}}{(3 \times 2 \times 1) \times \cancel{(3 \times 2 \times 1)}}$

$\binom{6}{3} = \frac{\cancel{6}^1 \times 5 \times 4}{\cancel{3 \times 2 \times 1}_1} = 1 \times 5 \times 4 = 20$

Since the selection of black balls and red balls are independent events, the total number of ways to select 2 black balls and 3 red balls is the product of the number of ways for each selection.

Total number of ways = (Number of ways to select 2 black balls) $\times$ (Number of ways to select 3 red balls)

Total number of ways = $\binom{5}{2} \times \binom{6}{3}$

Total number of ways = $10 \times 20$

Total number of ways = 200

---

The total number of ways in which 2 black and 3 red balls can be selected is 200.

Given:

Total number of courses available = 9

Number of courses to be chosen by the student = 5

Number of compulsory courses = 2

---

To Find:

The number of ways a student can choose the programme.

---

Solution:

The student must choose 5 courses in total, and 2 specific courses are compulsory.

This means the student has already selected the 2 compulsory courses.

Number of courses already chosen = 2

Number of additional courses the student needs to choose = Total courses required - Compulsory courses

Number of additional courses to choose = $5 - 2 = 3$

The compulsory courses are from the 9 available courses. So, the remaining courses available for the student to choose from are:

Number of remaining courses available = Total courses available - Compulsory courses

Number of remaining courses available = $9 - 2 = 7$

The student needs to choose 3 additional courses from these 7 remaining courses.

The number of ways to do this is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$, where $n$ is the number of available items and $k$ is the number of items to choose.

Number of ways to choose 3 courses from the remaining 7 courses = $\binom{7}{3}$

$\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!}$

$\binom{7}{3} = \frac{7 \times 6 \times 5 \times \cancel{4 \times 3 \times 2 \times 1}}{(3 \times 2 \times 1) \times \cancel{(4 \times 3 \times 2 \times 1)}}$

$\binom{7}{3} = \frac{7 \times \cancel{6}^1 \times 5}{\cancel{3 \times 2 \times 1}_1} = 7 \times 1 \times 5 = 35$

Therefore, the number of ways a student can choose the programme is 35.

---

The student can choose the programme in 35 ways.

Given:

The given word is INVOLUTE.

Letters in the word: I, N, V, O, L, U, T, E

Total number of letters = 8

Vowels in the word: I, O, U, E

Number of vowels = 4

Consonants in the word: N, V, L, T

Number of consonants = 4

We need to form words consisting of 3 vowels and 2 consonants.

---

To Find:

The number of words that can be formed with 3 vowels and 2 consonants from the letters of the word INVOLUTE.

---

Solution:

The process involves two steps: first selecting the letters and then arranging them to form words.

Step 1: Selection of letters

We need to select 3 vowels from the 4 available vowels. The number of ways to do this is given by the combination formula $\binom{n}{k}$.

Number of ways to select 3 vowels from 4 = $\binom{4}{3}$

$\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = \frac{4 \times \cancel{3 \times 2 \times 1}}{(\cancel{3 \times 2 \times 1}) \times 1}$

$\binom{4}{3} = \frac{4}{1} = 4$

We also need to select 2 consonants from the 4 available consonants.

Number of ways to select 2 consonants from 4 = $\binom{4}{2}$

$\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3 \times \cancel{2 \times 1}}{(2 \times 1) \times \cancel{(2 \times 1)}}$

$\binom{4}{2} = \frac{\cancel{12}^6}{\cancel{2}_1} = 6$

The total number of ways to select a group of 3 vowels and 2 consonants is the product of the number of ways to select vowels and the number of ways to select consonants.

Total number of ways to select 3 vowels and 2 consonants = $\binom{4}{3} \times \binom{4}{2} = 4 \times 6 = 24$

Step 2: Arrangement of selected letters

Once a group of 3 vowels and 2 consonants is selected, we have a total of $3+2=5$ letters.

These 5 letters can be arranged in $5!$ ways to form words.

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

Total number of words formed

The total number of words is the product of the number of ways to select the letters and the number of ways to arrange them.

Total number of words = (Number of ways to select letters) $\times$ (Number of ways to arrange letters)

Total number of words = $24 \times 120$

Total number of words = 2880

---

Thus, 2880 words, with or without meaning, each of 3 vowels and 2 consonants, can be formed from the letters of the word INVOLUTE.

Given:

Number of girls in the group = 4

Number of boys in the group = 7

Total number of members in the group = $4 + 7 = 11$

Size of the team to be selected = 5 members

---

To Find:

The number of ways a team of 5 members can be selected under the following conditions:

(i) The team has no girl.

(ii) The team has at least one boy and one girl.

(iii) The team has at least 3 girls.

---

Solution:

(i) Number of ways the team has no girl:

If the team has no girl, it means the team consists entirely of boys.

We need to select 5 boys from the 7 available boys.

This can be done in $\binom{7}{5}$ ways.

$\binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7!}{5!2!}$

$\binom{7}{5} = \frac{7 \times 6 \times \cancel{5 \times 4 \times 3 \times 2 \times 1}}{(\cancel{5 \times 4 \times 3 \times 2 \times 1}) \times (2 \times 1)}$

$\binom{7}{5} = \frac{7 \times \cancel{6}^3}{\cancel{2}_1} = 7 \times 3 = 21$

So, there are 21 ways to select a team with no girl.

---

(ii) Number of ways the team has at least one boy and one girl:

The condition "at least one boy and one girl" means the team cannot be all boys or all girls.

The total number of ways to select a team of 5 members from the 11 members (4 girls + 7 boys) is $\binom{11}{5}$.

$\binom{11}{5} = \frac{11!}{5!(11-5)!} = \frac{11!}{5!6!}$

$\binom{11}{5} = \frac{11 \times 10 \times 9 \times 8 \times 7 \times \cancel{6!}}{(5 \times 4 \times 3 \times 2 \times 1) \times \cancel{6!}}$

$\binom{11}{5} = \frac{11 \times \cancel{10}^1 \times \cancel{9}^3 \times \cancel{8}^2 \times 7}{\cancel{5} \times \cancel{4} \times \cancel{3} \times \cancel{2} \times 1} = 11 \times 1 \times 3 \times 2 \times 7 = 462$

The number of ways the team has no girl (all boys) was calculated in part (i) as 21.

The number of ways the team has no boy (all girls) would be selecting 5 girls from 4 girls, which is $\binom{4}{5}$.

$\binom{4}{5} = 0$ (since we cannot select 5 items from a group of 4).

The number of ways the team has at least one boy and one girl is the total number of ways minus the ways with no girl and minus the ways with no boy.

Number of ways (at least one boy and one girl) = Total ways - Ways (no girl) - Ways (no boy)

Number of ways (at least one boy and one girl) = $462 - 21 - 0 = 441$

So, there are 441 ways to select a team with at least one boy and one girl.

---

(iii) Number of ways the team has at least 3 girls:

The condition "at least 3 girls" means the team can have 3 girls or 4 girls (since there are only 4 girls available).

We consider the possible cases for the composition of the 5-member team:

Case 1: Team has 3 girls and 2 boys.

Number of ways to select 3 girls from 4 = $\binom{4}{3} = 4$

Number of ways to select 2 boys from 7 = $\binom{7}{2} = \frac{7!}{2!5!} = \frac{7 \times 6}{2 \times 1} = 21$

Number of ways for Case 1 = $\binom{4}{3} \times \binom{7}{2} = 4 \times 21 = 84$

Case 2: Team has 4 girls and 1 boy.

Number of ways to select 4 girls from 4 = $\binom{4}{4} = 1$

Number of ways to select 1 boy from 7 = $\binom{7}{1} = 7$

Number of ways for Case 2 = $\binom{4}{4} \times \binom{7}{1} = 1 \times 7 = 7$

The total number of ways the team has at least 3 girls is the sum of the ways in Case 1 and Case 2.

Total ways (at least 3 girls) = Ways (3 girls, 2 boys) + Ways (4 girls, 1 boy)

Total ways (at least 3 girls) = $84 + 7 = 91$

So, there are 91 ways to select a team with at least 3 girls.

Given:

The word is AGAIN.

The letters in the word are A, G, A, I, N.

---

To Find:

1. The total number of words that can be formed using all the letters of AGAIN.

2. The 50^th word when these words are arranged in dictionary order.

---

Solution:

The letters of the word AGAIN are A, G, A, I, N.

There are 5 letters in total ($n=5$).

The letter 'A' repeats 2 times.

The letters G, I, N appear once.

1. Total number of words formed:

The number of permutations of $n$ objects where $n_1$ objects are of one type, $n_2$ are of another type, and so on, is given by $\frac{n!}{n_1! n_2! \dots}$.

Here, $n=5$, $n_A=2$, $n_G=1$, $n_I=1$, $n_N=1$.

Total number of words = $\frac{5!}{2!1!1!1!}$

= $\frac{5!}{2!}$

= $\frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1}$

= $\frac{120}{2} = 60$

Thus, 60 different words can be formed using the letters of the word AGAIN.

---

2. 50^th word in dictionary order:

To find the 50^th word, we list the distinct letters in alphabetical order: A, G, I, N.

We arrange the words in dictionary order:

Words starting with A:

If the first letter is A, the remaining letters are A, G, I, N. These are 4 letters with 'A' repeated once among the remaining letters (total letters were 5, one A is fixed, one A is left). The number of ways to arrange these 4 letters is $\frac{4!}{1!1!1!1!} = 4! = 24$.

Number of words starting with A = $\frac{4!}{2!} = \frac{24}{2} = 12$. (Correction: The remaining letters are A, G, I, N. The list of letters is {A, G, I, N} not {A, G, I, N}. Let's recheck. The letters are A, A, G, I, N. If the first letter is A, the remaining are A, G, I, N. These 4 letters (one A, one G, one I, one N) can be arranged in $4!$ ways.)

Let's restart the dictionary ordering part correctly.

The distinct letters in alphabetical order are A, G, I, N.

Words starting with A:

Fix 'A' as the first letter. The remaining letters are A, G, I, N. These 4 letters can be arranged among themselves in $4!$ ways.

Number of words starting with A = $4! = 4 \times 3 \times 2 \times 1 = 24$.

These are the first 24 words.

Words starting with G:

Fix 'G' as the first letter. The remaining letters are A, A, I, N. These 4 letters contain 'A' repeated 2 times. The number of ways to arrange these letters is $\frac{4!}{2!}$.

Number of words starting with G = $\frac{4!}{2!} = \frac{24}{2} = 12$.

Cumulative count = $24 (A) + 12 (G) = 36$.

Words starting with I:

Fix 'I' as the first letter. The remaining letters are A, A, G, N. These 4 letters contain 'A' repeated 2 times. The number of ways to arrange these letters is $\frac{4!}{2!}$.

Number of words starting with I = $\frac{4!}{2!} = \frac{24}{2} = 12$.

Cumulative count = $36 (A, G) + 12 (I) = 48$.

The first 48 words start with A, G, or I. The 50^th word must start with the next letter in alphabetical order, which is N.

Words starting with N:

Fix 'N' as the first letter. The remaining letters are A, A, G, I. These 4 letters contain 'A' repeated 2 times. We need to find the words formed by arranging A, A, G, I in alphabetical order, starting from the 49^th word.

Arrange A, A, G, I in alphabetical order: A, A, G, I.

The words starting with N, in dictionary order, will be formed by arranging A, A, G, I alphabetically after N.

The arrangements of A, A, G, I in alphabetical order are:

1^st arrangement: AAGI

2^nd arrangement: AAIG

3^rd arrangement: AG AI (Incorrect ordering)

Let's list the permutations of A, A, G, I alphabetically:

A A G I

A A I G

A G A I

A G I A

A I A G

A I G A

G A A I

G A I A

G I A A

I A A G

I A G A

I G A A

Total $\frac{4!}{2!} = 12$ arrangements.

The words starting with N, in dictionary order:

1^st word starting with N (49^th word overall): N + (1^st arrangement of A, A, G, I) = NAAGI

2^nd word starting with N (50^th word overall): N + (2^nd arrangement of A, A, G, I) = NAAIG

The 50^th word is NAAIG.

---

The total number of words formed is 60.

The 50^th word in dictionary order is NAAIG.

Given:

The digits are 1, 2, 0, 2, 4, 2, 4.

There are 7 digits in total.

The frequency of each digit is:

Digit 1: 1 time

Digit 2: 3 times

Digit 4: 2 times

Digit 0: 1 time

---

To Find:

The number of numbers greater than 1000000 that can be formed using these digits.

---

Solution:

A number greater than 1000000 formed using 7 digits must be a 7-digit number. This means the first digit cannot be 0.

First, let's find the total number of distinct 7-digit numbers that can be formed using the given digits. This is the total number of permutations of these 7 digits where some digits are repeated.

The formula for permutations with repetitions is $\frac{n!}{n_1! n_2! \dots n_k!}$, where $n$ is the total number of items, and $n_i$ is the frequency of each distinct item.

Total number of arrangements of the 7 digits = $\frac{7!}{3! \times 2! \times 1! \times 1!}$ (for digits 2, 4, 1, and 0 respectively)

= $\frac{7!}{3!2!}$

= $\frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1) \times (2 \times 1)}$

= $\frac{5040}{6 \times 2} = \frac{5040}{12} = 420$

These 420 arrangements include numbers that start with 0. A number starting with 0 is effectively a 6-digit number and is not greater than 1000000. We need to subtract these arrangements.

Number of arrangements starting with 0:

If 0 is fixed as the first digit, we need to arrange the remaining 6 digits: 1, 2, 2, 2, 4, 4.

Among these 6 digits, the digit '2' appears 3 times and the digit '4' appears 2 times.

Number of arrangements of the remaining 6 digits = $\frac{6!}{3! \times 2! \times 1!}$ (for digits 2, 4, and 1 respectively)

= $\frac{6!}{3!2!}$

= $\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1) \times (2 \times 1)}$

= $\frac{720}{6 \times 2} = \frac{720}{12} = 60$

These 60 arrangements are the numbers formed using the given digits that start with 0.

The number of 7-digit numbers that can be formed using the given digits is the total number of arrangements minus the number of arrangements starting with 0.

Number of 7-digit numbers = Total arrangements - Arrangements starting with 0

= $420 - 60 = 360$

Any 7-digit number formed using these digits will start with a non-zero digit (1, 2, or 4). The smallest possible first digit is 1. The smallest 7-digit number that can be formed is by placing 1 first and arranging the remaining digits (0, 2, 2, 2, 4, 4) in ascending order: 1022244.

Since $1022244 > 1000000$, all 7-digit numbers formed using these digits are greater than 1000000.

Therefore, the number of numbers greater than 1000000 that can be formed is equal to the number of distinct 7-digit numbers formed.

Number of numbers greater than 1000000 = 360.

---

The number of numbers greater than 1000000 that can be formed using the digits 1, 2, 0, 2, 4, 2, 4 is 360.

Given:

Number of girls = 5

Number of boys = 3

Total number of people to be seated = $5 + 3 = 8$

---

To Find:

The number of ways to seat 5 girls and 3 boys in a row such that no two boys sit together.

---

Solution:

To ensure that no two boys sit together, we must place the boys in the spaces between the girls.

First, arrange the 5 girls in a row. The number of ways to arrange 5 distinct girls is $5!$.

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

When 5 girls are seated, they create 6 possible spaces where the boys can be placed so that no two boys are adjacent. Let G represent a girl and underscore represent a possible space for a boy:

_ G _ G _ G _ G _ G _

There are 6 spaces available for the 3 boys.

We need to choose 3 of these 6 spaces and arrange the 3 boys in those chosen spaces.

The number of ways to choose 3 spaces from 6 is $\binom{6}{3}$.

The number of ways to arrange 3 boys in these 3 chosen spaces is $3!$.

Alternatively, the number of ways to place 3 distinct boys in 3 distinct spaces out of 6 is the number of permutations of 6 items taken 3 at a time, denoted as $P(6, 3)$.

$P(6, 3) = \frac{6!}{(6-3)!} = \frac{6!}{3!}$

$P(6, 3) = \frac{6 \times 5 \times 4 \times \cancel{3!}}{\cancel{3!}}$

$P(6, 3) = 6 \times 5 \times 4 = 120$

The total number of ways to seat the 5 girls and 3 boys such that no two boys are together is the product of the number of ways to arrange the girls and the number of ways to place the boys in the spaces between the girls.

Total number of ways = (Ways to arrange girls) $\times$ (Ways to place boys)

Total number of ways = $5! \times P(6, 3)$

Total number of ways = $120 \times 120$

Total number of ways = 14400

---

There are 14400 ways to seat 5 girls and 3 boys in a row so that no two boys are together.

Given:

The given word is DAUGHTER.

The letters in the word are D, A, U, G, H, T, E, R.

Total number of distinct letters = 8

Vowels in the word: A, U, E

Number of vowels = 3

Consonants in the word: D, G, H, T, R

Number of consonants = 5

We need to form words consisting of 2 vowels and 3 consonants.

---

To Find:

The number of words that can be formed with 2 vowels and 3 consonants from the letters of the word DAUGHTER.

---

Solution:

The formation of a word involves two steps: first selecting the required letters from the available letters, and then arranging the selected letters to form a word.

Step 1: Selection of letters

We need to select 2 vowels from the 3 available vowels. The number of ways to do this is given by the combination formula $\binom{n}{k}$.

Number of ways to select 2 vowels from 3 = $\binom{3}{2}$

$\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3!}{2!1!}$

$\binom{3}{2} = \frac{3 \times \cancel{2 \times 1}}{\cancel{(2 \times 1)} \times 1}$

$\binom{3}{2} = \frac{3}{1} = 3$

We need to select 3 consonants from the 5 available consonants.

Number of ways to select 3 consonants from 5 = $\binom{5}{3}$

$\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!}$

$\binom{5}{3} = \frac{5 \times 4 \times \cancel{3 \times 2 \times 1}}{\cancel{(3 \times 2 \times 1)} \times (2 \times 1)}$

$\binom{5}{3} = \frac{5 \times \cancel{4}^2}{\cancel{2}_1} = 5 \times 2 = 10$

The total number of ways to select a group of 2 vowels and 3 consonants is the product of the number of ways to select vowels and the number of ways to select consonants.

Total number of ways to select 2 vowels and 3 consonants = $\binom{3}{2} \times \binom{5}{3} = 3 \times 10 = 30$

Step 2: Arrangement of selected letters

Once a group of 2 vowels and 3 consonants is selected, we have a total of $2+3=5$ letters. Since all the letters in the word DAUGHTER are distinct, the selected 5 letters will also be distinct.

These 5 distinct letters can be arranged in $5!$ ways to form words.

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

Total number of words formed

The total number of words is the product of the number of ways to select the letters and the number of ways to arrange them.

Total number of words = (Number of ways to select letters) $\times$ (Number of ways to arrange letters)

Total number of words = $30 \times 120$

Total number of words = 3600

---

Thus, 3600 words, with or without meaning, each of 2 vowels and 3 consonants, can be formed from the letters of the word DAUGHTER.

Given:

The given word is EQUATION.

The letters in the word are E, Q, U, A, T, I, O, N.

Total number of letters = 8.

Vowels in the word: E, U, A, I, O

Number of vowels = 5

Consonants in the word: Q, T, N

Number of consonants = 3

All the letters are distinct.

---

To Find:

The number of words that can be formed using all the letters of the word EQUATION such that the vowels occur together and the consonants occur together.

---

Solution:

To solve this problem, we treat the group of vowels as a single block and the group of consonants as another single block. The letters within each block can be arranged among themselves.

The vowels are E, U, A, I, O. There are 5 distinct vowels.

The number of ways to arrange these 5 vowels among themselves is $5!$.

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

The consonants are Q, T, N. There are 3 distinct consonants.

The number of ways to arrange these 3 consonants among themselves is $3!$.

$3! = 3 \times 2 \times 1 = 6$

Now, we have two blocks: the block of vowels (V) and the block of consonants (C).

These two blocks can be arranged in two ways: (V block, C block) or (C block, V block).

The number of ways to arrange these 2 blocks is $2!$.

$2! = 2 \times 1 = 2$

The total number of words formed such that the vowels occur together and the consonants occur together is the product of the number of ways to arrange the vowels within their block, the number of ways to arrange the consonants within their block, and the number of ways to arrange the two blocks.

Total number of words = (Ways to arrange vowels) $\times$ (Ways to arrange consonants) $\times$ (Ways to arrange blocks)

Total number of words = $5! \times 3! \times 2!$

Total number of words = $120 \times 6 \times 2$

Total number of words = $120 \times 12$

Total number of words = 1440

---

Thus, 1440 words can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together.

Given:

Number of boys = 9

Number of girls = 4

Size of the committee to be formed = 7

---

(i) Exactly 3 girls:

If the committee consists of exactly 3 girls, then the number of boys in the committee must be $7 - 3 = 4$.

Number of ways to select 3 girls from 4 is $\binom{4}{3}$.

Number of ways to select 4 boys from 9 is $\binom{9}{4}$.

Total number of ways to form the committee with exactly 3 girls is $\binom{4}{3} \times \binom{9}{4}$.

Calculation:

$\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1) \times 1} = 4$

$\binom{9}{4} = \frac{9!}{4!(9-4)!} = \frac{9!}{4!5!} = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(4 \times 3 \times 2 \times 1) \times (5 \times 4 \times 3 \times 2 \times 1)} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = \frac{3024}{24} = 126$

Total ways = $4 \times 126 = 504$.

Therefore, the number of ways to form the committee with exactly 3 girls is 504.

---

(ii) At least 3 girls:

At least 3 girls means the committee can have 3 girls or 4 girls (since there are only 4 girls available).

Case 1: Exactly 3 girls and 4 boys.

Number of ways = $\binom{4}{3} \times \binom{9}{4} = 4 \times 126 = 504$ (from part i).

Case 2: Exactly 4 girls and 3 boys.

Number of ways to select 4 girls from 4 is $\binom{4}{4}$.

Number of ways to select 3 boys from 9 is $\binom{9}{3}$.

Total number of ways for Case 2 = $\binom{4}{4} \times \binom{9}{3}$.

Calculation for Case 2:

$\binom{4}{4} = \frac{4!}{4!(4-4)!} = \frac{4!}{4!0!} = 1$ (since $0! = 1$)

$\binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1) \times (6 \times 5 \times 4 \times 3 \times 2 \times 1)} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84$

Total ways for Case 2 = $1 \times 84 = 84$.

Total number of ways to form the committee with at least 3 girls = (Ways for Case 1) + (Ways for Case 2).

Total ways = $504 + 84 = 588$.

Therefore, the number of ways to form the committee with at least 3 girls is 588.

---

(iii) At most 3 girls:

At most 3 girls means the committee can have 0 girls, 1 girl, 2 girls, or 3 girls.

Since the committee size is 7 and there are only 4 girls, the number of girls can be 0, 1, 2, 3, or 4. The constraint "at most 3 girls" limits the number of girls to 0, 1, 2, or 3.

The corresponding number of boys will be $7 - (\text{number of girls})$.

Case 1: 0 girls and 7 boys.

Number of ways = $\binom{4}{0} \times \binom{9}{7}$.

Calculation for Case 1:

$\binom{4}{0} = 1$

$\binom{9}{7} = \binom{9}{9-7} = \binom{9}{2} = \frac{9 \times 8}{2 \times 1} = \frac{72}{2} = 36$

Ways for Case 1 = $1 \times 36 = 36$.

Case 2: 1 girl and 6 boys.

Number of ways = $\binom{4}{1} \times \binom{9}{6}$.

Calculation for Case 2:

$\binom{4}{1} = 4$

$\binom{9}{6} = \binom{9}{9-6} = \binom{9}{3} = 84$ (from part ii)

Ways for Case 2 = $4 \times 84 = 336$.

Case 3: 2 girls and 5 boys.

Number of ways = $\binom{4}{2} \times \binom{9}{5}$.

Calculation for Case 3:

$\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = \frac{12}{2} = 6$

$\binom{9}{5} = \binom{9}{9-5} = \binom{9}{4} = 126$ (from part i)

Ways for Case 3 = $6 \times 126 = 756$.

Case 4: 3 girls and 4 boys.

Number of ways = $\binom{4}{3} \times \binom{9}{4} = 504$ (from part i).

Total number of ways to form the committee with at most 3 girls = (Ways for Case 1) + (Ways for Case 2) + (Ways for Case 3) + (Ways for Case 4).

Total ways = $36 + 336 + 756 + 504$.

Sum = $372 + 756 + 504 = 1128 + 504 = 1632$.

Therefore, the number of ways to form the committee with at most 3 girls is 1632.

Given:

The word is EXAMINATION.

---

The total number of letters in the word EXAMINATION is 11.

The letters and their frequencies are:

A: 2 times

E: 1 time

I: 2 times

M: 1 time

N: 2 times

O: 1 time

T: 1 time

X: 1 time

---

The different permutations of the letters of the word are listed as in a dictionary.

In a dictionary, words are ordered alphabetically.

The distinct letters in EXAMINATION in alphabetical order are A, E, I, M, N, O, T, X.

The words starting with a letter that comes alphabetically before E will appear before the first word starting with E.

The only letter alphabetically before E in the word EXAMINATION is A.

Therefore, we need to count the number of words that start with A.

---

If a word starts with A, the remaining 10 letters are E, X, A, M, I, N, T, I, O, N from the original word EXAMINATION.

The set of remaining letters to be arranged is {A, E, I, I, M, N, N, O, T, X}.

The number of these 10 letters is 10.

The frequencies of these 10 letters are:

A: 1 time

E: 1 time

I: 2 times

M: 1 time

N: 2 times

O: 1 time

T: 1 time

X: 1 time

The number of distinct permutations of these 10 letters is given by the formula for permutations with repetitions:

Number of permutations = $\frac{\text{(Total number of letters)}!}{\text{(Frequency of repeating letter 1)}! \times \text{(Frequency of repeating letter 2)}! \times ...}$

In this case, the number of words starting with A is the number of permutations of the remaining 10 letters where I repeats 2 times and N repeats 2 times:

Number of words starting with A = $\frac{10!}{2! \times 2!}$

Calculating the value:

$\frac{10!}{2! \times 2!} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1) \times (2 \times 1)}$

$= \frac{3,628,800}{4}$

$= 907,200$

Thus, there are 907,200 words starting with A.

---

These words starting with A are listed before the first word starting with E in the dictionary order.

Therefore, the number of words in the list before the first word starting with E is 907,200.

Given:

Available digits are 0, 1, 3, 5, 7, and 9.

We need to form a 6-digit number using these digits without repetition.

The number must be divisible by 10.

---

A number is divisible by 10 if and only if its unit digit (the last digit) is 0.

In this case, the last digit of the 6-digit number must be 0.

Let the 6-digit number be represented by six places:

$\_$ $\_$ $\_$ $\_$ $\_$ $\_$

The last place must be filled with the digit 0.

$\_$ $\_$ $\_$ $\_$ $\_$ $\underline{0}$

---

Now we have 5 remaining places to fill using the remaining 5 digits from the set {0, 1, 3, 5, 7, 9}, which are {1, 3, 5, 7, 9}.

The first digit (leftmost place) cannot be 0 for a 6-digit number, but since 0 is already used in the last place and the remaining digits {1, 3, 5, 7, 9} are all non-zero, any arrangement of these 5 digits in the remaining 5 places will result in a valid 6-digit number.

We need to arrange the 5 distinct digits {1, 3, 5, 7, 9} in the remaining 5 positions.

The number of ways to arrange 5 distinct items in 5 positions is given by the number of permutations of 5 items taken 5 at a time, which is $P(5, 5)$ or $5!$.

---

Calculation:

Number of ways = $5!$

$5! = 5 \times 4 \times 3 \times 2 \times 1$

$5! = 20 \times 6$

$5! = 120$

So, there are 120 ways to fill the first five places with the digits {1, 3, 5, 7, 9} when the last digit is fixed as 0.

---

Thus, the total number of 6-digit numbers that can be formed from the digits 0, 1, 3, 5, 7, and 9 without repetition and are divisible by 10 is 120.

Given:

Number of vowels in the English alphabet = 5

Number of consonants in the English alphabet = 21

We need to form words with two different vowels and two different consonants.

---

To Find:

The number of words that can be formed with two different vowels and two different consonants.

---

Solution:

To form a word with two different vowels and two different consonants, we need to perform the following steps:

1. Select 2 different vowels from the 5 vowels.

2. Select 2 different consonants from the 21 consonants.

3. Arrange the selected 4 letters (2 vowels and 2 consonants) to form a word.

---

Number of ways to select 2 different vowels from 5 is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Number of ways to select vowels = $\binom{5}{2}$.

Calculation:

$\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4 \times 3!}{2 \times 1 \times 3!} = \frac{5 \times 4}{2} = 10$.

There are 10 ways to choose 2 different vowels.

---

Number of ways to select 2 different consonants from 21 is given by the combination formula.

Number of ways to select consonants = $\binom{21}{2}$.

Calculation:

$\binom{21}{2} = \frac{21!}{2!(21-2)!} = \frac{21!}{2!19!} = \frac{21 \times 20 \times 19!}{2 \times 1 \times 19!} = \frac{21 \times 20}{2} = 21 \times 10 = 210$.

There are 210 ways to choose 2 different consonants.

---

Once we have selected 2 vowels and 2 consonants, we have a total of $2 + 2 = 4$ different letters.

These 4 distinct letters can be arranged in $4!$ ways to form a word.

Calculation:

$4! = 4 \times 3 \times 2 \times 1 = 24$.

There are 24 ways to arrange the 4 chosen letters.

---

The total number of words that can be formed is the product of the number of ways to select the vowels, the number of ways to select the consonants, and the number of ways to arrange the selected letters.

Total number of words = (Ways to select vowels) $\times$ (Ways to select consonants) $\times$ (Ways to arrange letters).

Total number of words = $\binom{5}{2} \times \binom{21}{2} \times 4!$.

Total number of words = $10 \times 210 \times 24$.

$10 \times 210 = 2100$.

$2100 \times 24 = 50400$.

---

Therefore, the number of words with two different vowels and two different consonants that can be formed from the English alphabet is 50400.

Given:

Total number of questions = 12.

Number of questions in Part I = 5.

Number of questions in Part II = 7.

Number of questions to be attempted = 8.

Constraint: Student must select at least 3 questions from each part.

---

To Find:

The number of ways a student can select 8 questions satisfying the given constraints.

---

Solution:

Let $n_1$ be the number of questions selected from Part I and $n_2$ be the number of questions selected from Part II.

The total number of questions attempted is $n_1 + n_2 = 8$.

The constraints are:

$n_1 \geq 3$ (at least 3 from Part I)

$n_2 \geq 3$ (at least 3 from Part II)

---

Also, $n_1$ cannot exceed the number of questions in Part I (5), so $n_1 \leq 5$.

And $n_2$ cannot exceed the number of questions in Part II (7), so $n_2 \leq 7$.

We need to find integer pairs $(n_1, n_2)$ such that:

$n_1 + n_2 = 8$

$3 \leq n_1 \leq 5$

$3 \leq n_2 \leq 7$

---

Let's list the possible values for $n_1$ that satisfy $3 \leq n_1 \leq 5$ and find the corresponding $n_2$ using $n_2 = 8 - n_1$. Then we check if $n_2$ satisfies $3 \leq n_2 \leq 7$.

Case 1: $n_1 = 3$

$n_2 = 8 - 3 = 5$. Check constraint for $n_2$: $3 \leq 5 \leq 7$. This case is valid.

Number of ways to select 3 questions from Part I (5 questions) is $\binom{5}{3}$.

Number of ways to select 5 questions from Part II (7 questions) is $\binom{7}{5}$.

Total ways for Case 1 = $\binom{5}{3} \times \binom{7}{5}$.

$\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10$.

$\binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7!}{5!2!} = \frac{7 \times 6}{2 \times 1} = 21$.

Ways for Case 1 = $10 \times 21 = 210$.

---

Case 2: $n_1 = 4$

$n_2 = 8 - 4 = 4$. Check constraint for $n_2$: $3 \leq 4 \leq 7$. This case is valid.

Number of ways to select 4 questions from Part I (5 questions) is $\binom{5}{4}$.

Number of ways to select 4 questions from Part II (7 questions) is $\binom{7}{4}$.

Total ways for Case 2 = $\binom{5}{4} \times \binom{7}{4}$.

$\binom{5}{4} = \frac{5!}{4!(5-4)!} = \frac{5!}{4!1!} = 5$.

$\binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.

Ways for Case 2 = $5 \times 35 = 175$.

---

Case 3: $n_1 = 5$

$n_2 = 8 - 5 = 3$. Check constraint for $n_2$: $3 \leq 3 \leq 7$. This case is valid.

Number of ways to select 5 questions from Part I (5 questions) is $\binom{5}{5}$.

Number of ways to select 3 questions from Part II (7 questions) is $\binom{7}{3}$.

Total ways for Case 3 = $\binom{5}{5} \times \binom{7}{3}$.

$\binom{5}{5} = \frac{5!}{5!(5-5)!} = \frac{5!}{5!0!} = 1$ (since $0! = 1$).

$\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.

Ways for Case 3 = $1 \times 35 = 35$.

---

The possible combinations of questions satisfying the conditions are (3 from Part I, 5 from Part II), (4 from Part I, 4 from Part II), and (5 from Part I, 3 from Part II).

The total number of ways to select the questions is the sum of the ways in these valid cases.

Total ways = (Ways for Case 1) + (Ways for Case 2) + (Ways for Case 3).

Total ways = $210 + 175 + 35 = 420$.

---

Therefore, the number of ways a student can select the questions is 420.

Given:

Total number of cards in a deck = 52.

Number of Kings in a deck = 4.

Number of non-King cards in a deck = $52 - 4 = 48$.

We need to form a 5-card combination.

Constraint: Each selection of 5 cards must have exactly one king.

---

To Find:

The number of ways to form a 5-card combination with exactly one king.

---

Solution:

To form a 5-card combination with exactly one king, we need to make two independent selections:

1. Select exactly 1 king from the 4 available kings.

2. Select the remaining $5 - 1 = 4$ cards from the 48 non-king cards.

---

The number of ways to select 1 king from 4 kings is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Number of ways to select 1 king = $\binom{4}{1}$.

Calculation for selecting kings:

$\binom{4}{1} = \frac{4!}{1!(4-1)!} = \frac{4!}{1!3!} = \frac{4 \times 3!}{1 \times 3!} = 4$.

There are 4 ways to choose 1 king.

---

The number of ways to select 4 cards from the 48 non-king cards is given by the combination formula.

Number of ways to select 4 non-kings = $\binom{48}{4}$.

Calculation for selecting non-kings:

$\binom{48}{4} = \frac{48!}{4!(48-4)!} = \frac{48!}{4!44!} = \frac{48 \times 47 \times 46 \times 45 \times 44!}{4 \times 3 \times 2 \times 1 \times 44!} = \frac{48 \times 47 \times 46 \times 45}{24}$.

$\binom{48}{4} = \frac{48}{24} \times 47 \times 46 \times 45 = 2 \times 47 \times 46 \times 45$.

$\binom{48}{4} = 94 \times (46 \times 45)$.

$46 \times 45 = 2070$.

$\binom{48}{4} = 94 \times 2070 = 194580$.

There are 194580 ways to choose 4 non-king cards.

---

The total number of 5-card combinations with exactly one king is the product of the number of ways to select 1 king and the number of ways to select 4 non-king cards.

Total number of ways = (Ways to select 1 king) $\times$ (Ways to select 4 non-kings).

Total number of ways = $\binom{4}{1} \times \binom{48}{4} = 4 \times 194580$.

Total number of ways = $778320$.

---

Therefore, the number of 5-card combinations out of a deck of 52 cards with exactly one king is 778320.

Given:

Number of men = 5

Number of women = 4

Total number of people = $5 + 4 = 9$.

The people are to be seated in a row, so there are 9 positions.

Constraint: Women must occupy the even places.

---

To Find:

The number of possible arrangements where women occupy the even places.

---

Solution:

There are 9 positions in the row. The positions are numbered 1, 2, 3, 4, 5, 6, 7, 8, 9.

The even places are the positions with even numbers: 2nd, 4th, 6th, and 8th.

Number of even places = 4.

The odd places are the positions with odd numbers: 1st, 3rd, 5th, 7th, and 9th.

Number of odd places = 5.

---

According to the constraint, the 4 women must occupy the 4 even places.

The number of ways to arrange the 4 distinct women in the 4 distinct even places is the number of permutations of 4 items taken 4 at a time, which is $P(4,4)$ or $4!$.

Number of ways to arrange women = $4!$

Calculation for women's arrangements:

$4! = 4 \times 3 \times 2 \times 1 = 24$.

---

The remaining 5 places are the odd places (1st, 3rd, 5th, 7th, 9th).

The remaining 5 people are the 5 men.

These 5 distinct men must occupy the 5 distinct odd places.

The number of ways to arrange the 5 distinct men in the 5 distinct odd places is the number of permutations of 5 items taken 5 at a time, which is $P(5,5)$ or $5!$.

Number of ways to arrange men = $5!$

Calculation for men's arrangements:

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.

---

The total number of such arrangements is the product of the number of ways to arrange the women in the even places and the number of ways to arrange the men in the odd places, as these two arrangements are independent of each other.

Total number of arrangements = (Ways to arrange women) $\times$ (Ways to arrange men)

Total number of arrangements = $4! \times 5!$

Total number of arrangements = $24 \times 120$

Calculation:

$24 \times 120 = 2880$

---

Therefore, the number of such arrangements possible is 2880.

Given:

Total number of students in the class = $25$.

Number of students to be chosen for the excursion party = $10$.

There is a group of $3$ students who have a condition: either all $3$ join, or none of the $3$ join.

---

To Find:

The number of ways the excursion party can be chosen under the given condition.

---

Solution:

The class is divided into two groups for this problem: the group of $3$ special students and the group of the remaining students.

Number of special students = $3$.

Number of remaining students = $25 - 3 = 22$.

The condition gives us two mutually exclusive cases for forming the party:

---

Case 1: The group of 3 students join the excursion party.

In this case, all $3$ special students are included in the party. The number of ways to choose these $3$ students is $\binom{3}{3} = 1$.

Since the total number of students in the party must be $10$, we need to choose the remaining $10 - 3 = 7$ students from the other $22$ students.

The number of ways to choose $7$ students from the $22$ remaining students is $\binom{22}{7}$.

Calculation for $\binom{22}{7}$:

$\binom{22}{7} = \frac{22!}{7!(22-7)!} = \frac{22!}{7!15!} = \frac{22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}$

$\binom{22}{7} = \frac{859541760}{5040} = 170544$.

The number of ways for Case 1 is $\binom{3}{3} \times \binom{22}{7} = 1 \times 170544 = 170544$.

---

Case 2: The group of 3 students do not join the excursion party.

In this case, none of the $3$ special students are included in the party. These $3$ students are excluded from the selection pool.

The party of $10$ must be chosen entirely from the remaining $22$ students.

The number of ways to choose $10$ students from the $22$ remaining students is $\binom{22}{10}$.

Calculation for $\binom{22}{10}$:

$\binom{22}{10} = \frac{22!}{10!(22-10)!} = \frac{22!}{10!12!} = \frac{22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13}{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}$

$\binom{22}{10} = \frac{2346549004800}{3628800} = 646646$.

The number of ways for Case 2 is $\binom{22}{10} = 646646$.

---

Since these two cases are mutually exclusive, the total number of ways to form the excursion party is the sum of the ways in Case 1 and Case 2.

Total number of ways = Ways for Case 1 + Ways for Case 2

Total number of ways = $170544 + 646646 = 817190$.

---

Therefore, the number of ways the excursion party can be chosen is 817190.

Given:

The word is ASSASSINATION.

---

To Find:

The number of distinct arrangements of the letters of the word ASSASSINATION such that all the S’s are together.

---

Solution:

The word ASSASSINATION has a total of 13 letters.

Let's count the frequency of each letter:

A: 3 times

S: 4 times

I: 2 times

N: 2 times

T: 1 time

O: 1 time

Total letters = $3 + 4 + 2 + 2 + 1 + 1 = 13$.

---

We are asked to find the number of arrangements where all the S’s are together.

To handle the constraint that all S’s must be together, we treat the group of four S’s (SSSS) as a single unit or block.

Now, the distinct units/letters we need to arrange are:

The block (SSSS)

The remaining letters: A, A, A, I, I, N, N, T, O

The total number of units/letters to be arranged is the block (SSSS) plus the other 9 letters ($13 - 4 = 9$).

Total number of units/letters = $1 + 9 = 10$.

---

Among these 10 units/letters, some are repeated:

The letter A appears 3 times.

The letter I appears 2 times.

The letter N appears 2 times.

The block (SSSS) appears 1 time (as a single unit).

The letters T and O appear 1 time each.

---

The number of distinct arrangements of these 10 units/letters is given by the formula for permutations with repetitions:

Number of arrangements = $\frac{\text{(Total number of units)}!}{\text{(Frequency of A)}! \times \text{(Frequency of I)}! \times \text{(Frequency of N)}!}$

Number of arrangements = $\frac{10!}{3! \times 2! \times 2!}$

Calculation:

$10! = 3,628,800$

$3! = 3 \times 2 \times 1 = 6$

$2! = 2 \times 1 = 2$

The denominator is $3! \times 2! \times 2! = 6 \times 2 \times 2 = 24$.

Number of arrangements = $\frac{3,628,800}{24}$

$\frac{3,628,800}{24} = 151,200$.

Since the four S's within the block (SSSS) are identical, their arrangement among themselves does not create distinct permutations of the block, so we do not multiply by $4!$ for arranging the S's within the block; it is $\frac{4!}{4!} = 1$.

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Therefore, the number of ways to arrange the letters of the word ASSASSINATION so that all the S’s are together is 151,200.