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Could you please help me solve the following problem:

a committee of 4 people is chosen from 5 married couples. Find how many ways the committee can be chosen if

(a)everyone is equally eligible.

(b)the committee should include at least one woman?

Let the first choice be a woman. There are 5 possibilities.

Choose another 3 people from the remaining 9 people. No repetition, order does't matter.

Number of possibilities = 9!/(3!(9-3)!) = 9!/(3!6!) = 7×8×9/3! = 84

There are 5×84 = 420 ways to choose the committee.

http://www.mathsisfun.com/combinatorics/combinations-permutations.html

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