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

Concept: The number of ways to choose r items from n items is nCr = n! / (r!(n−r)!). When a selection must satisfy a case-based constraint (here, the number of men in the team), split the selection into the mutually exclusive cases the constraint allows, compute each case as (ways to choose the men) × (ways to choose the women), and add the case totals — this is the addition principle applied over a partition of the sample space.

Application: The team has 9 members drawn from 5 men and 10 women, with at most 3 men allowed. So the number of men can be 0, 1, 2, or 3, with the remaining seats filled by 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

Adding the four case totals: 10 + 225 + 1200 + 2100 = 3535.

Cross-check: Using the complement — the total unrestricted ways to pick a team of 9 from all 15 people is 15C9 = 5005. Teams that violate the constraint have 4 or 5 men: (5C4 × 10C5) + (5C5 × 10C4) = (5 × 252) + (1 × 210) = 1260 + 210 = 1470. Subtracting: 5005 − 1470 = 3535, which matches the case-by-case total.

So 3535 is the number of valid ways to form the team.