Problem Analysis

We are asked to find the number of ways to form a 5-person committee from a group of 7 men and 6 women, with the condition that there must be more men than women in the committee.

1. Identifying Possible Committee Structures

For a committee of 5, let 'M' be the number of men and 'W' be the number of women. The condition is $M > W$ and $M + W = 5$. This gives us three possible cases:

• Case 1: 3 Men and 2 Women (3 > 2)
• Case 2: 4 Men and 1 Woman (4 > 1)
• Case 3: 5 Men and 0 Women (5 > 0)

The total number of ways is the sum of the ways each of these cases can be formed.

2. Calculating Ways for Each Case

We will use the combination formula $C(n, k) = \frac{n!}{k!(n-k)!}$ since the order in which members are chosen does not matter.

Case 1: 3 Men from 7, and 2 Women from 6

Number of ways = $C(7, 3) \times C(6, 2)$

$C(7, 3) = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$

$C(6, 2) = \frac{6!}{2!4!} = \frac{6 \times 5}{2 \times 1} = 15$

Total for Case 1 = $35 \times 15 = 525$

Case 2: 4 Men from 7, and 1 Woman from 6

Number of ways = $C(7, 4) \times C(6, 1)$

$C(7, 4) = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$

$C(6, 1) = \frac{6!}{1!5!} = 6$

Total for Case 2 = $35 \times 6 = 210$

Case 3: 5 Men from 7, and 0 Women from 6

Number of ways = $C(7, 5) \times C(6, 0)$

$C(7, 5) = \frac{7!}{5!2!} = \frac{7 \times 6}{2 \times 1} = 21$

$C(6, 0) = 1$

Total for Case 3 = $21 \times 1 = 21$

3. Summing the Cases

The total number of ways to form the committee is the sum of the ways for each valid case:

Total Ways = 525 + 210 + 21 = 756