Problem Statement
From 8 men and 5 women, how many committees of 5 can be formed with at least 2 women?
Explanation
Sum cases. 2 women: C(5,2)×C(8,3)=10×56=560. 3 women: C(5,3)×C(8,2)=10×28=280. 4 women: C(5,4)×C(8,1)=5×8=40. 5 women: C(5,5)×C(8,0)=1×1=1. Total 560+280+40+1=881? That seems off; we miscounted because we only have 5 women total, correct combinations are fine. Recompute: 560 + 280 + 40 + 1 = 881. None of the options match. We must have made an arithmetic slip earlier in option design. Fix counts: For 2 women: C(5,2)=10, C(8,3)=56 → 560. For 3 women: 10×28=280. For 4 women: 5×8=40. For 5 women: 1×1=1. Sum is 881. Provide correct option 881.
Code Solution
SolutionRead Only
Σ = 560 + 280 + 40 + 1 = 881
Practice Sets
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