(a) (i) The number of ways to choose 6 books from 18 is given by the combination formula \(\binom{18}{6}\). Calculating this gives:

\(\binom{18}{6} = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 18564\)

(ii) To include the Harry Potter book, choose 5 more books from the remaining 17. This is \(\binom{17}{5}\):

\(\binom{17}{5} = \frac{17 \times 16 \times 15 \times 14 \times 13}{5 \times 4 \times 3 \times 2 \times 1} = 6188\)

(b) (i) The total number of ways to arrange 8 people (5 boys and 3 girls) is \(8!\):

\(8! = 40320\)

(ii) Treat the 5 boys as a single unit. This gives 4 units to arrange (3 girls + 1 unit of boys). The number of ways to arrange these 4 units is \(4!\). Within the boys' unit, the 5 boys can be arranged in \(5!\) ways. Therefore, the total number of arrangements is:

\(5! \times 4! = 120 \times 24 = 2880\)

(i) To choose 3 men from 6 and 2 women from 4, we use combinations:

\(\binom{6}{3} \times \binom{4}{2} = 20 \times 6 = 120\).

(ii) More men than women means either 3 men and 2 women, 4 men and 1 woman, or 5 men and 0 women. Calculate each case:

• 3 men and 2 women: \(\binom{6}{3} \times \binom{4}{2} = 120\)
• 4 men and 1 woman: \(\binom{6}{4} \times \binom{4}{1} = 15 \times 4 = 60\)
• 5 men and 0 women: \(\binom{6}{5} \times \binom{4}{0} = 6 \times 1 = 6\)

Total: \(120 + 60 + 6 = 186\).

(iii) If one particular woman and one particular man cannot be together, calculate the total ways without restriction and subtract the restricted cases:

Total ways for 3 men and 2 women: \(120\).

Restricted case (both are in the committee): Choose 2 more men from the remaining 5 and 1 more woman from the remaining 3: \(\binom{5}{2} \times \binom{3}{1} = 10 \times 3 = 30\).

Subtract restricted from total: \(120 - 30 = 90\).

(i) To find the number of ways to punch exactly 2 holes in a column with 8 positions, we use the combination formula: \(\binom{8}{2} = 28\).

(ii) The number of different patterns of holes in a column is the sum of combinations for 1, 2, 3, and 4 holes: \(\binom{8}{1} + \binom{8}{2} + \binom{8}{3} + \binom{8}{4} = 8 + 28 + 56 + 70 = 162\).

(iii) Since each column can have 162 different patterns, and there are 4 columns, the total number of different possible key cards is \(162^4 = 688,747,536\).

(a) To choose a team of 5 people with exactly one adult, select 1 adult from 5 and 4 others from the remaining 7 (3 boys and 4 girls):

\(\binom{5}{1} \times \binom{7}{4} = 5 \times 35 = 175\)

(b) Consider the cases where the team includes at least 2 boys and at least 1 girl:

1. 2 boys, 1 girl, 2 adults: \(\binom{3}{2} \times \binom{4}{1} \times \binom{5}{2} = 3 \times 4 \times 10 = 120\)
2. 2 boys, 2 girls, 1 adult: \(\binom{3}{2} \times \binom{4}{2} \times \binom{5}{1} = 3 \times 6 \times 5 = 90\)
3. 2 boys, 3 girls: \(\binom{3}{2} \times \binom{4}{3} = 3 \times 4 = 12\)
4. 3 boys, 1 girl, 1 adult: \(\binom{3}{3} \times \binom{4}{1} \times \binom{5}{1} = 1 \times 4 \times 5 = 20\)
5. 3 boys, 2 girls: \(\binom{3}{3} \times \binom{4}{2} = 1 \times 6 = 6\)

Summing these cases gives the total number of ways:

\(120 + 90 + 12 + 20 + 6 = 248\)

Since Mr and Mrs Khan and Abad must be included, we have already chosen 3 people. Therefore, we need to choose 17 more people from the remaining 23 members of the club.

The number of ways to choose 17 people from 23 is given by the combination formula:

\(\binom{23}{17} = \frac{23!}{17!(23-17)!} = \frac{23!}{17!6!}\)

Calculating this gives:

\(\binom{23}{17} = 100947\)

Thus, there are 100947 ways to choose the 20 people.