In how many ways can a team of 9 be selected from 5 men and 10 women such that…

Concept: When a selection is capped by the count taken from one sub-group (here, at most 3 of the 5 men), split the count into mutually exclusive cases by the exact number chosen from that sub-group. In each case, choosing the men and choosing the women are independent decisions, so the case total is the product of the two combinations; by the sum rule, the overall total is the sum of the case totals across every allowed case.

Applying to this problem: The team has 9 members with at most 3 men, so the number of men can be 0, 1, 2, or 3 (with the rest of the team made up of women):

1. 0 men and 9 women: 5C0 × 10C9 = 1 × 10 = 10

2. 1 man and 8 women: 5C1 × 10C8 = 5 × 45 = 225

3. 2 men and 7 women: 5C2 × 10C7 = 10 × 120 = 1200

4. 3 men and 6 women: 5C3 × 10C6 = 10 × 210 = 2100

Total = 10 + 225 + 1200 + 2100 = 3535.

Cross-check: Count the complement instead. The unrestricted ways to pick any 9 of the 15 people is 15C9 = 5005. Subtract the excluded cases where more than 3 men are chosen (4 or 5 men): 5C4 × 10C5 = 5 × 252 = 1260, and 5C5 × 10C4 = 1 × 210 = 210, together 1470. 5005 − 1470 = 3535, confirming the case-sum result above.