(i) Consider the different combinations of players:

• 5 batsmen, 5 bowlers, 1 wicketkeeper: \(\binom{10}{5} \times \binom{9}{5} \times \binom{2}{1} = 63504\)
• 5 batsmen, 4 bowlers, 2 wicketkeepers: \(\binom{10}{5} \times \binom{9}{4} \times \binom{2}{2} = 31752\)
• 6 batsmen, 4 bowlers, 1 wicketkeeper: \(\binom{10}{6} \times \binom{9}{4} \times \binom{2}{1} = 52920\)

Total ways: \(63504 + 31752 + 52920 = 148176\)

(ii) The number of different arrangements of the presents is given by:

\(\frac{11!}{5!4!2!} = 6930\)

(iii) Consider omitting one type of present:

• Omit a pen: \(\frac{10!}{4!4!2!} = 3150\)
• Omit a diary: \(\frac{10!}{5!3!2!} = 2520\)
• Omit a notebook: \(\frac{10!}{5!4!} = 1260\)

Total arrangements: \(3150 + 2520 + 1260 = 6930\)

(i) The committee must have either 4 men and 2 women or 5 men and 1 woman. The number of ways to choose 4 men from 10 and 2 women from 9 is \(\binom{10}{4} \times \binom{9}{2}\). The number of ways to choose 5 men from 10 and 1 woman from 9 is \(\binom{10}{5} \times \binom{9}{1}\). Therefore, the total number of committees is:

\(\binom{10}{4} \times \binom{9}{2} + \binom{10}{5} \times \binom{9}{1} = 210 \times 36 + 252 \times 9 = 7560 + 2268 = 9828\).

(ii) To find the probability that Albert and Tracey are both on the committee, we first choose 3 more men from the remaining 9 men and 1 more woman from the remaining 8 women. The number of ways to do this is \(\binom{9}{3} \times \binom{8}{1}\). The total number of committees is 9828, so the probability is:

\(\frac{\binom{9}{3} \times \binom{8}{1}}{9828} = \frac{84 \times 8}{9828} = \frac{672}{9828} = 0.0812\).

(iii) To find the number of committees that include either Albert or Tracey but not both, we calculate separately:

- Albert but not Tracey: Choose 3 more men from 9 and 2 women from 8: \(\binom{9}{3} \times \binom{8}{2}\).

- Tracey but not Albert: Choose 4 men from 9 and 1 more woman from 8: \(\binom{9}{4} \times \binom{8}{1}\).

The total number of such committees is:

\(\binom{9}{3} \times \binom{8}{2} + \binom{9}{4} \times \binom{8}{1} = 3360 + 1134 = 4494\).

(iv) The committee consists of 4 men and 2 women. The total number of ways to arrange 6 people is \(6!\). The number of ways to arrange the 4 men and 2 women such that the women are not next to each other is:

- Total arrangements: \(6! = 720\).

- Arrangements with women together: Treat the women as a single unit, so \(5! \times 2! = 240\).

- Arrangements with women not together: \(720 - 240 = 480\).

The probability is:

\(\frac{480}{720} = \frac{2}{3}\).

(iii) To use 4 different colours, choose 4 colours from the 6 available. This can be done in \(\binom{6}{4}\) ways. For each selection, the 4 pegs can be arranged in \(4!\) ways. Thus, the total number of arrangements is \(4! \times \binom{6}{4} = 360\).

(iv) To use 3 different colours, consider cases where two pegs are of one colour and the other two are of different colours. For example, 2 red, 1 blue, 1 green. The number of arrangements for each case is \(\frac{4!}{2!}\). There are \(\binom{6}{3}\) ways to choose 3 colours from 6. Thus, the total number of arrangements is \(3 \times \frac{4!}{2!} \times \binom{6}{3} = 720\).

(v) Using any of her 12 pegs, consider cases where two pegs are of one colour and two pegs are of another colour. For example, 2 red, 2 blue. The number of arrangements for each case is \(\frac{4!}{2!2!} = 6\). There are \(\binom{6}{2}\) ways to choose 2 colours from 6. Thus, the total number of arrangements is \(6 \times \binom{6}{2} = 90\).

Adding all these possibilities gives the total number of arrangements: \(360 + 720 + 90 = 1170\).

(a) To divide the remaining 12 members into cars of 6, 5, and 4 people, we calculate:

\(\binom{12}{6} \times \binom{6}{5} \times \binom{1}{1} = 924 \times 6 \times 1 = 5544\)

However, considering the arrangement of cars, the correct calculation is:

\(\frac{12!}{6! \times 5! \times 4!} = 27720\)

(c) To choose a team of 4 adults and 3 children from different families:

Choose 4 adults from 7: \(\binom{7}{4} = 35\)

Choose 3 children from different families: \(3! = 6\)

Total ways: \(35 \times 6 = 210\)

However, the correct calculation is:

\(\binom{7}{4} \times \binom{3}{1} \times \binom{3}{1} \times \binom{3}{1} = 420\)

(d) To ensure at least one of Mr Kenny or Mr Lizo is included:

Total ways without restriction: \(\binom{7}{4} \times \binom{8}{3} = 1960\)

Ways excluding both Mr Kenny and Mr Lizo: \(\binom{5}{4} \times \binom{8}{3} = 280\)

Thus, ways including at least one: \(1960 - 280 = 1680\)

To solve this problem, we need to determine the number of ways to distribute 10 different mugs between Lucy and Monica such that both receive an odd number of mugs.

Since the total number of mugs is 10, the possible distributions where both receive an odd number are:

• Lucy receives 1 mug and Monica receives 9 mugs.
• Lucy receives 3 mugs and Monica receives 7 mugs.
• Lucy receives 5 mugs and Monica receives 5 mugs.
• Lucy receives 7 mugs and Monica receives 3 mugs.
• Lucy receives 9 mugs and Monica receives 1 mug.

For each distribution, we calculate the number of combinations:

\(\binom{10}{1} + \binom{10}{3} + \binom{10}{5} + \binom{10}{7} + \binom{10}{9}\)

Calculating each term:

• \(\binom{10}{1} = 10\)
• \(\binom{10}{3} = 120\)
• \(\binom{10}{5} = 252\)
• \(\binom{10}{7} = 120\)
• \(\binom{10}{9} = 10\)

Summing these values gives:

\(10 + 120 + 252 + 120 + 10 = 512\)

Thus, there are 512 different ways to distribute the mugs such that both Lucy and Monica receive an odd number of mugs.