For a social work, 7 men and 6 women gave their nominations. The committee is formed to select 5 people from the nominated persons in such a way that atleast 3 men are there in the final team. Find the number of ways in which the people can be selected.

Given:

Total number of men = 7

Total number of women = 6

The total number of people need to select = 5

The total number of men will be in the final team = 3

Formula used:

The number of ways of selecting r things out of n = ⁿC_r

Also, nCr = \(\frac{n!}{r!×(n-r)!}\)

Calculation:

According to the question, we may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only)

The required number of ways to select 3 men and 2 women = ⁷C₃ × ⁶C₂

⇒ \(\frac{7!}{3!×(7-3)!} × \frac{6!}{2!×(6-2)!}\)

⇒ \(\frac{(7\times 6\times 5)}{(3\times2\times1)}\)×\(\frac{(6\times 5)}{(2\times1)}\)

⇒ 35 × 15

⇒ 525

The required number of ways to select 4 men and 1 woman = 7C4 × 6C1

⇒ \(\frac{7!}{4!×(7-4)!} × \frac{6!}{1!×(6-1)!}\)

⇒ \(\frac{(7\times 6\times 5)}{(3\times2\times1)}\)× 6

⇒ 35 × 6

⇒ 210

The required number of ways to select 5 men only = 7C5

⇒ \(\frac{7!}{5! × (7-5)!}\)

⇒ (7 × 6)/(2 × 1) = 21

∴ Total number of ways = 525 + 210 + 21 = 756

∴​ The people can be selected in a total of 756 ways.