Questions

To solve this problem, we need to calculate the number of ways to select a committee of 3 people such that at least 2 of them are men.

Let's break it down into cases:

Case 1: 2 men and 1 woman

1. Selecting 2 men out of 5:

   $$\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$$

2. Selecting 1 woman out of 4:

   $$\binom{4}{1} = \frac{4!}{1!(4-1)!} = \frac{4}{1} = 4$$

• Total ways to form a committee with 2 men and 1 woman: $$10 \times 4 = 40$$

Case 2: 3 men

1. Selecting 3 men out of 5: $$\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$$

• Total ways to form a committee with 3 men: $$10$$

Total number of ways:

• Adding the ways from both cases: $$40 + 10 = 50$$

Conclusion:

The number of ways to form a committee such that at least 2 men are on the committee is $\mathbf{50}$.

Thus, the correct answer is: a) 50