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

We need to calculate \( m \) and \( n \). Step 1: Calculate the number of ways to form the committee with at least 6 males. The possible cases are: - 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: \[ \binom{8}{8} \times \binom{5}{3} = 1 \times 10 = 10 \] Thus, the total number of ways to form the committee with at least 6 males is: \[ m = 28 + 40 + 10 = 78. \] Step 2: Calculate the number of ways to form the committee with at least 3 females. The possible cases 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: \[ \binom{8}{6} \times \binom{5}{5} = 28 \times 1 = 28 \] Thus, the total number of ways to form the committee with at least 3 females is: \[ n = 10 + 40 + 28 = 78. \] Step 3: Conclusion. Since both \( m \) and \( n \) are 78, the correct answer is \( m = n = 78 \).