The number of committees of size 8 that can be formed from 10 men and 10 women…

Concept: When a committee has an 'at least' constraint on one subgroup, split the selection into mutually exclusive cases by the exact count from that subgroup — here women counts of 5, 6, 7, and 8, with the remaining seats filled by men. Each case is counted using the combination formula C(n, r) = n! / (r! · (n − r)!) for the men and women choices, multiplied together (multiplication principle) since the two choices are independent within a case. Because the cases do not overlap, their counts are added (sum rule) to get the total.

Application: List every valid split of the committee by women count, compute each case, then add them.

1. Valid women counts for 'at least 5 out of 8' are w = 5, 6, 7, 8; the men count in each case is 8 − w.

2. w = 5 (5 women, 3 men): C(10,5) × C(10,3) = 252 × 120 = 30240.

3. w = 6 (6 women, 2 men): C(10,6) × C(10,2) = 210 × 45 = 9450.

4. w = 7 (7 women, 1 man): C(10,7) × C(10,1) = 120 × 10 = 1200.

5. w = 8 (8 women, 0 men): C(10,8) × C(10,0) = 45 × 1 = 45.

6. Sum the four cases: 30240 + 9450 + 1200 + 45 = 40935.

Cross-check: Using complementary counting, the unrestricted count of size-8 committees from 20 people is C(20,8) = 125970. The excluded cases (fewer than 5 women, i.e. w = 0,1,2,3,4) total 45 + 1200 + 9450 + 30240 + 44100 = 85035. Subtracting, 125970 − 85035 = 40935, which matches the case-sum above.

Answer: 40935