4 women and 6 men work in the accounting department. In how

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4 women and 6 men work in the accounting department. In how [#permalink]

08 Jul 2008, 07:25

4 women and 6 men work in the accounting department. In how many ways can a committee of 3 be formed if it has to include only one woman?

Please explain!

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Re: Combinatorics - Committee [#permalink]

08 Jul 2008, 07:32

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Committee of 3 will include 1 woman and 2 men as per the question.

Total Women = 4
Number of ways to select 1 woman out of 4 = 4C1 = 4

Total Men = 6
Number of ways to select 2 men out of 6 = 6C2 = 15

Total number of ways = 4 * 15 = 60

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Re: Combinatorics - Committee [#permalink]

08 Jul 2008, 07:37

excellent solution!

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Re: Combinatorics - Committee [#permalink]

08 Jul 2008, 07:38

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just wanted a confirmation

the reason to multiply 4 with 15 is "select 1 woman AND 2 men" => 4*15

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Re: Combinatorics - Committee [#permalink]

08 Jul 2008, 07:41

Thanks ahijit. I appreciate it. +1
Thanks parag.

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Re: Combinatorics - Committee [#permalink]

08 Jul 2008, 07:55

# of ways 1 woman can be selected out of 4* # of ways 2 men can be selected out of 6= 4C1 * 6C2 = 4 * 15 = 60

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Re: Combinatorics - Committee [#permalink]

08 Jul 2008, 08:22

hi0parag wrote:

just wanted a confirmation

the reason to multiply 4 with 15 is "select 1 woman AND 2 men" => 4*15

Yes you are absolutely correct.

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Re: Combinatorics - Committee [#permalink]

08 Jul 2008, 09:02

stingraybullray wrote:

4 women and 6 men work in the accounting department. In how many ways can a committee of 3 be formed if it has to include only one woman?

Please explain!

WMM = 4 * 6C2 = 4*15 =60

Re: Combinatorics - Committee [#permalink] 08 Jul 2008, 09:02

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4 women and 6 men work in the accounting department. In how

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4 women and 6 men work in the accounting department. In how

4 women and 6 men work in the accounting department. In how

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