1. **Problem statement:** We need to find the number of possible committees of 5 people chosen from 6 men and 4 women under three different conditions. 2. **(a) Committee with 3 men and 2 women:** - Number of ways to choose 3 men from 6: $\binom{6}{3}$ - Number of ways to choose 2 women from 4: $\binom{4}{2}$ - Total number of committees: $\binom{6}{3} \times \binom{4}{2}$ - Calculate: $$\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$$ $$\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6$$ - So, total committees = $20 \times 6 = 120$ 3. **(b) Committee composed of men only:** - Choose 5 men from 6 men: $\binom{6}{5}$ - Calculate: $$\binom{6}{5} = \frac{6!}{5!(6-5)!} = 6$$ - So, total committees = 6 4. **(c) Committee with a majority of women:** - Majority means more women than men, so more than 2.5 women, i.e., at least 3 women. - Possible women counts: 3 women + 2 men, 4 women + 1 man, 5 women + 0 men - Number of ways: - 3 women and 2 men: $\binom{4}{3} \times \binom{6}{2}$ - 4 women and 1 man: $\binom{4}{4} \times \binom{6}{1}$ - 5 women and 0 men: $\binom{4}{5} = 0$ (not possible as we only have 4 women) - Calculate: $$\binom{4}{3} = 4, \quad \binom{6}{2} = \frac{6 \times 5}{2} = 15$$ $$\binom{4}{4} = 1, \quad \binom{6}{1} = 6$$ - Total committees = $4 \times 15 + 1 \times 6 + 0 = 60 + 6 = 66$ **Final answers:** - (a) 120 - (b) 6 - (c) 66