Permutation And Combination

Answer: Option A
Solution:
Taking all person of same nationality as one person, then we will have only three people.
These three person can be arranged themselves in 3! Ways.
8 Indians can be arranged themselves in 8! Way.
4 American can be arranged themselves in 4! Ways.
4 Englishman can be arranged themselves in 4! Ways.
Hence, required number of ways = 3! 8! 4! 4! Ways.

Answer: Option D
Solution:
To find the number of permutations of the letters in the word “APPLE,” we need to account for the repeated letters.

The word “APPLE” consists of 5 letters: A, P, P, L, E. Notice that the letter **P** is repeated twice.

The general formula for the number of permutations of a set of objects where some objects are repeated is:

\[
\text{Number of permutations} = \frac{n!}{k_1! \cdot k_2! \cdot \ldots \cdot k_r!}
\]

where:
– \(n\) is the total number of objects (letters in this case),
– \(k_1, k_2, \ldots, k_r\) are the frequencies of the repeated objects.

For the word “APPLE”:
– Total number of letters, \(n = 5\),
– The letter **P** is repeated 2 times, so \(k_1 = 2\),
– All other letters (A, L, E) are distinct, so their frequencies are 1.

Step 1: Apply the formula

\[
\text{Number of permutations} = \frac{5!}{2!}
\]

Step 2: Calculate the factorials

– \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\),
– \(2! = 2 \times 1 = 2\).

Step 3: Compute the result

\[
\text{Number of permutations} = \frac{120}{2} = 60
\]

Thus, the number of distinct permutations of the letters in the word “APPLE” is 60.

Answer: Option D
Solution:
To solve these two parts of the problem, let’s break them down step by step.

Word: **ALLAHABAD**

The letters in “ALLAHABAD” are: A, L, L, A, H, A, B, A, D. We have 9 letters in total, and we need to consider some repetitions:
– A appears 4 times,
– L appears 2 times,
– H, B, and D each appear 1 time.

We will first calculate the total number of distinct permutations of the word without any restrictions, and then address the specific conditions for each part.

—

Part (a): When vowels occupy the even positions**

First, let’s identify the vowels and consonants in the word:
– Vowels: A, A, A, A (4 vowels),
– Consonants: L, L, H, B, D (5 consonants).

The even positions in a 9-letter word are positions 2, 4, 6, and 8. There are 4 even positions, and we are required to place the 4 vowels in these positions.

Step 1: Arrange the vowels in the even positions

Since there are 4 vowels (A, A, A, A), and they are all identical, there is only **1 way** to arrange them in the 4 even positions.

Step 2: Arrange the consonants in the remaining positions

Now, we need to arrange the 5 consonants (L, L, H, B, D) in the remaining 5 positions. The two L’s are identical, so the number of distinct ways to arrange them is:

\[
\frac{5!}{2!} = \frac{120}{2} = 60
\]

Step 3: Total number of arrangements

The total number of distinct arrangements when vowels occupy the even positions is the product of the two results:

\[
1 \times 60 = 60
\]

So, the number of different words that can be formed when vowels occupy the even positions is **60**.

—

Part (b): Both L’s do not occur together

Now, we need to find the number of distinct arrangements where the two L’s do not occur together.

Step 1: Total number of arrangements without any restriction

First, let’s calculate the total number of distinct arrangements of the letters in “ALLAHABAD” without any restriction. We use the formula for permutations of multiset:

\[
\text{Total permutations} = \frac{9!}{4! \cdot 2! \cdot 1! \cdot 1! \cdot 1!}
\]

Calculating the factorials:

\[
9! = 362880, \quad 4! = 24, \quad 2! = 2
\]

\[
\text{Total permutations} = \frac{362880}{24 \cdot 2} = \frac{362880}{48} = 7560
\]

So, the total number of distinct permutations is **7560**.

Step 2: Calculate the number of arrangements where L’s are together

Now, let’s consider the case where both L’s occur together. Treat the two L’s as a single unit. The letters we need to arrange are: (LL), A, A, A, A, H, B, D. This gives us 8 “letters” to arrange, where the 4 A’s are identical.

The number of distinct arrangements in this case is:

\[
\frac{8!}{4! \cdot 1! \cdot 1! \cdot 1!} = \frac{40320}{24} = 1680
\]

So, the number of arrangements where both L’s are together is **1680**.

Step 3: Subtract the number of arrangements where L’s are together

To find the number of arrangements where the two L’s do not occur together, subtract the number of arrangements where the L’s are together from the total number of arrangements:

\[
7560 – 1680 = 5880
\]

So, the number of distinct arrangements where the L’s do not occur together is **5880**.

—

Final Answers:

(a) The number of distinct words when vowels occupy the even positions is 60.

(b) The number of distinct words where the L’s do not occur together is 5880.

Answer: Option A
Solution:
To solve this problem, we will use the concept of complementary counting. We need to determine how many ways 10 examination papers can be arranged such that the best and the worst papers never come together.

Step 1: Total number of arrangements
The total number of ways to arrange all 10 examination papers, without any restrictions, is simply the number of permutations of 10 distinct papers:

\[
10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3,628,800
\]

Step 2: Arrangements where the best and the worst papers are together
Now, let’s count the number of arrangements where the best and the worst papers are together.

Treat the best and worst papers as a single “block.” This way, we now have 9 “objects” to arrange: the “block” of the best and worst papers, plus the 8 other papers.

The number of ways to arrange these 9 objects is:

\[
9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362,880
\]

Within the “block,” the best and worst papers can be arranged in 2 ways (best-first or worst-first). So, for each arrangement of the 9 objects, there are 2 ways to arrange the best and worst papers within the block.

Thus, the total number of arrangements where the best and worst papers are together is:

\[
9! \times 2 = 362,880 \times 2 = 725,760
\]

Step 3: Arrangements where the best and the worst papers are NOT together
To find the number of arrangements where the best and worst papers are **not** together, subtract the number of arrangements where they are together from the total number of arrangements:

\[
10! – 9! \times 2 = 3,628,800 – 725,760 = 2,903,040
\]

Final Answer:
The number of ways to arrange the 10 examination papers such that the best and worst papers never come together is 2,903,040.

Answer: Option A
Solution:
To solve this problem, we need to arrange 4 boys and 3 girls in a row such that they are seated alternately.

Step 1: Seating arrangement pattern
Since the boys and girls must sit alternately, we can follow one of two patterns:
1. Boy-Girl-Boy-Girl-Boy-Girl-Boy** (starting with a boy)
2. Girl-Boy-Girl-Boy-Girl-Boy** (starting with a girl)

However, since there are 4 boys and only 3 girls, the only valid pattern is **Boy-Girl-Boy-Girl-Boy-Girl-Boy**, where we start with a boy.

Step 2: Arrange the boys
There are 4 boys, and they can be arranged in the 4 available positions (1st, 3rd, 5th, and 7th) in any order. The number of ways to arrange the 4 boys is:

\[
4! = 4 \times 3 \times 2 \times 1 = 24
\]

Step 3: Arrange the girls
There are 3 girls, and they can be arranged in the 3 available positions (2nd, 4th, and 6th) in any order. The number of ways to arrange the 3 girls is:

\[
3! = 3 \times 2 \times 1 = 6
\]

Step 4: Total number of arrangements
To find the total number of arrangements, multiply the number of ways to arrange the boys by the number of ways to arrange the girls:

\[
4! \times 3! = 24 \times 6 = 144
\]

Final Answer:
The number of ways 4 boys and 3 girls can be seated in a row such that they are alternate is 144.

Answer: Option D
Solution:
Let’s break this down step by step.

We are tasked with forming a two-member committee consisting of one male and one female from a pool of:
– 5 males
– 3 females

Additionally, we have the restriction that **Ms. A** refuses to be on the committee if **Mr. B** is on the committee.

Step 1: Total number of committees without any restrictions
Without any restrictions, the committee can be formed by choosing one male from the 5 males and one female from the 3 females. The total number of ways to do this is:

\[
\text{Total committees} = \text{ways to choose 1 male} \times \text{ways to choose 1 female} = 5 \times 3 = 15
\]

Step 2: Account for the restriction
Now, we need to subtract the number of committees where **Ms. A** and **Mr. B** are both selected, since **Ms. A** refuses to be on the committee with **Mr. B**.

The number of ways to form a committee where both **Ms. A** and **Mr. B** are selected is:

\[
\text{Committees with Ms. A and Mr. B} = 1 \text{ way}
\] (there is exactly one way to choose **Ms. A** and **Mr. B**).

Step 3: Calculate the valid committees
Now, to find the total number of valid committees (where **Ms. A** is not on the committee with **Mr. B**), we subtract the invalid committees from the total committees:

\[
\text{Valid committees} = 15 – 1 = 14
\]

Final Answer:
The number of different ways the committee can be constituted, with one male and one female, and taking into account the restriction, is 14.

Answer: Option A
Solution:
To determine how many ways 2 students can be chosen from a class of 20 students, we use the formula for combinations, since the order of selection doesn’t matter.

The formula for combinations is:

\[
\text{Number of ways to choose } r \text{ objects from } n \text{ objects} = \binom{n}{r} = \frac{n!}{r!(n-r)!}
\]

In this case, \(n = 20\) (total number of students) and \(r = 2\) (number of students to be chosen). So, we need to calculate:

\[
\binom{20}{2} = \frac{20!}{2!(20-2)!} = \frac{20 \times 19}{2 \times 1} = \frac{380}{2} = 190
\]

Final Answer:
The number of ways to choose 2 students from a class of 20 students is 190.

Answer: Option D
Solution:
In this problem, we have 3 gentlemen and 3 ladies, and a voter needs to vote for two candidates, one from each gender. The question asks how many ways the voter can cast their vote, which means we need to consider different combinations of gentlemen and ladies.

Step 1: Choosing one gentleman and one lady
The voter needs to choose:
1. One gentleman from the 3 available gentlemen, and
2. One lady from the 3 available ladies.

The number of ways to choose one gentleman from 3 is:

\[
\text{Ways to choose 1 gentleman} = 3
\]

The number of ways to choose one lady from 3 is:

\[
\text{Ways to choose 1 lady} = 3
\]

Step 2: Total number of ways
To find the total number of ways to cast a vote, multiply the number of ways to choose a gentleman by the number of ways to choose a lady:

\[
\text{Total number of ways} = 3 \times 3 = 9
\]

Final Answer:
The number of ways one can cast their vote is 9.

Answer: Option A
Solution:
To solve this problem, we need to find how many ways a student can choose 8 questions from Part A (which has 10 questions) and 5 questions from Part B (which also has 10 questions).

This is a problem of combinations, as the order of the questions chosen doesn’t matter. We can apply the combination formula:

\[
\binom{n}{r} = \frac{n!}{r!(n-r)!}
\]

where \(n\) is the total number of questions and \(r\) is the number of questions to be chosen.

Step 1: Choose 8 questions from Part A
There are 10 questions in Part A, and the student needs to choose 8. The number of ways to choose 8 questions from Part A is:

\[
\binom{10}{8} = \frac{10!}{8!(10-8)!} = \frac{10!}{8!2!} = \frac{10 \times 9}{2 \times 1} = 45
\]

Step 2: Choose 5 questions from Part B
Similarly, there are 10 questions in Part B, and the student needs to choose 5. The number of ways to choose 5 questions from Part B is:

\[
\binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252
\]

Step 3: Total number of ways
Since the choices from Part A and Part B are independent, the total number of ways the student can choose the questions is the product of the two:

\[
\text{Total number of ways} = \binom{10}{8} \times \binom{10}{5} = 45 \times 252 = 11340
\]

Final Answer:
The student can choose the questions in 11,340 ways.

Answer: Option B
Solution:
To solve this problem, we need to find how many triangles can be formed by selecting 3 points from a total of 10 points, out of which 4 points are collinear.

Step 1: Total number of triangles
To form a triangle, we need to choose 3 points from the 10 points. The total number of ways to choose 3 points from 10 is given by the combination formula:

\[
\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120
\]

So, the total number of ways to choose 3 points from 10 is 120. However, this count includes triangles that might be degenerate (i.e., where the 3 points are collinear, and hence do not form a triangle).

Step 2: Subtract degenerate triangles
Degenerate triangles occur when the 3 points chosen are collinear. We are told that 4 points are collinear. The number of ways to choose 3 collinear points from these 4 is:

\[
\binom{4}{3} = \frac{4 \times 3 \times 2}{3 \times 2 \times 1} = 4
\]

These 4 ways represent degenerate triangles that do not actually form valid triangles.

Step 3: Calculate the number of valid triangles
To find the number of valid triangles, we subtract the number of degenerate triangles from the total number of triangles:

\[
\text{Valid triangles} = \binom{10}{3} – \binom{4}{3} = 120 – 4 = 116
\]

Final Answer:
The number of triangles that can be formed by the points as vertices is 116.

Answer: Option A
Solution:
To solve this problem, we need to find the number of people in a party based on the number of handshakes.

When every person shakes hands with every other person, the total number of handshakes can be represented by the number of ways to choose 2 people from the total number of people, since a handshake involves exactly 2 people. This is a combination problem, and the number of ways to choose 2 people from \(n\) people is given by:

\[
\binom{n}{2} = \frac{n(n-1)}{2}
\]

We are given that the number of handshakes is 105, so we set up the equation:

\[
\frac{n(n-1)}{2} = 105
\]

Step 1: Solve for \(n\)
Multiply both sides of the equation by 2 to eliminate the denominator:

\[
n(n-1) = 210
\]

Now, expand the equation:

\[
n^2 – n = 210
\]

Rearrange it to form a quadratic equation:

\[
n^2 – n – 210 = 0
\]

Step 2: Solve the quadratic equation
We solve the quadratic equation using the quadratic formula:

\[
n = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
\]

For the equation \(n^2 – n – 210 = 0\), the coefficients are:
– \(a = 1\),
– \(b = -1\),
– \(c = -210\).

Substitute these values into the quadratic formula:

\[
n = \frac{-(-1) \pm \sqrt{(-1)^2 – 4(1)(-210)}}{2(1)}
\] \[
n = \frac{1 \pm \sqrt{1 + 840}}{2}
\] \[
n = \frac{1 \pm \sqrt{841}}{2}
\] \[
n = \frac{1 \pm 29}{2}
\]

Step 3: Find the possible values of \(n\)
We have two possible solutions for \(n\):

\[
n = \frac{1 + 29}{2} = \frac{30}{2} = 15
\] or
\[
n = \frac{1 – 29}{2} = \frac{-28}{2} = -14
\]

Since the number of people cannot be negative, we discard \(n = -14\), leaving us with \(n = 15\).

Final Answer:
The number of people in the party is 15.

Answer: Option B
Solution
Let’s break down the problem step by step to calculate the minimum number of matches in the next World Cup.

Step 1: Group Stage
– There are 12 teams divided equally into 2 groups, so each group will have 6 teams.
– In each group, every team will play against every other team. This means that in each group, the number of matches is the number of ways to choose 2 teams from 6. This is given by the combination formula \(\binom{n}{2}\), where \(n\) is the number of teams in the group.

\[
\text{Matches in one group} = \binom{6}{2} = \frac{6 \times 5}{2} = 15
\]

Since there are two groups, the total number of matches in the group stage is:

\[
\text{Total matches in the group stage} = 15 \times 2 = 30
\]

Step 2: Qualification for the Next Round
– From each group, the top 3 teams will qualify for the next round. This means 3 teams from each group will qualify, and 6 teams will advance to the next round.

Step 3: Round-Robin Stage
– In this round, the 6 qualifying teams (3 from each group) will play against each other. This is again a round-robin format, where each team plays against every other team once.
– The number of matches in a round-robin format with 6 teams is the number of ways to choose 2 teams from 6, i.e., \(\binom{6}{2}\):

\[
\text{Matches in the round-robin stage} = \binom{6}{2} = \frac{6 \times 5}{2} = 15
\]

Step 4: Semifinal Round
– The top 4 teams from the round-robin stage will qualify for the semifinals.
– In the semifinals, 2 teams will play against each other, and the winners will advance to the final. Since it’s the best of three matches, the minimum number of matches in the semifinals will be 3 matches.

Step 5: Final Round
– The 2 winners of the semifinals will play in the final, and since it’s the best of three matches, the minimum number of matches in the final will be 3 matches.

Total Number of Matches
Now, we add up all the matches:
– Group stage: 30 matches,
– Round-robin stage: 15 matches,
– Semifinals: 3 matches,
– Final: 3 matches.

The total number of matches is:

\[
\text{Total matches} = 30 + 15 + 3 + 3 = 51
\]

Final Answer:
The minimum number of matches in the next World Cup will be 51.

Answer: Option A
Solution:
To solve this problem, we need to determine the number of ways 10 people, including two brothers, can be seated around a round table such that exactly one person sits between the two brothers.

Step 1: Fix one brother’s position
Since the seating is around a **round table**, we can fix one of the brothers at any position, as rotating the seating does not create distinct arrangements. Fixing one brother’s position eliminates the rotational symmetry of the problem.

So, let’s fix one brother in a specific seat.

Step 2: Position the second brother
Now, we need to place the second brother such that there is exactly one person sitting between the two brothers. There are 2 possible positions for the second brother: either the seat to the immediate left or the seat to the immediate right of the fixed brother, with one person sitting between them.

So, there are **2 choices** for the second brother’s position.

Step 3: Arrange the remaining people
After placing the two brothers, there are 8 remaining people who need to be seated in the remaining 8 seats. These 8 people can be arranged in any order, and the number of ways to arrange 8 people in 8 seats is \(8!\).

Step 4: Calculate the total number of arrangements
The total number of ways to arrange the people with exactly one person sitting between the brothers is:

\[
\text{Total arrangements} = 1 \times 2 \times 8! = 2 \times 8!
\]

We know that \(8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40,320\).

Thus, the total number of arrangements is:

\[
2 \times 40,320 = 80,640
\]

Final Answer:
The total number of ways in which these 10 persons can be seated around a round table such that exactly one person sits between the brothers is 80,640.

Answer: Option A
Solution
To determine the rank of the word **KUBER** when the letters of the word are arranged in all possible orders in dictionary (lexicographical) order, we need to systematically count how many words come before “KUBER” in the list of all possible permutations.

Step 1: Find the total number of distinct permutations
The word **KUBER** consists of 5 distinct letters: K, U, B, E, and R. The total number of distinct permutations of these 5 letters is:

\[
5! = 5 \times 4 \times 3 \times 2 \times 1 = 120
\]

So, there are 120 distinct permutations of the word **KUBER**.

Step 2: Identify the dictionary order of the letters
To arrange the letters in dictionary order, we first list the letters in alphabetical order:

\[
B, E, K, R, U
\]

Step 3: Count how many words come before “KUBER”
We will start from the leftmost letter of the word “KUBER” and count how many words start with each possible letter that comes before the corresponding letter in “KUBER”.

1. **First letter (K):**
The letters before **K** in alphabetical order are **B** and **E**. For each of these letters, we can form words with the remaining 4 letters. So, for each of **B** and **E**, there are \(4!\) distinct permutations of the remaining letters.

Number of permutations starting with **B** or **E**:
\[
2 \times 4! = 2 \times 24 = 48
\]

2. **Second letter (U), after “K”:**
Now, we look at the remaining letters: **U, B, E, R**. The letter **U** in “KUBER” is preceded by **B**, **E**, and **R** in dictionary order. For each of **B**, **E**, and **R**, we can form words with the remaining 3 letters.

Number of permutations starting with **KB**, **KE**, or **KR**:
\[
3 \times 3! = 3 \times 6 = 18
\]

3. **Third letter (B), after “KU”:**
After fixing **KU**, the remaining letters are **B, E, R**. The letter **B** in “KUBER” is the first letter in alphabetical order among **B**, **E**, and **R**. Hence, there are no words that come before “KUBER” with “KU” as the prefix.

4. **Fourth letter (E), after “KUB”:**
Now, the remaining letters are **E, R**. The letter **E** in “KUBER” is the first letter in alphabetical order. Hence, there are no words that come before “KUBER” with “KUB” as the prefix.

5. Fifth letter (R), after “KUBE”:
The only remaining letter is R, and since it’s the last letter, there’s no possibility of any other word coming before “KUBER” with “KUBE” as the prefix.

Step 4: Calculate the rank
Now, we add up all the words that come before **KUBER**:

\[
\text{Words before “KUBER”} = 48 + 18 + 0 + 0 = 66
\]

Since KUBER is the next word, its rank is \(66 + 1 = 67\).

Final Answer:
The rank of the word **KUBER** in the dictionary order is 67.

Answer: Option B
Solution:

To solve this problem, let’s break it down step by step.

Step 1: Understanding the problem
– The lock has 4 rings.
– Each ring contains 9 non-zero digits. These digits are the numbers from 1 to 9.
– You need to set a 4-digit code to open the lock, with each digit corresponding to one of the rings.

Step 2: Determine the choices for each ring
– For each ring, you can choose any of the 9 digits (since the digits are non-zero, they range from 1 to 9).
– So, for each ring, there are 9 possible choices for the digit.

Step 3: Calculate the total number of possible codes
Since there are 9 choices for each of the 4 rings, the total number of possible codes is:

\[
9 \times 9 \times 9 \times 9 = 9^4
\]

Step 4: Calculate \(9^4\)
We now compute \(9^4\):

\[
9^4 = 9 \times 9 \times 9 \times 9 = 81 \times 81 = 6561
\]

Final Answer:
The maximum number of codes that can be formed to open the lock is 6561.

Answer: Option C
Solution:
To determine how many ways the 10 students can be seated in two rows for the Mock test, with the condition that each row can only have one set of papers (either Code A or Code B), and each row must have a different set of papers, we need to follow these steps:

Step 1: Seating the Students in the Rows
– There are 10 students in total, and they are to be seated in two rows, each having 5 students.
– The number of ways to arrange the 10 students in two rows of 5 students each is to first choose 5 students for the first row and then seat the remaining 5 in the second row.

1. Choosing 5 students for the first row:** The number of ways to select 5 students from 10 is given by the combination formula \(\binom{10}{5}\), which is:

\[
\binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252
\]

After choosing 5 students for the first row, the remaining 5 students automatically go to the second row.

– Arranging the students in each row: Once the students are assigned to the rows, each row can have the students arranged in any order. Since each row has 5 students, the number of ways to arrange the 5 students in each row is \(5!\):

\[
5! = 5 \times 4 \times 3 \times 2 \times 1 = 120
\]

So, the total number of ways to arrange the students in both rows is:

\[
\text{Total ways to arrange students} = \binom{10}{5} \times 5! \times 5! = 252 \times 120 \times 120
\]

Step 2: Assigning Paper Codes to the Rows
– Each row can only have one set of paper (either Code A or Code B), and each row must have a different set. So, there are 2 choices for assigning the paper sets to the rows: one row gets Code A, and the other gets Code B.

Step 3: Calculate the Total Number of Ways
Now, we calculate the total number of ways by multiplying the results from both steps:

\[
\text{Total number of ways} = \binom{10}{5} \times 5! \times 5! \times 2
\] \[
\text{Total number of ways} = 252 \times 120 \times 120 \times 2
\]

Let’s calculate this step by step:

\[
252 \times 120 = 30240
\] \[
30240 \times 120 = 3628800
\] \[
3628800 \times 2 = 7257600
\]

Final Answer:
The total number of ways the 10 students can be arranged in the two rows, with the different paper codes assigned to each row, is 7,257,600.

Answer: Option B
Solution:
To determine how many ways 4 prizes can be given to 3 boys, where each boy is eligible for all the prizes, we need to consider the following:

Step 1: Understanding the problem
Each of the 4 prizes can be given to any of the 3 boys, and there is no restriction on how many prizes each boy can receive. This means that a boy can receive more than one prize, and it’s possible for some boys to receive no prizes at all.

Step 2: Assigning prizes
For each of the 4 prizes, we have 3 possible choices (since there are 3 boys, and any boy can receive any prize). So, for each prize, there are 3 options for who can receive it.

Thus, the total number of ways to assign the 4 prizes is:

\[
3 \times 3 \times 3 \times 3 = 3^4
\]

Step 3: Calculate the total number of ways
Now, calculate \(3^4\):

\[
3^4 = 3 \times 3 \times 3 \times 3 = 81
\]

Final Answer:
The total number of ways to give 4 prizes to 3 boys is 81.

Answer: Option B
Solution:
To find the number of ways in which 8064 can be resolved as the product of two factors, we need to determine how many distinct pairs of factors (a, b) satisfy \( 8064 = a \times b \), where \(a\) and \(b\) are integers.

Step 1: Prime Factorization of 8064
We begin by performing the prime factorization of 8064.

1. 8064 is even, so divide by 2:
\[
8064 \div 2 = 4032
\] 2. 4032 is even, so divide by 2 again:
\[
4032 \div 2 = 2016
\] 3. 2016 is even, so divide by 2 again:
\[
2016 \div 2 = 1008
\] 4. 1008 is even, so divide by 2 again:
\[
1008 \div 2 = 504
\] 5. 504 is even, so divide by 2 again:
\[
504 \div 2 = 252
\] 6. 252 is even, so divide by 2 again:
\[
252 \div 2 = 126
\] 7. 126 is even, so divide by 2 again:
\[
126 \div 2 = 63
\] 8. 63 is not divisible by 2, so divide by 3 (next smallest prime):
\[
63 \div 3 = 21
\] 9. 21 is divisible by 3, so divide by 3:
\[
21 \div 3 = 7
\] 10. 7 is prime.

Thus, the prime factorization of 8064 is:

\[
8064 = 2^7 \times 3^2 \times 7
\]

Step 2: Finding the Number of Divisors
The number of divisors of a number is determined by adding 1 to each of the exponents in its prime factorization and then multiplying the results.

For \( 8064 = 2^7 \times 3^2 \times 7^1 \), the number of divisors is:

\[
(7 + 1) \times (2 + 1) \times (1 + 1) = 8 \times 3 \times 2 = 48
\]

Thus, 8064 has 48 divisors.

Step 3: Pairing the Divisors
The divisors of 8064 come in pairs that multiply to 8064. For example, if \( d \) is a divisor of 8064, then \( \frac{8064}{d} \) is also a divisor. The pairs of divisors are:

\[
(d, \frac{8064}{d})
\]

Since each pair \( (d, \frac{8064}{d}) \) corresponds to one factorization, and there are 48 divisors, the number of ways to write 8064 as the product of two factors is:

\[
\frac{48}{2} = 24
\]

Thus, there are **24 distinct ways** to resolve 8064 as the product of two factors.

Final Answer:
The number of ways in which 8064 can be resolved as the product of two factors is 24.

Answer: Option A
Solution:
To determine the rank of the word LABOUR when the letters of the word are arranged in all possible ways in dictionary (lexicographical) order, we need to systematically count how many words come before **LABOUR** in the list of all possible permutations of its letters.

Step 1: List the letters in alphabetical order
First, we list the letters of LABOUR in alphabetical order:

\[
A, B, L, O, R, U
\]

Step 2: Find the number of permutations before “LABOUR”
We will now go letter by letter, counting how many words come before LABOUR.

1. First letter: L
The first letter in LABOUR is L. We count how many words start with letters that come before **L** in alphabetical order:

– Letters before L are A and B.

For each of these letters, the remaining 5 positions can be filled with any of the remaining 5 letters. The number of such permutations is \(5! = 120\).

So, there are \(2 \times 120 = 240\) words that start with A or B.

2. Second letter: A (after “L”)
Now, we fix **L** in the first position and look at the second letter, which is **A** in **LABOUR**. We count how many words start with **LA** but have a second letter that comes before **A**.

There are no letters before **A** among the remaining letters **B, O, R, U**. So, there are no words before **LA**.

3. Third letter: B (after “LA”)
Now, we fix **LA** and look at the third letter, which is **B** in **LABOUR**. We count how many words start with **LAB** but have a third letter that comes before **B**.

– Letters before **B** among the remaining letters **O, R, U** are none.

So, there are no words before **LAB**.

4. Fourth letter: O (after “LAB”)
Next, we fix **LAB** and look at the fourth letter, which is **O** in **LABOUR**. We count how many words start with **LAB** but have a fourth letter that comes before **O**.

– Letters before **O** among the remaining letters **R, U** are none.

So, there are no words before **LABO**.

5. Fifth letter: R (after “LABO”)
Now, we fix **LABO** and look at the fifth letter, which is **R** in **LABOUR**. We count how many words start with **LABO** but have a fifth letter that comes before **R**.

– Letters before **R** among the remaining letter **U** is none.

So, there are no words before **LABOR**.

6. Sixth letter: U (after “LABOR”)
Finally, we fix **LABOR** and look at the sixth letter, which is **U** in **LABOUR**. Since **U** is the last letter, there are no words before **LABOUR**.

Step 3: Calculate the rank
Now, we sum up all the words that come before **LABOUR**:

– Number of words starting with **A** or **B**: 240
– Number of words starting with **LA**, **LAB**, **LABO**, and **LABOR**: 0

Thus, the total number of words that come before **LABOUR** is \(240\).

Since **LABOUR** itself is the next word in the list, its rank is:

\[
\text{Rank of LABOUR} = 240 + 1 = 241
\]

Final Answer:
The rank of the word **LABOUR** in the dictionary order is 241.

Answer: Option B
Solution:
To solve this problem, let’s break it down step by step.

Step 1: Seating the Girls
There are 6 girls, and they are sitting in the front row. Since the order of the girls in the front row matters, the number of ways to arrange the 6 girls in the front row is:

\[
6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720
\]

Step 2: Arranging the Boys
There are 20 boys standing behind the 6 girls, and the two corner positions in the back row are reserved for the two tallest boys. Let’s calculate how the boys can be arranged:

1. **Placing the Tallest Boys in the Corner Positions:**
The two tallest boys must occupy the two corner positions. There are 2 corner positions, and the two tallest boys can be arranged in these two positions in \(2!\) ways:

\[
2! = 2 \times 1 = 2
\]

2. **Arranging the Remaining 18 Boys:**
After placing the two tallest boys in the corner positions, there are 18 remaining boys who need to be arranged in the remaining 18 positions. The number of ways to arrange these 18 boys is:

\[
18! = 18 \times 17 \times 16 \times \cdots \times 1
\]

Step 3: Total Number of Arrangements
Now, we combine all the possible arrangements. The total number of ways to arrange the students is the product of the arrangements for the girls and the arrangements for the boys:

\[
\text{Total arrangements} = 6! \times 2! \times 18!
\]

Substitute the values:

\[
\text{Total arrangements} = 720 \times 2 \times 18!
\]

Final Answer:
The total number of ways to arrange the students for the class photograph is \( 720 \times 2 \times 18! \).

Answer: Option A
Solution:
To solve this problem, we need to find the number of 5-letter words that can be formed from 10 different letters, where **at least one letter is repeated**.

Step 1: Calculate the total number of 5-letter words (with no restrictions)
When forming a 5-letter word using 10 different letters, the number of possible words (where letters can be repeated) is:

\[
10^5 = 10 \times 10 \times 10 \times 10 \times 10 = 100,000
\]

Step 2: Calculate the number of 5-letter words with **no repetition** of letters
If we want to form a word where no letter is repeated, we choose 5 distinct letters from the 10 available letters. The number of ways to do this is:

– For the first letter, there are 10 choices.
– For the second letter, there are 9 remaining choices.
– For the third letter, there are 8 remaining choices.
– For the fourth letter, there are 7 remaining choices.
– For the fifth letter, there are 6 remaining choices.

Thus, the number of 5-letter words with no repetition of letters is:

\[
10 \times 9 \times 8 \times 7 \times 6 = 30,240
\]

Step 3: Calculate the number of words with **at least one letter repeated**
To find the number of 5-letter words with at least one letter repeated, we subtract the number of words with no repetition from the total number of words:

\[
\text{Words with at least one repeated letter} = 10^5 – 10 \times 9 \times 8 \times 7 \times 6
\]

Substitute the values:

\[
\text{Words with at least one repeated letter} = 100,000 – 30,240 = 69,760
\]

Final Answer:
The number of 5-letter words that have at least one letter repeated is 69,760.

Answer: Option B
Solution:
Given there are 12 numbered chairs, such that chairs numbered 1 to 8 should be occupied.
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
The various combinations of chairs that ensure that no two men are sitting together are listed.
(1, 3, 5, __ ), The fourth chair can be 5,6,10,11 or 12, hence 5 ways.
(1, 4, 8, __ ), The fourth chair can be 6,10,11 or 12 hence 4 ways.
(1, 5, 8, __ ), the fourth chair can be 10,11 or 12 hence 3 ways.
(1, 6, 8, __ ), the fourth chair can be 10,11 or 12 hence 3 ways.
(1, 8, 10, 12) is also one of the combinations.
Hence, 16 such combinations exist.
In case of each these combinations we can make the four men inter arrange in 4! ways.
Hence, the required result =16 × 4! = 384

Answer: Option C
Solution:
We are tasked with finding how many words can be formed by rearranging the letters of the word “ASCENT” such that A and T occupy the first and last positions, respectively.

Step-by-step breakdown:

1. Fix the positions of A and T:
– The first position must be occupied by A.
– The last position must be occupied by T.

2. Rearrange the remaining letters:
– The remaining letters after fixing A in the first position and T in the last position are S, C, E, and N.
– These 4 letters need to be rearranged in the middle positions.

3. Count the possible arrangements:
– The number of ways to arrange 4 distinct letters (S, C, E, N) in 4 positions is given by the factorial of 4, which is:

\[
4! = 4 \times 3 \times 2 \times 1 = 24
\]

Thus, the total number of words that can be formed with A in the first position and T in the last position is 24.

Answer: Option C
Solution:
We are given two equations involving permutations and combinations:

1. \( ^6P_r = 360 \)
2. \( ^rC_r = 15 \)

We need to find the value of \( r \).

Step 1: Express the permutation equation

The formula for permutations is given by:

\[
^nP_r = \frac{n!}{(n – r)!}
\]

For \( ^6P_r \), the formula becomes:

\[
^6P_r = \frac{6!}{(6 – r)!} = 360
\]

So, we have:

\[
\frac{6!}{(6 – r)!} = 360
\]

First, calculate \( 6! \):

\[
6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720
\]

Thus, the equation becomes:

\[
\frac{720}{(6 – r)!} = 360
\]

Now solve for \( (6 – r)! \):

\[
(6 – r)! = \frac{720}{360} = 2
\]

This means:

\[
(6 – r)! = 2
\]

The factorial of 2 is 2, so:

\[
6 – r = 2
\]

Thus:

\[
r = 4
\]

Step 2: Verify using the combination equation

The formula for combinations is:

\[
^rC_r = \frac{r!}{r!(r – r)!} = 1
\]

However, we are given \( ^rC_r = 15 \). Therefore, the equation is:

\[
^rC_r = \binom{r}{r} = 15
\]

We can check if \( r = 4 \) satisfies this.

Answer: Option D
Solution:
We are tasked with determining how many different ways six distinct rings can be worn on four fingers of one hand.

Step-by-step breakdown:

1. Assigning Rings to Fingers:
– There are 6 distinct rings, and we have 4 fingers to wear them on.
– Each ring can be placed on any of the 4 fingers, and it is possible for multiple rings to be worn on the same finger.

2. Number of Choices for Each Ring:
– For each of the 6 rings, there are 4 possible fingers to choose from.

Thus, the total number of ways to assign the rings to the fingers is given by:

\[
4 \times 4 \times 4 \times 4 \times 4 \times 4 = 4^6
\]

3. Calculation of \(4^6\):

\[
4^6 = 4096
\]

Final Answer:
Therefore, the total number of ways in which the six different rings can be worn on four fingers is 4096.

Answer: Option A
Solution:
To determine how many triangles can be formed by joining 7 non-collinear points, we need to calculate how many ways we can select 3 points from the 7, since a triangle is formed by three non-collinear points.

Step-by-step breakdown:

1. Number of ways to choose 3 points from 7:

The number of ways to choose 3 points from 7 is given by the combination formula:

\[
^nC_r = \frac{n!}{r!(n – r)!}
\]

Here, \(n = 7\) (the number of points) and \(r = 3\) (the number of points required to form a triangle). So, we calculate:

\[
^7C_3 = \frac{7!}{3!(7 – 3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = \frac{210}{6} = 35
\]

Final Answer:
Therefore, the number of triangles that can be drawn by joining these 7 non-collinear points is 35.

Answer: Option A
Solution:
We are tasked with forming a committee of 5 members from 6 men and 4 ladies, with the condition that the committee must include at least one lady.

Step-by-step breakdown:

1. Total ways to form a committee of 5:
First, let’s calculate the total number of ways to form a committee of 5 members from the 10 people (6 men + 4 ladies) without any restriction.

The total number of ways to choose 5 people from 10 is given by the combination formula:

\[
^{10}C_5 = \frac{10!}{5!(10 – 5)!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252
\]

2. Committees with no ladies (only men):
Now, let’s calculate how many ways we can form a committee of 5 using only the 6 men. This is the number of ways to choose 5 people from the 6 men:

\[
^6C_5 = \frac{6!}{5!(6 – 5)!} = \frac{6}{1} = 6
\]

3. Committees with at least one lady:
To find the number of committees with at least one lady, we subtract the number of all-male committees from the total number of committees:

\[
\text{Committees with at least one lady} = ^{10}C_5 – ^6C_5 = 252 – 6 = 246
\]

Final Answer:

The number of ways to form a committee of 5 people that includes at least one lady is 246.

Answer: Option D
Solution:
We are given a question paper with three sections, each containing a different number of questions:
– Section 1: 4 questions
– Section 2: 5 questions
– Section 3: 6 questions

The conditions are:
– A candidate must attempt at least one question from each section.
– A candidate need not attempt all the questions in any section.

Step-by-step breakdown:

1. For Section 1 (4 questions):
– The candidate must attempt at least 1 question.
– The total number of ways to attempt questions in this section can be calculated by considering all possible non-empty subsets of the 4 questions.
– The total number of subsets of 4 questions is \( 2^4 = 16 \) (since for each question, the candidate can either attempt it or not).
– However, the candidate must attempt at least 1 question, so we subtract the empty subset (the case where no questions are attempted). Thus, the number of ways to attempt questions in Section 1 is:

\[
16 – 1 = 15
\]

2. For Section 2 (5 questions):
– Similarly, the total number of subsets of 5 questions is \( 2^5 = 32 \).
– Subtracting the empty subset (no questions attempted), the number of ways to attempt questions in Section 2 is:

\[
32 – 1 = 31
\]

3. **For Section 3 (6 questions):**
– The total number of subsets of 6 questions is \( 2^6 = 64 \).
– Subtracting the empty subset, the number of ways to attempt questions in Section 3 is:

\[
64 – 1 = 63
\]

4. Total number of ways to attempt questions:
– Since the candidate must attempt at least one question from each section, the total number of ways to attempt questions across all sections is the product of the number of ways for each section:

\[
15 \times 31 \times 63
\]

Let’s calculate this step-by-step:

\[
15 \times 31 = 465
\]

\[
465 \times 63 = 29295
\]

Final Answer:

Therefore, the total number of ways a candidate can attempt the questions is 29,295.

Answer: Option A
Solution:
We are given that in a hockey championship, 153 matches are played, and each pair of teams plays one match with each other. We need to determine the number of teams, \( n \), participating in the championship.

Step-by-step breakdown:

1.Understanding the situation:
– Each pair of teams plays exactly one match.
– The number of ways to select 2 teams out of \( n \) teams is given by the combination formula \( ^nC_2 \), which counts the number of ways to choose 2 teams from \( n \) teams.

2. Formula for combinations:
The formula for the number of ways to choose 2 teams out of \( n \) is:

\[
^nC_2 = \frac{n(n – 1)}{2}
\]

This represents the total number of matches played in the tournament.

3. Set up the equation:
We are told that 153 matches are played, so:

\[
\frac{n(n – 1)}{2} = 153
\]

4. Solve the equation:
Multiply both sides by 2 to eliminate the fraction:

\[
n(n – 1) = 306
\]

Now, expand and rearrange the equation:

\[
n^2 – n – 306 = 0
\]

5. Solve the quadratic equation:
We can solve this quadratic equation using the quadratic formula. The quadratic formula is:

\[
n = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
\]

For the equation \( n^2 – n – 306 = 0 \), the coefficients are:
– \( a = 1 \)
– \( b = -1 \)
– \( c = -306 \)

Plugging these values into the quadratic formula:

\[
n = \frac{-(-1) \pm \sqrt{(-1)^2 – 4(1)(-306)}}{2(1)}
\]

Simplifying:

\[
n = \frac{1 \pm \sqrt{1 + 1224}}{2}
\]

\[
n = \frac{1 \pm \sqrt{1225}}{2}
\]

\[
n = \frac{1 \pm 35}{2}
\]

So, we have two possible solutions:

\[
n = \frac{1 + 35}{2} = \frac{36}{2} = 18
\] or
\[
n = \frac{1 – 35}{2} = \frac{-34}{2} = -17
\]

Since the number of teams cannot be negative, we discard \( n = -17 \).

Final Answer:

Therefore, the number of teams participating in the championship is 18.

Answer: Option B
Solution:
We are given a box containing 10 balls, where 3 are red and 7 are blue. We are tasked with finding how many ways a random sample of 6 balls can be drawn from the box under the following conditions:
– At most 2 red balls are included in the sample.
– No sample contains all 6 balls of the same color.

Step-by-step breakdown:

1. Total number of red and blue balls:
– Red balls = 3
– Blue balls = 7

2. Conditions to satisfy:
– At most 2 red balls.
– No sample has all 6 balls of the same color (i.e., not all 6 balls can be red or all 6 balls blue).

3. Possible scenarios for drawing 6 balls:
Since the sample should have at most 2 red balls, the possible combinations of red and blue balls in the sample are:
– 0 red balls and 6 blue balls
– 1 red ball and 5 blue balls
– 2 red balls and 4 blue balls

4.Case 1: 0 red balls and 6 blue balls
– We cannot have all 6 balls blue, because this would violate the condition that no sample has all balls of the same color.
– Thus, this case does not contribute any valid samples.

5.Case 2: 1 red ball and 5 blue balls
– We choose 1 red ball from the 3 available red balls: \( ^3C_1 = 3 \)
– We choose 5 blue balls from the 7 available blue balls: \( ^7C_5 = \frac{7 \times 6}{2 \times 1} = 21 \)
– Total number of ways for this case: \( 3 \times 21 = 63 \)

6. Case 3: 2 red balls and 4 blue balls
– We choose 2 red balls from the 3 available red balls: \( ^3C_2 = \frac{3 \times 2}{2 \times 1} = 3 \)
– We choose 4 blue balls from the 7 available blue balls: \( ^7C_4 = \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} = 35 \)
– Total number of ways for this case: \( 3 \times 35 = 105 \)

Total number of ways:
Adding the valid cases together:

\[
63 \text{ (from Case 2)} + 105 \text{ (from Case 3)} = 168
\]

Final Answer:
Therefore, the total number of ways to draw a random sample of 6 balls that satisfies the given conditions is 168.

Answer: Option D
Solution:
We are given a situation with 8 crew members, where:

– 3 particular crew members can only sit on the left side.
– 2 particular crew members can only sit on the right side.
– We need to arrange 4 crew members on each side, with 4 on the left and 4 on the right.

Step-by-step breakdown:

1. Assigning the 3 particular members to the left side:
– Since 3 particular crew members must sit on the left side, they are automatically assigned to the left side. This means we need to select 1 additional crew member from the remaining 3 members (since there are 8 crew members in total, and 5 members have already been assigned their seating constraints).

2. Choosing the 1 additional member for the left side:
– The remaining crew members are 3 that haven’t been assigned yet. Out of these 3 members, we need to select 1 to sit on the left side. The number of ways to do this is:

\[
^3C_1 = 3
\]

3. Assigning the 2 particular members to the right side:
– The 2 particular crew members who must sit on the right side are automatically assigned to the right side.

4. Choosing the 2 additional members for the right side:
– After assigning the 2 particular crew members to the right side, we need to select 2 additional crew members from the remaining 2 members (after assigning the left side). Since all 2 remaining members must go to the right side, there is only 1 way to do this:

\[
^2C_2 = 1
\]

5. Arranging the crew members on each side:
– After assigning the crew members to their respective sides, we now need to arrange them. The 4 crew members on the left side can be arranged in:

\[
4! = 4 \times 3 \times 2 \times 1 = 24
\]

Similarly, the 4 crew members on the right side can be arranged in:

\[
4! = 24
\]

Total number of arrangements:
To find the total number of ways to arrange the crew, we multiply the number of ways to assign crew members to the left and right sides and the number of arrangements for each side:

\[
3 \times 1 \times 24 \times 24 = 1728
\]

Final Answer:
Therefore, the total number of ways in which the crew can be arranged so that 4 members sit on each side is 1728.

Answer: Option C
Solution:
We are given a man positioned at the origin (0,0) on a coordinate plane. The man can take steps of unit measure in one of four directions: North (N), East (E), West (W), or South (S). We are tasked with finding the number of ways he can reach the point (5,6) while covering the shortest possible distance.

Step-by-step breakdown:

1. Understanding the problem:
– The man starts at the origin, (0,0), and needs to reach the point (5,6).
– The shortest path to (5,6) requires exactly:
– 5 steps to the East (E), since the x-coordinate needs to increase from 0 to 5.
– 6 steps to the North (N), since the y-coordinate needs to increase from 0 to 6.

We cannot take any steps to the West (W) or South (S), because that would increase the distance from the target point.

2. Total steps:
To reach (5,6), the man must take a total of:
– 5 steps to the East (E),
– 6 steps to the North (N).

In total, he will take \(5 + 6 = 11\) steps.

3. Counting the number of ways:
To determine how many distinct ways the man can reach the point (5,6), we need to figure out how to arrange these 5 East steps and 6 North steps in a sequence of 11 steps.

The problem boils down to selecting 5 positions for the East (E) steps out of the 11 available steps (the remaining 6 positions will automatically be assigned to the North (N) steps). The number of ways to do this is given by the combination formula:

\[
\text{Number of ways} = ^{11}C_5 = \frac{11!}{5!(11 – 5)!} = \frac{11!}{5!6!}
\]

4. Calculation:

First, calculate \( 11! \), \( 5! \), and \( 6! \):
\[
11! = 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 39916800
\] \[
5! = 5 \times 4 \times 3 \times 2 \times 1 = 120
\] \[
6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720
\]

Now, plug these into the combination formula:

\[
^{11}C_5 = \frac{39916800}{120 \times 720} = \frac{39916800}{86400} = 462
\]

Final Answer:
Therefore, the number of ways the man can reach the point (5,6) covering the shortest possible distance is 462.

Answer: Option D
Solution:
We are given 6 equally spaced points labeled \(A, B, C, D, E, F\) on a circle with radius \(R\). The task is to determine how many convex pentagons of distinctly different areas can be drawn using these points as vertices.

Step-by-step breakdown:

1.Selecting 5 points from the 6 points:
– A pentagon requires 5 distinct vertices. Out of the 6 points, we need to choose 5 points to form a pentagon.
– The number of ways to choose 5 points from 6 is given by the combination formula \( ^6C_5 \):

\[
^6C_5 = \frac{6!}{5!(6-5)!} = 6
\]

Therefore, there are 6 possible ways to select 5 points from the 6 points.

2. Nature of the convex pentagons:
– Since the points are equally spaced on the circle, any selection of 5 points will form a convex pentagon. A convex polygon is one where all its interior angles are less than 180° (which will always be the case here because all the points lie on the circumference of a circle).

3. Distinct areas of pentagons:
– The key observation here is that the area of a pentagon inscribed in a circle depends on the distances between the points and the overall shape of the pentagon. As the points are equally spaced, each selection of 5 points results in a distinct pentagon with a unique set of side lengths and angles.
– For each selection of 5 points, the configuration of the pentagon is different, which will result in a distinctly different area. The areas of these pentagons are influenced by the specific arrangement of the points, such as how far apart they are and the overall shape of the pentagon.

4.Conclusion:
– Since each of the 6 selections of 5 points forms a convex pentagon, and each selection of points results in a distinct pentagon with a unique area, the number of distinct convex pentagons with different areas is simply the number of ways to select 5 points from 6, which is 6.

Final Answer:
The number of convex pentagons of distinctly different areas that can be drawn is 6.

Answer: Option B
Solution:
We are given an examination paper with two groups, each containing 4 questions. A candidate is required to attempt 5 questions, but the condition is that no more than 3 questions can be selected from any one group.

We need to determine how many ways the candidate can select 5 questions under these constraints.

Step-by-step breakdown:

1.Total number of questions:
– There are 4 questions in each group, so the total number of questions in the paper is \(4 + 4 = 8\).

2. Selection constraints:
– The candidate needs to select exactly 5 questions.
– The candidate cannot select more than 3 questions from any one group. This means the possible combinations of questions from the two groups are:
– 3 questions from Group 1 and 2 questions from Group 2.
– 2 questions from Group 1 and 3 questions from Group 2.

3. Case 1: 3 questions from Group 1 and 2 questions from Group 2
– From Group 1 (which has 4 questions), the number of ways to select 3 questions is given by \( ^4C_3 \):

\[
^4C_3 = \frac{4!}{3!(4 – 3)!} = 4
\]

– From Group 2 (which also has 4 questions), the number of ways to select 2 questions is given by \( ^4C_2 \):

\[
^4C_2 = \frac{4!}{2!(4 – 2)!} = \frac{4 \times 3}{2 \times 1} = 6
\]

– Therefore, the total number of ways to select 3 questions from Group 1 and 2 questions from Group 2 is:

\[
4 \times 6 = 24
\]

4. **Case 2: 2 questions from Group 1 and 3 questions from Group 2**
– From Group 1, the number of ways to select 2 questions is \( ^4C_2 \):

\[
^4C_2 = 6
\]

– From Group 2, the number of ways to select 3 questions is \( ^4C_3 \):

\[
^4C_3 = 4
\]

– Therefore, the total number of ways to select 2 questions from Group 1 and 3 questions from Group 2 is:

\[
6 \times 4 = 24
\]

5. Total number of ways:
The total number of ways to select 5 questions, considering both cases, is the sum of the results from Case 1 and Case 2:

\[
24 + 24 = 48
\]

Final Answer:
The number of ways in which 5 questions can be selected is 48.

Answer: Option B
Solution:
We are given that after every get-together, every person present shakes hands with every other person. The total number of handshakes at the party is 105, and we need to find how many persons were present.

Step-by-step breakdown:

1. Understanding the problem:
– Each handshake involves two people, so the number of handshakes is the number of ways to choose 2 people from the total number of people.
– If there are \( n \) persons present, the number of handshakes can be calculated using the combination formula \( ^nC_2 \), which gives the number of ways to choose 2 people from \( n \):

\[
^nC_2 = \frac{n(n-1)}{2}
\]

We are told that there were 105 handshakes, so:

\[
\frac{n(n-1)}{2} = 105
\]

2. Solve for \( n \):
Multiply both sides of the equation by 2 to eliminate the fraction:

\[
n(n-1) = 210
\]

Now expand the left side:

\[
n^2 – n = 210
\]

Rearrange the equation to form a quadratic equation:

\[
n^2 – n – 210 = 0
\]

3. Solve the quadratic equation:
We can solve this quadratic equation using the quadratic formula. The quadratic formula is:

\[
n = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
\]

For the equation \( n^2 – n – 210 = 0 \), the coefficients are:
– \( a = 1 \)
– \( b = -1 \)
– \( c = -210 \)

Substituting these values into the quadratic formula:

\[
n = \frac{-(-1) \pm \sqrt{(-1)^2 – 4(1)(-210)}}{2(1)}
\]

Simplifying:

\[
n = \frac{1 \pm \sqrt{1 + 840}}{2}
\]

\[
n = \frac{1 \pm \sqrt{841}}{2}
\]

\[
n = \frac{1 \pm 29}{2}
\]

4. Calculate the possible values of \( n \):
– \( n = \frac{1 + 29}{2} = \frac{30}{2} = 15 \)
– \( n = \frac{1 – 29}{2} = \frac{-28}{2} = -14 \)

Since the number of people cannot be negative, we discard \( n = -14 \).

Final Answer:
Therefore, the number of persons present at the party is 15.

Answer: Option A
Solution:
To determine how many diagonals can be drawn in a pentagon, we can use the formula for the number of diagonals in a polygon with \( n \) sides:

\[
\text{Number of diagonals} = \frac{n(n – 3)}{2}
\]

For a pentagon, the number of sides \( n = 5 \). Plugging this into the formula:

\[
\text{Number of diagonals} = \frac{5(5 – 3)}{2} = \frac{5 \times 2}{2} = 5
\]

Final Answer:
Therefore, a pentagon has 5 diagonals.

Answer: Option B
Solution:
We are given a group of 5 women and 6 men, and we need to select a committee of 4 members, consisting of 3 women and 1 man.

Step-by-step breakdown:

1. Choosing 3 women:
From the 5 women, we need to select 3 women. The number of ways to choose 3 women from 5 is given by the combination formula:

\[
^5C_3 = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10
\]

2. Choosing 1 man:
From the 6 men, we need to select 1 man. The number of ways to choose 1 man from 6 is:

\[
^6C_1 = 6
\]

3. Total number of ways:
To form the committee, we multiply the number of ways to select 3 women and 1 man:

\[
\text{Total ways} = ^5C_3 \times ^6C_1 = 10 \times 6 = 60
\]

Final Answer:
Therefore, the number of different ways to form a committee of 3 women and 1 man is 60.

Answer: Option B
Solution:
We are asked to find the total number of distinct vehicle numbers that can be formed using two letters followed by two numbers, with the condition that the two letters must be distinct.

Step-by-step breakdown:

1. Choosing the letters:
– The English alphabet contains 26 letters.
– For the first letter, we can choose any of the 26 letters.
– For the second letter, since the letters must be distinct, we can choose any of the remaining 25 letters.

So, the number of ways to select the two distinct letters is:

\[
26 \times 25 = 650
\]

2. Choosing the numbers:
– The vehicle number also includes two digits, and each digit can be any number from 0 to 9, so there are 10 possible choices for each digit.
– The digits are not required to be distinct, so for the first number, we can choose any of the 10 digits, and for the second number, we can also choose any of the 10 digits.

So, the number of ways to select the two digits is:

\[
10 \times 10 = 100
\]

3. Total number of distinct vehicle numbers:
To find the total number of distinct vehicle numbers, we multiply the number of ways to choose the letters and the number of ways to choose the digits:

\[
650 \times 100 = 65,000
\]

Final Answer:
The total number of distinct vehicle numbers that can be formed is 65,000.

Answer: Option D
Solution:
We are given a railway compartment with 2 rows of seats facing each other, with 5 seats in each row, making a total of 10 seats. The passengers have specific seating preferences:
– 4 passengers wish to sit facing forward.
– 3 passengers wish to sit facing towards the rear.
– 3 passengers are indifferent and can sit in any available seat.

We need to determine the number of ways in which these 10 passengers can be seated, given their preferences.

Step-by-step breakdown:

1. Seats Arrangement:
– There are 5 seats facing forward and 5 seats facing towards the rear.
– We need to assign the passengers to these seats based on their preferences.

2. Seating the 4 passengers who want to face forward:
– Out of the 5 seats facing forward, we need to choose 4 seats for the passengers who want to sit facing forward. The number of ways to choose 4 seats from 5 is given by \( ^5C_4 \):

\[
^5C_4 = \frac{5!}{4!(5-4)!} = 5
\]

– After selecting the 4 seats, we can arrange the 4 passengers who wish to face forward in those 4 seats. The number of ways to arrange 4 passengers in 4 seats is \( 4! \):

\[
4! = 4 \times 3 \times 2 \times 1 = 24
\]

3. Seating the 3 passengers who want to face towards the rear:
– Out of the 5 seats facing the rear, we need to choose 3 seats for the passengers who want to sit facing the rear. The number of ways to choose 3 seats from 5 is \( ^5C_3 \):

\[
^5C_3 = \frac{5!}{3!(5-3)!} = 10
\]

– After selecting the 3 seats, we can arrange the 3 passengers who wish to face towards the rear in those 3 seats. The number of ways to arrange 3 passengers in 3 seats is \( 3! \):

\[
3! = 3 \times 2 \times 1 = 6
\]

4. Seating the 3 indifferent passengers:
– There are 3 remaining seats (one in the forward row and two in the rear row) that the 3 indifferent passengers can sit in. The number of ways to arrange the 3 indifferent passengers in these 3 seats is \( 3! \):

\[
3! = 3 \times 2 \times 1 = 6
\]

5. Total number of seating arrangements:
To find the total number of ways to seat the 10 passengers, we multiply all the individual results:

\[
\text{Total ways} = ^5C_4 \times 4! \times ^5C_3 \times 3! \times 3! = 5 \times 24 \times 10 \times 6 \times 6
\]

Simplifying:

\[
\text{Total ways} = 5 \times 24 \times 10 \times 36 = 43200
\]

Final Answer:
The total number of ways the 10 passengers can be seated is 43,200.

Answer: Option C
Solution:
We are asked to distribute 3 prizes among 5 students, with the condition that no student gets more than one prize.

Step-by-step breakdown:

1. Selecting the students who will receive prizes:
– We need to select 3 students from the 5 to receive the prizes. The number of ways to choose 3 students from 5 is given by the combination formula \( ^5C_3 \):

\[
^5C_3 = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10
\]

2. Assigning the prizes:
– After selecting the 3 students, we need to assign the 3 distinct prizes to these 3 students. The number of ways to assign 3 distinct prizes to 3 students is \( 3! \):

\[
3! = 3 \times 2 \times 1 = 6
\]

3. Total number of ways:
The total number of ways to distribute the prizes is the product of the number of ways to select the students and the number of ways to assign the prizes:

\[
\text{Total ways} = ^5C_3 \times 3! = 10 \times 6 = 60
\]

Final Answer:
The total number of ways to distribute the 3 prizes among the 5 students is 60.

Answer: Option C
Solution:
We are given that a teacher has 6 students and takes 2 students at a time to the zoo. The teacher does this as often as possible, without repeating any pair of students. We need to determine how many times the teacher goes to the zoo.

Step-by-step breakdown:

1. Understanding the problem:
– The teacher is taking 2 students at a time, and no pair of students should be repeated.
– There are 6 students, and we need to find how many distinct pairs can be formed from these 6 students.

2. Number of ways to form pairs:
– The number of ways to select 2 students from 6 is given by the combination formula \( ^6C_2 \), which represents the number of ways to choose 2 objects from 6:

\[
^6C_2 = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15
\]

Therefore, there are 15 distinct pairs of students.

3. Conclusion:
Since the teacher can take each pair of students to the zoo only once, the teacher will go to the zoo 15 times, once for each distinct pair.

Final Answer:
The teacher goes to the zoo 15 times.

Answer: Option C
Solution:
We are given a family consisting of:
– A grandfather,
– 5 sons,
– 1 daughter,
– 8 grandchildren.

We are asked to determine the number of ways in which the family can be seated in a row for dinner, given the following constraints:
1. The 8 grandchildren must occupy the 4 seats at each end of the row.
2. The grandfather refuses to have a grandchild seated on either side of him.

Step-by-step breakdown:

1. Seating the grandchildren:
– There are 4 seats at each end of the row, and the 8 grandchildren must occupy these 8 seats.
– The number of ways to arrange 8 grandchildren in these 8 seats is \( 8! \).

\[
8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320
\]

2. Seating the grandfather:
– The grandfather cannot sit next to a grandchild, so he must sit in one of the 6 remaining seats that are not adjacent to the grandchildren (the 4 middle seats).
– There are 6 possible positions for the grandfather to sit.

3. Seating the remaining family members (5 sons and 1 daughter):
– After seating the grandfather, we have 5 remaining family members (5 sons and 1 daughter) to be seated in the remaining 5 seats.
– The number of ways to arrange these 6 family members in the 6 available seats is \( 6! \).

\[
6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720
\]

4. Total number of arrangements:
The total number of ways to seat the family is the product of:
– The number of ways to arrange the 8 grandchildren,
– The number of ways to seat the grandfather,
– The number of ways to arrange the 6 remaining family members.

Therefore, the total number of ways is:

\[
8! \times 6 \times 6! = 40320 \times 6 \times 720
\]

First, calculate \( 6 \times 720 \):

\[
6 \times 720 = 4320
\]

Now, multiply by \( 8! \):

\[
40320 \times 4320 = 174182400
\]

Final Answer:
The number of ways in which the family can be seated is 174,182,400.

Answer: Option C
Solution:
We are asked to determine in how many ways 5 different toys can be packed into 3 identical boxes such that no box is empty, and any box may hold all the toys.

Step-by-step breakdown:

This problem can be modeled as a **partitions problem** where we need to partition 5 distinct objects (the toys) into 3 non-empty groups (the boxes), and the boxes are identical, meaning the arrangement of the groups does not matter.

This is essentially asking for the number of **set partitions** of 5 distinct toys into 3 non-empty subsets.

The number of ways to partition \( n \) distinct objects into \( k \) non-empty subsets is given by the Stirling number of the second kind**, \( S(n, k) \), which counts the ways to partition \( n \) objects into \( k \) non-empty subsets. After finding \( S(5, 3) \), we do not need to multiply by any additional factor since the boxes are identical.

Step 1: Find the Stirling number of the second kind \( S(5, 3) \)
The Stirling number \( S(5, 3) \) counts the number of ways to partition 5 distinct objects into 3 non-empty subsets. This value can be found using the recurrence relation or from known values in Stirling number tables.

The value of \( S(5, 3) \) is **25**.

Step 2: Conclusion
Thus, the number of ways to pack 5 different toys into 3 identical boxes such that no box is empty is **25**.

Final Answer:
There are 25 ways to pack the toys into the boxes.

Answer: Option D
Solution:
We are asked to determine how many outcomes there are when six fair coins are tossed simultaneously, such that at most three of the coins turn up heads.

Step-by-step breakdown:

When six coins are tossed, each coin has two possible outcomes: heads (H) or tails (T). The total number of possible outcomes for tossing six coins is \( 2^6 = 64 \).

Now, we are interested in the number of outcomes where **at most three coins show heads**. This means we need to count the number of outcomes where there are 0, 1, 2, or 3 heads.

We can use the **binomial coefficient** \( ^nC_k \) to determine the number of ways to get \( k \) heads out of \( n \) coins, where \( n = 6 \).

1.0 heads (all tails):
The number of ways to have 0 heads (i.e., all tails) is \( ^6C_0 \):
\[
^6C_0 = 1
\]

2. 1 head:
The number of ways to have 1 head is \( ^6C_1 \):
\[
^6C_1 = 6
\]

3. 2 heads:
The number of ways to have 2 heads is \( ^6C_2 \):
\[
^6C_2 = \frac{6 \times 5}{2 \times 1} = 15
\]

4. 3 heads:
The number of ways to have 3 heads is \( ^6C_3 \):
\[
^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20
\]

Total outcomes with at most 3 heads:
To find the total number of outcomes with at most 3 heads, we sum the results for 0, 1, 2, and 3 heads:

\[
^6C_0 + ^6C_1 + ^6C_2 + ^6C_3 = 1 + 6 + 15 + 20 = 42
\]

Final Answer:
Therefore, there are 42 outcomes where at most three of the coins turn up heads.

Answer: Option A
Solution:
We are tasked with determining how many distinct four-letter initials can be formed using the English alphabet, with the condition that the last letter is always a consonant.

Step-by-step breakdown:

1. Total number of letters in the English alphabet:
The English alphabet consists of 26 letters, which are:
– 5 vowels: A, E, I, O, U
– 21 consonants: all other letters except for the vowels.

2. Choices for the first three letters:
Since there is no restriction on the first three letters, each of them can be any of the 26 letters (vowels or consonants).

Therefore, the number of ways to choose the first, second, and third letters is:

\[
26 \times 26 \times 26 = 26^3
\]

3. Choice for the last letter:
The last letter must be a consonant, and there are 21 consonants in the English alphabet.

Therefore, the number of ways to choose the last letter is 21.

4. Total number of distinct initials:
To find the total number of distinct four-letter initials, we multiply the number of choices for the first three letters by the number of choices for the last letter:

\[
26^3 \times 21
\]

Now calculate \( 26^3 \):

\[
26^3 = 26 \times 26 \times 26 = 17576
\]

Therefore, the total number of distinct initials is:

\[
17576 \times 21 = 368096
\]

Final Answer:
The total number of distinct four-letter initials that can be formed, such that the last letter is always a consonant, is 368,096.

Answer: Option B
Solution:
To find how many times the digit 7 appears when listing all integers from 1 to 1000, we need to examine each place value (hundreds, tens, and ones) individually.

Step-by-step breakdown:

1. Consider numbers from 000 to 999:
Since we are dealing with 1000 numbers (from 000 to 999), we treat the number 1000 separately at the end. We will focus on the three-digit numbers first, with leading zeroes if needed (e.g., 007, 056, etc.).

2. Counting the 7s in the hundreds place:
In the hundreds place, the digit 7 will appear in any number from 700 to 799. These are the 100 numbers: 700, 701, 702, …, 799.

So, there are 100 occurrences of the digit 7 in the hundreds place.

3. Counting the 7s in the tens place:
In the tens place, the digit 7 will appear in numbers like 70x, 71x, 72x, …, 79x, where x can be any digit from 0 to 9. This pattern repeats for each group of 100 numbers:

– 070-079, 170-179, 270-279, …, 970-979

So, there are 10 numbers in each set where the tens digit is 7, and there are 10 sets (one for each hundred: 000-099, 100-199, …, 900-999).

Therefore, there are:

\[
10 \times 10 = 100
\] occurrences of the digit 7 in the tens place.

4. **Counting the 7s in the ones place:**
In the ones place, the digit 7 will appear in numbers like x07, x17, x27, …, x97, where x can be any digit from 0 to 9. This pattern repeats for each group of 10 numbers:

– 007, 017, 027, …, 097, 107, 117, …, 997

So, there are 1 number in each set of 10 where the ones digit is 7, and there are 100 sets (one for each group of 10: 000-009, 010-019, …, 990-999).

Therefore, there are:

\[
100 \times 1 = 100
\] occurrences of the digit 7 in the ones place.

5. Consider the number 1000:
The number 1000 does not contain the digit 7, so it does not contribute any occurrences of 7.

Total number of times the digit 7 appears:
Now, we sum the occurrences of 7 in the hundreds, tens, and ones places:

\[
100 \text{ (hundreds place)} + 100 \text{ (tens place)} + 100 \text{ (ones place)} = 300
\]

Final Answer:
The digit 7 will be written 300 times when listing the integers from 1 to 1000.

Answer: Option C
Solution:
We are tasked with finding the number of ways the letters of the word **EDUCATION** can be rearranged such that the relative positions of the vowels and consonants remain the same as in the original word.

Step-by-step breakdown:

1. dentify the vowels and consonants in EDUCATION:

The word **EDUCATION** consists of the following letters:
– Vowels: E, U, A, I, O (5 vowels)
– Consonants: D, C, T, N (4 consonants)

The vowels and consonants must maintain their relative positions in the rearranged word.

2. Fixed positions for vowels and consonants:

– The positions of the vowels in the original word are the 1st, 3rd, 5th, 7th, and 9th positions. These positions must remain filled with vowels.
– The positions of the consonants are the 2nd, 4th, 6th, and 8th positions. These positions must remain filled with consonants.

3. Rearranging the vowels:

There are 5 vowels (E, U, A, I, O). These vowels must remain in the 5 positions where the vowels originally appeared (1st, 3rd, 5th, 7th, and 9th). The number of ways to rearrange these 5 vowels in these 5 positions is:

\[
5! = 5 \times 4 \times 3 \times 2 \times 1 = 120
\]

4. Rearranging the consonants:

There are 4 consonants (D, C, T, N). These consonants must remain in the 4 positions where the consonants originally appeared (2nd, 4th, 6th, and 8th). The number of ways to rearrange these 4 consonants in these 4 positions is:

\[
4! = 4 \times 3 \times 2 \times 1 = 24
\]

5. Total number of ways to rearrange the letters:

Since the vowels and consonants are rearranged independently, the total number of ways to rearrange the letters while keeping the relative positions of vowels and consonants the same is the product of the number of ways to rearrange the vowels and consonants:

\[
5! \times 4! = 120 \times 24 = 2880
\]

Final Answer:
The number of ways the letters of the word EDUCATION can be rearranged such that the relative position of the vowels and consonants remains the same is 2880.

Answer: Option B
Solution:
We are tasked with determining how many ways a team of 8 students can travel in two cars: one car has a seating capacity of 5 and the other has a seating capacity of 4.

Step-by-step breakdown:

1. Choose students for the first car:
The car with 5 seats will have 5 students. We need to choose 5 students from the 8 students for this car. The number of ways to choose 5 students from 8 is given by the combination formula:

\[
^8C_5 = \frac{8!}{5!(8-5)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56
\]

2. Assigning remaining students to the second car:
After selecting 5 students for the first car, the remaining 3 students will automatically go into the second car, which can seat 4 students. The number of ways to arrange the 3 remaining students in the second car (which has 4 seats, but only 3 students) is simply the number of ways to arrange 3 students, which is:

\[
3! = 3 \times 2 \times 1 = 6
\]

3. Total number of ways to assign students to the cars:
Since the cars are distinguishable (one can seat 5 and the other 4), the total number of ways to assign the students is the product of the number of ways to choose 5 students for the first car and the number of ways to assign the remaining 3 students to the second car:

\[
^8C_5 \times 3! = 56 \times 6 = 336
\]

Final Answer:
The total number of ways the 8 students can travel in the two cars is 336.

Answer: Option A
Solution:
The problem asks for the number of different possible ways to present 10 cups of tea to the expert, where 5 cups are made by adding milk first and 5 cups are made by adding tea leaves first.

This is a combinatorics problem, where we need to determine how to arrange 5 cups of one type (e.g., milk first) and 5 cups of another type (e.g., tea leaves first) among 10 cups total.

We can think of this as a problem of choosing 5 positions out of 10 to place the cups where milk was added first (or equivalently, tea leaves were added first). The remaining 5 positions will automatically be for the other type.

The number of ways to do this is given by the binomial coefficient:

\[
\binom{10}{5}
\]

This represents the number of ways to choose 5 positions out of 10.

The formula for the binomial coefficient is:

\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\]

In our case, \( n = 10 \) and \( k = 5 \), so we calculate:

\[
\binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10!}{5!5!}
\]

Now, calculate the factorials:

– \( 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5! \)
– \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)

So,

\[
\binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = \frac{30240}{120} = 252
\]

Thus, the number of different possible ways of presenting the 10 cups to the expert is \( \boxed{252} \).

Answer: Option A
Solution:
To solve this problem, we need to form a committee of 7 members with a simple majority of men and at least 1 woman from a shortlist of 9 men and 6 women. The committee must satisfy two conditions:

1. **A simple majority of men**: A simple majority in a committee of 7 members means there must be more men than women. So, we need at least 4 men (since 4 is greater than 3, which would be the number of women in that case).
2. **At least 1 woman**: This means the committee cannot consist of only men.

Step 1: Determine the possible compositions of the committee

Since the committee must have at least 4 men, the valid combinations of men and women on the committee are:

– 4 men and 3 women
– 5 men and 2 women
– 6 men and 1 woman
– 7 men and 0 women (but this case is excluded because it doesn’t satisfy the “at least 1 woman” condition).

Thus, the valid compositions are:
– 4 men, 3 women
– 5 men, 2 women
– 6 men, 1 woman

Step 2: Count the number of ways to form each type of committee

Case 1: 4 men and 3 women
– The number of ways to choose 4 men from 9 men is \( \binom{9}{4} \).
– The number of ways to choose 3 women from 6 women is \( \binom{6}{3} \).

The total number of ways to form a committee with 4 men and 3 women is:

\[
\binom{9}{4} \times \binom{6}{3} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \times \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 126 \times 20 = 2520
\]

Case 2: 5 men and 2 women
– The number of ways to choose 5 men from 9 men is \( \binom{9}{5} \).
– The number of ways to choose 2 women from 6 women is \( \binom{6}{2} \).

The total number of ways to form a committee with 5 men and 2 women is:

\[
\binom{9}{5} \times \binom{6}{2} = \frac{9 \times 8 \times 7 \times 6 \times 5}{5 \times 4 \times 3 \times 2 \times 1} \times \frac{6 \times 5}{2 \times 1} = 126 \times 15 = 1890
\]

Case 3: 6 men and 1 woman
– The number of ways to choose 6 men from 9 men is \( \binom{9}{6} \).
– The number of ways to choose 1 woman from 6 women is \( \binom{6}{1} \).

The total number of ways to form a committee with 6 men and 1 woman is:

\[
\binom{9}{6} \times \binom{6}{1} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \times 6 = 84 \times 6 = 504
\]

Step 3: Total number of ways

Now, to find the total number of ways to form the committee, we sum the results of all three cases:

\[
2520 + 1890 + 504 = 4914
\]

Thus, the total number of ways to form the committee is \( \boxed{4914} \).