Question

The number of ways a committee of 8 members can be formed from a group of 10 men and 8 women such that the committee contains at most 5 men and at least 5 women is:

• A. \( 8061 \)
• B. \( 8612 \)
• C. \( 6082 \)
• D. \( 8271 \)

Hint:

When forming committees with constraints, consider each valid case separately and sum the individual possibilities. Use combinations (\( ^nC_r \)) for selection problems.

Answer 1

The Correct Option is A

Step 1: Understanding the Selection Constraints
We need to form a committee of 8 members where:
- At most 5 men are selected.
- At least 5 women are selected.
This means the possible distributions of men (M) and women (W) are: \[ (5M,3W), \quad (4M,4W), \quad (3M,5W), \quad (2M,6W), \quad (1M,7W), \quad (0M,8W) \] Step 2: Compute Combinations for Each Case Using the combination formula: \[ \text{Ways to select } r \text{ elements from } n \text{ elements:} \quad ^nC_r = \frac{n!}{r!(n-r)!} \] For each case: 1. Case (5M, 3W) \[ ^{10}C_5 \times ^8C_3 = \frac{10!}{5!(10-5)!} \times \frac{8!}{3!(8-3)!} = 252 \times 56 = 14112 \] 2. Case (4M, 4W) \[ ^{10}C_4 \times ^8C_4 = 210 \times 70 = 14700 \] 3. Case (3M, 5W) \[ ^{10}C_3 \times ^8C_5 = 120 \times 56 = 6720 \] 4. Case (2M, 6W) \[ ^{10}C_2 \times ^8C_6 = 45 \times 28 = 1260 \] 5. Case (1M, 7W) \[ ^{10}C_1 \times ^8C_7 = 10 \times 8 = 80 \] 6. Case (0M, 8W) \[ ^{10}C_0 \times ^8C_8 = 1 \times 1 = 1 \] Step 3: Compute Total Valid Committees \[ 6720 + 1260 + 80 + 1 = 8061 \] Thus, the correct answer is 8061.