In how many ways can a team of 11 be selected from 5 men and 11 women such…

Concept

When a selection must satisfy an “at most” constraint on one group, split the counting into disjoint cases by the exact number taken from that group, and count each case with the combination rule nCr = n! / (r!(n−r)!). Since the cases cannot overlap, the total number of ways is the sum of the individual cases (rule of sum).

Application

The team of 11 must contain at most 3 men out of the 5 available, so the remaining seats are filled by women (11 available). The valid, mutually exclusive cases are 0, 1, 2, or 3 men, each paired with enough women to make the team size 11:

1. 0 men + 11 women: 5C0 × 11C11 = 1 × 1 = 1

2. 1 man + 10 women: 5C1 × 11C10 = 5 × 11 = 55

3. 2 men + 9 women: 5C2 × 11C9 = 10 × 55 = 550

4. 3 men + 8 women: 5C3 × 11C8 = 10 × 165 = 1650

Adding the four disjoint cases (rule of sum): 1 + 55 + 550 + 1650 = 2256.

Cross-check

Verify via the complement. The total ways to pick any team of 11 from all 16 people is 16C11 = 4368. The excluded cases are those with more than 3 men — 4 men (5C4 × 11C7 = 5 × 330 = 1650) or 5 men (5C5 × 11C6 = 1 × 462 = 462), totalling 2112. So 4368 − 2112 = 2256, confirming the case-wise sum.

So the number of ways is 2256.