A committee of 3 people is to be chosen from 5 men and 4 women. How many different committees include at least one woman?
- A60
- B64
- C70
- D74
- E80
Counting with a restriction: complement first
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Correct answer: D
“At least one” is the signal to count the complement: all committees minus the committees with no women.
Total committees of 3 from 9 people:
C(9,3) = 9! / (3! × 6!) = (9 × 8 × 7) / 6 = 84.
All-male committees (zero women) chosen from the 5 men:
C(5,3) = 10.
Committees with at least one woman: 84 − 10 = 74.
Direct counting also works but takes three cases: exactly 1 woman is C(4,1)·C(5,2) = 4 × 10 = 40; exactly 2 women is C(4,2)·C(5,1) = 6 × 5 = 30; exactly 3 women is C(4,3) = 4. Total 40 + 30 + 4 = 74. Same answer, three times the work, and three more places to slip.
Trap (E), 80, comes from C(9,3) − C(4,1), subtracting one woman instead of the all-male case. Trap (C), 70, is C(9,3) − C(5,3) miscomputed as 84 − 14.