In how many ways a team of 9 must be selected from 6 men and 10 women , such…

Concept: When a selection problem splits into mutually exclusive cases, the total count is the SUM of the ways to complete each case (the addition principle). Each case is counted using the combination formula C(n, r) = n! / (r!(n − r)!) — the number of ways to choose r items from n distinct items when order does not matter.

Application: The team has 9 members drawn from 6 men and 10 women, with at most 3 men allowed. So the number of men selected can only be 0, 1, 2, or 3 — each choice fixes how many women fill the remaining seats.

1. 0 men, 9 women: C(6,0) × C(10,9) = 1 × 10 = 10

2. 1 man, 8 women: C(6,1) × C(10,8) = 6 × 45 = 270

3. 2 men, 7 women: C(6,2) × C(10,7) = 15 × 120 = 1800

4. 3 men, 6 women: C(6,3) × C(10,6) = 20 × 210 = 4200

5. Add the four mutually exclusive cases: 10 + 270 + 1800 + 4200 = 6280

Cross-check: by the symmetry identity C(n, r) = C(n, n − r), C(10,9) = C(10,1) = 10 and C(10,8) = C(10,2) = 45, which match the values used above — confirming each case count independently.

So the total number of ways to select the team is 6280.