First, calculate the total number of ways to select 9 people from 14 without any restrictions:

\(\binom{14}{9} = 2002\).

Next, calculate the number of ways where the two particular people (say T and M) are both included in the group. If T and M are both in the group, we need to select 7 more people from the remaining 12:

\(\binom{12}{7} = 792\).

Subtract the restricted cases from the total:

\(2002 - 792 = 1210\).

Alternatively, consider the cases where neither or only one of T or M is in the group:

• Neither T nor M in the group: \(\binom{12}{9} = 220\).
• Only T in the group: \(\binom{12}{8} = 495\).
• Only M in the group: \(\binom{12}{8} = 495\).

Summing these cases gives:

\(220 + 495 + 495 = 1210\).

(a) To form a committee of 6 people with 3 men and 3 women from 6 men and 8 women, we use combinations:

The number of ways to choose 3 men from 6 is given by \(^6C_3\).

The number of ways to choose 3 women from 8 is given by \(^8C_3\).

Thus, the total number of ways is \(^6C_3 \times ^8C_3 = 20 \times 56 = 1120\).

(b) We need to form a committee of 6 people with no more than 2 brothers. We consider the following cases:

1. 0 brothers: Choose 0 brothers from 3 and 6 others from the remaining 11 people (3 non-brothers + 8 women): \(^3C_0 \times ^{11}C_6 = 1 \times 462 = 462\).
2. 1 brother: Choose 1 brother from 3 and 5 others from the remaining 11 people: \(^3C_1 \times ^{11}C_5 = 3 \times 462 = 1386\).
3. 2 brothers: Choose 2 brothers from 3 and 4 others from the remaining 11 people: \(^3C_2 \times ^{11}C_4 = 3 \times 330 = 990\).

Adding these cases gives the total number of ways: \(462 + 1386 + 990 = 2838\).

Let T, S, and G represent train, social network, and games apps respectively. We need to select 5 apps such that there is at least one of each type.

Consider the following cases:

1. 1 train, 1 social, 3 games: The number of ways is given by:

   \(3 \times 6 \times \binom{14}{3} = 6552\)

2. 1 train, 3 social, 1 game: The number of ways is given by:

   \(3 \times \binom{6}{3} \times 14 = 840\)

3. 3 train, 1 social, 1 game: The number of ways is given by:

   \(\binom{3}{3} \times 6 \times 14 = 84\)

4. 2 train, 2 social, 1 game: The number of ways is given by:

   \(\binom{3}{2} \times \binom{6}{2} \times 14 = 630\)

5. 2 train, 1 social, 2 games: The number of ways is given by:

   \(\binom{3}{2} \times 6 \times \binom{14}{2} = 1638\)

6. 1 train, 2 social, 2 games: The number of ways is given by:

   \(3 \times \binom{6}{2} \times \binom{14}{2} = 4095\)

Summing all these possibilities gives the total number of ways:

\(6552 + 840 + 84 + 630 + 1638 + 4095 = 13839\)

To solve this problem, we need to distribute 9 different games among Wainah (W), Jingyi (J), and Hebe (H) such that each receives an odd number of games. The possible distributions of games are (1, 1, 7), (1, 7, 1), (7, 1, 1), (1, 3, 5), (1, 5, 3), (3, 1, 5), (3, 5, 1), (5, 1, 3), (5, 3, 1), and (3, 3, 3).

We calculate the number of ways for each distribution:

1. For (1, 1, 7):
   egin{align*} 9C_1 imes 8C_1 imes 1C_1 &= 72 ext{(This is multiplied by 3 for permutations of (1, 1, 7))} ext{Total} &= 72 imes 3 = 216 ext{ways} ext{(for each permutation)} ext{Total for all permutations} &= 216
2. For (1, 3, 5):
   egin{align*} 9C_1 imes 8C_3 imes 5C_1 &= 504 ext{(This is multiplied by 6 for permutations of (1, 3, 5))} ext{Total} &= 504 imes 6 = 3024 ext{ways} ext{(for each permutation)} ext{Total for all permutations} &= 3024
3. For (3, 3, 3):
   egin{align*} 9C_3 imes 6C_3 imes 3C_3 &= 1680 ext{(This is multiplied by 1 as all are equal)} ext{Total} &= 1680 ext{ways} ext{(for each permutation)} ext{Total for all permutations} &= 1680

Adding all these possibilities gives the total number of ways:

egin{align*} 216 + 3024 + 1680 &= 4920 ext{ways} ext{(total)} ext{Thus, the total number of ways is } 4920. ext{This matches the mark scheme.}

Since the club president and the club treasurer must be included, we need to choose the remaining 43 members from the other 48 members (50 total members minus the 2 already chosen).

The number of ways to choose 43 members from 48 is given by the combination formula:

\(\binom{48}{43} = \binom{48}{5}\)

Calculating \(\binom{48}{5}\):

\(\binom{48}{5} = \frac{48 \times 47 \times 46 \times 45 \times 44}{5 \times 4 \times 3 \times 2 \times 1} = 1,712,304\)

Therefore, there are 1,712,304 ways to choose the 45 members.