To solve this problem, we need to consider the different cases where at least one woman is included in the selection of three people.

1. Case 1: 1 woman and 2 men are chosen.

   The number of ways to choose 1 woman from 3 is given by the combination \(\binom{3}{1} = 3\).

   The number of ways to choose 2 men from 6 is given by the combination \(\binom{6}{2} = 15\).

   Thus, the total number of ways for this case is \(3 \times 15 = 45\).

2. Case 2: 2 women and 1 man are chosen.

   The number of ways to choose 2 women from 3 is given by the combination \(\binom{3}{2} = 3\).

   The number of ways to choose 1 man from 6 is given by the combination \(\binom{6}{1} = 6\).

   Thus, the total number of ways for this case is \(3 \times 6 = 18\).

3. Case 3: 3 women are chosen.

   The number of ways to choose 3 women from 3 is given by the combination \(\binom{3}{3} = 1\).

Adding all these cases together, the total number of ways to choose at least one woman is \(45 + 18 + 1 = 64\).

Alternatively, calculate the total number of ways to choose any 3 people from 9 (6 men + 3 women) without restriction: \(\binom{9}{3} = 84\).

Subtract the number of ways to choose only men (no women): \(\binom{6}{3} = 20\).

Thus, the number of ways to choose at least one woman is \(84 - 20 = 64\).

To solve this problem, we need to select 2 married couples and 1 additional person from the remaining passengers.

First, choose 2 married couples from the 3 available couples. The number of ways to do this is given by the combination formula:

\(\binom{3}{2} = 3\)

After choosing 2 married couples, there are 7 passengers left (5 unmarried and 2 from the remaining couple). We need to choose 1 additional person from these 7 passengers:

\(\binom{7}{1} = 7\)

Therefore, the total number of ways to choose the passengers is:

\(3 \times 7 = 21\)

(i) Since the captain, a midfield player, must be included, we need to choose 1 player from the 3 defence players and 1 player from the 5 forward players. The number of ways to do this is given by:

\(\binom{3}{1} \times \binom{5}{1} = 3 \times 5 = 15\)

(ii) If exactly one forward player must be included, we choose 1 player from the 5 forward players. Then, we choose 2 more players from the remaining 6 players (3 defence and 3 midfield). The number of ways to do this is given by:

\(\binom{5}{1} \times \binom{6}{2} = 5 \times 15 = 75\)

(a) (i) To find the number of different three-course meals, multiply the number of choices for each course. There are 3 choices for the starter, 5 choices for the main course, and 3 choices for the dessert. Thus, the total number of three-course meals is:

\(3 \times 5 \times 3 = 45\)

(a) (ii) If customers choose only two of the three courses, consider the combinations:

• Starter and Main: \(3 \times 5 = 15\)
• Starter and Dessert: \(3 \times 3 = 9\)
• Main and Dessert: \(5 \times 3 = 15\)

Summing these gives:

\(15 + 9 + 15 = 39\)

(b) To divide 14 people into groups of 5, 5, and 4, use combinations:

\(\binom{14}{5} \times \binom{9}{5} \times \binom{4}{4} = 252252\)

(a) To find the number of ways to select the committee with exactly one of Mr Lan or Mrs Lan, we consider two cases:

1. Mr Lan is on the committee, but Mrs Lan is not. We choose 4 more members from the remaining 12 people (6 men and 6 women): \\(\binom{12}{4} = 495 \\\)
2. Mrs Lan is on the committee, but Mr Lan is not. We choose 4 more members from the remaining 12 people (6 men and 6 women): \\(\binom{12}{4} = 495 \\\)

Thus, the total number of ways is \\(495 + 495 = 990 \\\).

(b) If Mrs Lan must be on the committee and there must be more women than men, we consider the following scenarios:

1. 2 women and 2 men in addition to Mrs Lan: \\(\binom{7}{2} \times \binom{6}{2} = 315 \\\)
2. 3 women and 1 man in addition to Mrs Lan: \\(\binom{7}{3} \times \binom{6}{1} = 210 \\\)
3. 4 women in addition to Mrs Lan: \\(\binom{7}{4} = 35 \\\)

Adding these gives the total number of ways: \\(315 + 210 + 35 = 560 \\\).