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Word Problems/Permutation & Combination

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Question
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?

Answer
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󭅍/3! = 84

There are 584 = 420 ways to choose the committee.

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

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Janet Yang

Expertise

Word problems are my favorite type of math questions! I would not feel comfortable answering questions that require specialized knowledge (Physics, Statistics, etc.) because I have not studied these in depth.

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I tutor students (fifth through twelfth grades) and am a Top Contributor on Yahoo!Answers with over 24,000 math solutions.

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Co-author of An Outline of Scientific Writing: For Researchers With English as a Foreign Language.

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I have a Bachelor's degree in Applied Mathematics from the University of California at Berkeley, and a Master of Business Administration degree from The Wharton School.

Past/Present Clients
George White Elementary School. Homework Help program at the Ridgewood Public Library, Ridgewood, NJ. Individual students.

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