Suppose the group of twelve consists of 5 men and 7 women. How many…
Suppose the group of twelve consists of 5 men and 7 women. How many five-person teams contain at least one man?
Answer: 771 — Concept: When forming a group of r people from a total of n people under an “at least one from a specific subgroup” constraint, the direct approach requires…
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Correct answer: 771
Concept: When forming a group of r people from a total of n people under an “at least one from a specific subgroup” constraint, the direct approach requires summing multiple cases. It is simpler to use the complement principle: (ways satisfying the constraint) = (total ways to choose r from n) − (ways that violate the constraint, i.e., contain none from that subgroup). The number of ways to choose r items from n distinct items, where order does not matter, is given by the combination formula C(n, r) = n! / (r! × (n − r)!).
Application:
Total people = 12 (5 men + 7 women). The team size required is 5, with no restriction other than order not mattering, so the total number of ways to form any 5-person team = C(12, 5) = (12 × 11 × 10 × 9 × 8) / (5 × 4 × 3 × 2 × 1) = 792.
A team with “no men” means all 5 members are chosen from the 7 women only: C(7, 5) = (7 × 6) / (2 × 1) = 21.
By the complement principle, the number of teams containing at least one man = Total teams − teams with no men = 792 − 21 = 771.
Cross-check: Summing directly over the number of men k = 1 to 5 in the team: C(5,1) × C(7,4) + C(5,2) × C(7,3) + C(5,3) × C(7,2) + C(5,4) × C(7,1) + C(5,5) × C(7,0) = 175 + 350 + 210 + 35 + 1 = 771, which matches the complement-method result.
Answer: 771.
