Since Ranuf is already in the coach, we need to choose 34 more members from the remaining 38 members (excluding Saed who is in the car).

The number of ways to choose 34 members from 38 is given by the combination formula:

\(^{38}C_{34} = \frac{38!}{34! \times (38-34)!} = \frac{38!}{34! \times 4!}\)

Calculating this gives:

\(^{38}C_{34} = \frac{38 \times 37 \times 36 \times 35}{4 \times 3 \times 2 \times 1} = 73815\)

Therefore, there are 73815 ways to choose the members who will travel in the coach.

(i) We need to choose a team of 7 people with at least 3 attackers, 2 defenders, and 1 midfielder. We consider the following cases:

• 3 attackers, 2 defenders, 2 midfielders: \(\binom{6}{3} \times \binom{5}{2} \times \binom{4}{2} = 1200\)
• 4 attackers, 2 defenders, 1 midfielder: \(\binom{6}{4} \times \binom{5}{2} \times \binom{4}{1} = 600\)
• 3 attackers, 3 defenders, 1 midfielder: \(\binom{6}{3} \times \binom{5}{3} \times \binom{4}{1} = 800\)

Summing these gives the total number of ways: \(1200 + 600 + 800 = 2600\).

(ii) To divide the team of 7 into a group of 4 and a group of 3, we choose 4 people out of 7: \(\binom{7}{4} = 35\).

(i) To satisfy the condition of having at least twice as many men as women, consider the following cases:

1. 4 men and 2 women: The number of ways is \(\binom{8}{4} \times \binom{4}{2} = 70 \times 6 = 420\).
2. 5 men and 1 woman: The number of ways is \(\binom{8}{5} \times \binom{4}{1} = 56 \times 4 = 224\).
3. 6 men and 0 women: The number of ways is \(\binom{8}{6} \times \binom{4}{0} = 28 \times 1 = 28\).

\(Total number of ways = 420 + 224 + 28 = 672.\)

(ii) First, calculate the total number of ways to choose 6 people from 12 (8 men + 4 women):

\(\binom{12}{6} = 924\).

Now, calculate the number of ways where the two particular men are together. Treat them as a single unit, so choose 4 more people from the remaining 10:

\(\binom{10}{4} = 210\).

Subtract the cases where the two men are together from the total:

\(924 - 210 = 714\).

To find the number of ways to divide 6 teenagers into groups of 3, 2, and 1, we use combinations:

First, choose 3 people for the first boat: \(\binom{6}{3} = 20\).

Next, choose 2 people for the second boat from the remaining 3: \(\binom{3}{2} = 3\).

The last person automatically goes into the third boat: \(\binom{1}{1} = 1\).

Therefore, the total number of ways is:

\(20 \times 3 \times 1 = 60\).

First, we need to choose a table for Rajid and Sue. There are 2 choices: either they sit at table X or table Y.

Once Rajid and Sue are seated at a table, Tan must sit at the other table. This leaves 5 friends to be divided into two groups of 3 and 2 to fill the remaining seats at each table.

We choose 2 additional friends to sit with Rajid and Sue. The number of ways to choose 2 friends from the remaining 5 is given by the combination:

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

Since Rajid and Sue can sit at either table, we multiply by 2:

\(10 \times 2 = 20\)

Thus, there are 20 ways to arrange the seating under the given conditions.