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Intern
Joined: 02 Jun 2009
Posts: 22

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couple [#permalink]

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02 Jun 2009, 02:50
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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

Kudos [?]: 8 [0], given: 0

CEO
Joined: 17 Nov 2007
Posts: 3583

Kudos [?]: 4718 [0], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Re: couple [#permalink]

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03 Jun 2009, 11:08
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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Kudos [?]: 4718 [0], given: 360

Senior Manager
Joined: 16 Jan 2009
Posts: 355

Kudos [?]: 241 [0], given: 16

Concentration: Technology, Marketing
GMAT 1: 700 Q50 V34
GPA: 3
WE: Sales (Telecommunications)
Re: couple [#permalink]

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03 Jun 2009, 12:40
80 = 5C3*(2C1)(2C1)(2C1)
IMO D
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Lahoosaher

Kudos [?]: 241 [0], given: 16

Senior Manager
Joined: 15 Jan 2008
Posts: 277

Kudos [?]: 49 [0], given: 3

Re: couple [#permalink]

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04 Jun 2009, 01:15
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

Kudos [?]: 49 [0], given: 3

Re: couple   [#permalink] 04 Jun 2009, 01:15
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