Question

A committee of 11 members is to be formed out of 8 males and 5 females. If \( m \) is the number of ways the committee is formed with at least 6 males and \( n \) is the number of ways with at least 3 females, then:

• A. \( n = m - 8 \)
• B. \( m = n = 78 \)
• C. \( m = n = 68 \)
• D. \( m + n = 68 \)

Hint:

For combinatorics problems involving committees, break down the problem by considering all possible cases and apply the binomial coefficient to calculate the number of ways.

Answer 1

The Correct Option is B

The objective is to determine the values of \( m \) and \( n \).

Step 1: Calculate \( m \), the number of ways to form a committee with at least 6 males. This involves summing the possibilities for the following compositions: - 6 males and 5 females: \[ \binom{8}{6} \times \binom{5}{5} = 28 \times 1 = 28 \] - 7 males and 4 females: \[ \binom{8}{7} \times \binom{5}{4} = 8 \times 5 = 40 \] - 8 males and 3 females: \[ \binom{8}{8} \times \binom{5}{3} = 1 \times 10 = 10 \] Therefore, \( m = 28 + 40 + 10 = 78 \). 

Step 2: Calculate \( n \), the number of ways to form a committee with at least 3 females. The possible committee compositions are: - 8 males and 3 females: \[ \binom{8}{8} \times \binom{5}{3} = 1 \times 10 = 10 \] - 7 males and 4 females: \[ \binom{8}{7} \times \binom{5}{4} = 8 \times 5 = 40 \] - 6 males and 5 females: \[ \binom{8}{6} \times \binom{5}{5} = 28 \times 1 = 28 \] Consequently, \( n = 10 + 40 + 28 = 78 \). 

Step 3: Conclusion. Since \( m = 78 \) and \( n = 78 \), the result is \( m = n = 78 \).