(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: \(120 - 30 = 90\).