In how many ways a team of 8 must be selected from 9 boys and 7 girls such…

Concept: When a selection is capped by an “at most k” restriction on one group, split the count into mutually exclusive cases by exactly how many members of that group are chosen. Each case's count is found independently by multiplying the ways to choose from each group, and because a team can fall into only one such case, the case counts are added (never multiplied) to get the total.

Applying this here: A team of 8 is formed from 9 boys and 7 girls with the girls capped at 3, so the number of girls chosen can only be 0, 1, 2 or 3, with the rest of the team filled by boys. Each split is counted as (ways to choose the girls) × (ways to choose the remaining boys):

1. 0 girls and 8 boys: 7C0 × 9C8 = 1 × 9 = 9

2. 1 girl and 7 boys: 7C1 × 9C7 = 7 × 36 = 252

3. 2 girls and 6 boys: 7C2 × 9C6 = 21 × 84 = 1764

4. 3 girls and 5 boys: 7C3 × 9C5 = 35 × 126 = 4410

Total: Adding the four case counts gives 9 + 252 + 1764 + 4410 = 6435.

Cross-check: Complementary counting confirms this: choosing at least 4 girls out of the same 16 people for an 8-member team covers the cases of 4, 5, 6 and 7 girls. Since the two complementary groups — at most 3 girls and at least 4 girls — must together account for every possible 8-member team, their counts should add up to the unrestricted total of choosing 8 from 16 people. Indeed, 6435 + 6435 = 12870, which equals 16C8, confirming the result.