The Committee With Constraints
A club has 6 girls and 5 boys, all distinct. A committee of 4 must be formed. The rules state: the committee must contain at least one girl and at least one boy, and two particular students - Ana (a girl) and Boris (a boy) - refuse to serve together. A student computes the number of valid committees by taking all C(11,4) committees, subtracting those with no girls and those with no boys, then subtracting committees containing both Ana and Boris. The teacher suspects this double-subtracts and asks for a careful count, plus a probability.
1. Count the valid committees correctly, explaining the inclusion–exclusion logic.
Read the worked answer
Total: C(11,4) = 330.
No girls (all boys): C(5,4) = 5. No boys (all girls): C(6,4) = 15. These are disjoint, so committees with at least one of each: 330 − 5 − 15 = 310.
Now remove committees containing both Ana and Boris. Fix them; choose 2 more from the remaining 9: C(9,2) = 36. All such committees already have a girl and a boy, so they are inside the 310. Valid: 310 − 36 = 274.
The student's method is actually correct here because the 36 committees with both Ana and Boris are contained in the 310 set, so no double subtraction occurs. A common error is subtracting them from 330 before removing single-sex committees, which would overcount removals.
2. Given a uniformly random valid committee, find the probability it contains Ana.
Read the worked answer
Count valid committees containing Ana. Ana is a girl, so we need 3 more members from the other 10, with at least one boy among the four, and excluding Boris.
Case split on Boris:
- Boris not in committee: choose 3 from the other 9 (5 boys minus Boris = 4 boys, 5 other girls). Total C(9,3) = 84. Subtract all-girl choices from the 5 other girls: C(5,3) = 10. Gives 74.
- Boris in committee: choose 2 more from the other 9 (excluding Ana and Boris): C(9,2) = 36. All have at least one boy (Boris), so 36.
Total containing Ana: 74 + 36 = 110.
Probability = 110/274 = 55/137 ≈ 0.401.