Understanding the Social Work Committee Selection Problem

The problem asks us to find the number of ways to select a committee of 5 people from a group of 7 men and 6 women. The key condition is that the committee must contain at least 3 men.

The total number of people nominated is $7 (\text{men}) + 6 (\text{women}) = 13$. We need to select a committee of 5.

The condition "at least 3 men" means the committee can have:

• Exactly 3 men
• Exactly 4 men
• Exactly 5 men

We need to calculate the number of ways for each case and sum them up to get the total number of ways to form the committee according to the given condition. We will use combinations $\binom{n}{k}$ to solve this, where $\binom{n}{k}$ represents the number of ways to choose $k$ items from a set of $n$ items without regard to the order.

The formula for combinations is:

\(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)

Case 1: Exactly 3 Men in the Committee

If the committee has exactly 3 men, then the remaining $5 - 3 = 2$ people must be women.

• Number of ways to select 3 men from 7 men: \(\binom{7}{3}\)
• Number of ways to select 2 women from 6 women: \(\binom{6}{2}\)

Total ways for Case 1 = \(\binom{7}{3} \times \binom{6}{2}\)

Calculating the combinations:

• \(\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35\)
• \(\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} = \frac{6 \times 5}{2 \times 1} = 15\)

Ways for Case 1 = \(35 \times 15 = 525\)

Case 2: Exactly 4 Men in the Committee

If the committee has exactly 4 men, then the remaining $5 - 4 = 1$ person must be a woman.

• Number of ways to select 4 men from 7 men: \(\binom{7}{4}\)
• Number of ways to select 1 woman from 6 women: \(\binom{6}{1}\)

Total ways for Case 2 = \(\binom{7}{4} \times \binom{6}{1}\)

Calculating the combinations:

• \(\binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35\)
• \(\binom{6}{1} = \frac{6!}{1!(6-1)!} = \frac{6!}{1!5!} = 6\)

Ways for Case 2 = \(35 \times 6 = 210\)

Case 3: Exactly 5 Men in the Committee

If the committee has exactly 5 men, then the remaining $5 - 5 = 0$ people must be women.

• Number of ways to select 5 men from 7 men: \(\binom{7}{5}\)
• Number of ways to select 0 women from 6 women: \(\binom{6}{0}\)

Total ways for Case 3 = \(\binom{7}{5} \times \binom{6}{0}\)

Calculating the combinations:

• \(\binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7!}{5!2!} = \frac{7 \times 6}{2 \times 1} = 21\)
• \(\binom{6}{0} = \frac{6!}{0!(6-0)!} = \frac{6!}{0!6!} = 1\) (By definition, $\binom{n}{0}=1$)

Ways for Case 3 = \(21 \times 1 = 21\)

Total Number of Ways to Select the Committee

To find the total number of ways to select the committee with at least 3 men, we add the number of ways from each case:

Total ways = (Ways for Case 1) + (Ways for Case 2) + (Ways for Case 3)

Total ways = \(525 + 210 + 21\)

Total ways = \(756\)

The number of ways to select the people for the committee with the condition of having at least 3 men is 756.