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A committee of 5 is to be chosen from 6 men and 5 women. In how many ways can it be done if it may have at most 2 women?

Competition practicePermutations & Combinations

A281
B462
C200
D275

Step-by-Step Solution

0 women:

^{6} C_{5} = 6

. 1 woman:

^{5} C_{1} \times^{6} C_{4} = 75

. 2 women:

^{5} C_{2} \times^{6} C_{3} = 200

. Total

281. (462 =^{11} C_{5}

has no limit.)

Answer: A) 281

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