To solve this problem, we need to divide 8 children into two groups of 3 and one group of 2.

First, choose 3 children out of 8 for the first group: \(\binom{8}{3}\).

Next, choose 3 children out of the remaining 5 for the second group: \(\binom{5}{3}\).

The remaining 2 children automatically form the last group: \(\binom{2}{2}\).

The total number of ways is given by multiplying these combinations:

\(\binom{8}{3} \times \binom{5}{3} \times \binom{2}{2} = 56 \times 10 \times 1 = 560\).

Consider two cases: twins in the team and twins out of the team.

Case 1: Twins in the team

If the twins are in the team, we need to choose 2 boys from 6 and 0 more girls from the remaining 5 girls.

The number of ways to choose 2 boys is given by:

\(\binom{6}{2} = 15\)

Case 2: Twins out of the team

If the twins are not in the team, we need to choose 2 girls from the remaining 5 girls and 2 boys from 6.

The number of ways to choose 2 girls is:

\(\binom{5}{2} = 10\)

The number of ways to choose 2 boys is:

\(\binom{6}{2} = 15\)

Thus, the total number of ways for this case is:

\(10 \times 15 = 150\)

Total number of ways

Adding both cases together, the total number of ways is:

\(15 + 150 = 165\)

(a) To form a committee with more men than women, consider the following cases:

1. 5 men and 0 women: \(\binom{8}{5} \times \binom{7}{0} = 56\)
2. 4 men and 1 woman: \(\binom{8}{4} \times \binom{7}{1} = 490\)
3. 3 men and 2 women: \(\binom{8}{3} \times \binom{7}{2} = 1176\)

Adding these gives the total number of ways: \(56 + 490 + 1176 = 1722\).

(b) To divide the 15 members into groups of 3, 5, and 7:

First, choose 3 people for the first group: \(\binom{15}{3}\).

Then, choose 5 people for the second group from the remaining 12: \(\binom{12}{5}\).

The remaining 7 people form the last group: \(\binom{7}{7} = 1\).

The total number of ways is \(\binom{15}{3} \times \binom{12}{5} = 455 \times 792 = 360360\).

We need to consider two cases: when Tom and Henry are both in the team and when both are not in the team.

Case 1: Tom and Henry are both in the team.

If Tom and Henry are in the team, we need to choose 1 more boy from the remaining 10 boys. This can be done in \(\binom{10}{1}\) ways. We also need to choose 3 girls from 9, which can be done in \(\binom{9}{3}\) ways.

The number of ways for this case is:

\(\binom{10}{1} \times \binom{9}{3} = 10 \times 84 = 840 \text{ ways}\)

Case 2: Tom and Henry are both not in the team.

If Tom and Henry are not in the team, we need to choose 3 boys from the remaining 10 boys. This can be done in \(\binom{10}{3}\) ways. We also need to choose 3 girls from 9, which can be done in \(\binom{9}{3}\) ways.

The number of ways for this case is:

\(\binom{10}{3} \times \binom{9}{3} = 120 \times 84 = 10080 \text{ ways}\)

Total number of ways:

Adding both cases together, we get:

\(840 + 10080 = 10920 \text{ ways}\)

To solve this problem, we need to distribute 9 pies among 3 people such that each person receives an odd number of pies. The possible distributions of odd numbers that sum to 9 are (1, 1, 7), (1, 3, 5), and (3, 3, 3).

1. For the distribution (1, 1, 7):

Choose 1 pie for the first person: \(\binom{9}{1}\)

Choose 1 pie for the second person from the remaining 8 pies: \(\binom{8}{1}\)

The third person gets the remaining 7 pies: \(\binom{7}{7}\)

Since the people are distinct, multiply by the number of ways to assign these distributions: \(\binom{3}{1}\)

Total ways for (1, 1, 7) = \(9 \times 8 \times 1 \times 3 = 216\)

2. For the distribution (1, 3, 5):

Choose 1 pie for the first person: \(\binom{9}{1}\)

Choose 3 pies for the second person from the remaining 8 pies: \(\binom{8}{3}\)

The third person gets the remaining 5 pies: \(\binom{5}{5}\)

Since the people are distinct, multiply by the number of ways to assign these distributions: \(3!\)

Total ways for (1, 3, 5) = \(9 \times 56 \times 1 \times 6 = 3024\)

3. For the distribution (3, 3, 3):

Choose 3 pies for the first person: \(\binom{9}{3}\)

Choose 3 pies for the second person from the remaining 6 pies: \(\binom{6}{3}\)

The third person gets the remaining 3 pies: \(\binom{3}{3}\)

Total ways for (3, 3, 3) = \(84 \times 20 \times 1 = 1680\)

Adding all the possibilities: \(216 + 3024 + 1680 = 4920\)