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Intern
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Joined: 02 Jun 2009
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couple [#permalink]

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New post 02 Jun 2009, 03:50
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If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

A. 20

B. 40

C. 50

D. 80

E. 120

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Re: couple [#permalink]

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New post 03 Jun 2009, 12:08
D

1) \(C^5_3\) - we choose 3 couples out of 5 couples.
2) \((C^2_1)^3\) - we choose one spouse for each couple.
3) total number = \(C^5_3*(C^2_1)^3 = 10*8 = 80\)
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Re: couple [#permalink]

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New post 03 Jun 2009, 13:40
80 = 5C3*(2C1)(2C1)(2C1)
IMO D
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Re: couple [#permalink]

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New post 04 Jun 2009, 02:15
i have a slightly diff approach which is simple..

when encountered with such questions.. i deduct the unwanted ways from teh total number of ways..

in this case, the total number of ways is 10c3 =>120.

unwanted ways = 5 * 8 (5 couples * select one from rest of the group ) = 40.

answer = 120 - 40 = 80

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Re: couple   [#permalink] 04 Jun 2009, 02:15
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