Question

From a group of 10 men and 8 women, a committee of 5 is to be formed such that it contains at least 3 women. How many such committees are possible?

• A. 3276
• B. 3500
• C. 3657
• D. 4012

Hint:

To find the number of committees with at least \(k\) women, sum over cases from \(k\) to the committee size, calculating combinations accordingly.

Answer 1

The Correct Option is A

To solve the problem of forming a committee of 5 members from a group of 10 men and 8 women, with the condition that the committee contains at least 3 women, we consider the following scenarios:

1. 3 Women and 2 Men:

   The number of ways to choose 3 women from 8 is given by the combination formula \( C(n, r) = \frac{n!}{r!(n-r)!} \).

   \( C(8, 3) = \frac{8!}{3!5!} = 56 \)

   The number of ways to choose 2 men from 10 is:

   \( C(10, 2) = \frac{10!}{2!8!} = 45 \)

   Total for this scenario: \( 56 \times 45 = 2520 \)

2. 4 Women and 1 Man:

   The number of ways to choose 4 women from 8:

   \( C(8, 4) = \frac{8!}{4!4!} = 70 \)

   The number of ways to choose 1 man from 10:

   \( C(10, 1) = 10 \)

   Total for this scenario: \( 70 \times 10 = 700 \)

3. 5 Women:

   The number of ways to choose 5 women from 8:

   \( C(8, 5) = \frac{8!}{5!3!} = 56 \)

   Total for this scenario: 56

Adding all these possibilities gives:

• Total committees with at least 3 women = 2520 + 700 + 56 = 3276

Therefore, the number of such committees possible is 3276.

Answer 2

Total men = 10, total women = 8, committee size = 5, with at least 3 women. Possible cases: - 3 women and 2 men - 4 women and 1 man - 5 women and 0 men Number of committees with 3 women and 2 men: \[ \binom{8}{3} \times \binom{10}{2} = 56 \times 45 = 2520 \] Number of committees with 4 women and 1 man: \[ \binom{8}{4} \times \binom{10}{1} = 70 \times 10 = 700 \] Number of committees with 5 women and 0 men: \[ \binom{8}{5} \times \binom{10}{0} = 56 \times 1 = 56 \] Total number of committees: \[ 2520 + 700 + 56 = 3276 \]