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M 06 No. 10

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Joined: 11 Apr 2009
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M 06 No. 10 [#permalink] New post 05 Jun 2009, 13:06
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A
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C
D
E

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If 4 people are selected from a group of 6 married couples, what is the probability that none of them would be married to each other?

A. 1/33
B. 2/33
C. 1/3
D. 16/33
E. 11/12
[Reveal] Spoiler: OA
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Re: M 06 No. 10 [#permalink] New post 17 Apr 2012, 07:35
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gmatprep09 wrote:
If 4 people are selected from a group of 6 married couples, what is the probability that none of them would be married to each other?

A. 1/33
B. 2/33
C. 1/3
D. 16/33
E. 11/12


Each couple can send only one "representative" to the committee. We can choose 4 couples (as there should be 4 members) to send that one "representatives" to the committee in C^4_6 # of ways.

But these 4 chosen couples can send two persons (either husband or wife): 2*2*2*2=2^4.

So # of ways to choose 4 people out 6 married couples so that none of them would be married to each other is: C^4_6*2^4.

Total # of ways to choose 4 people out of 12 is C^4_{12}.

P=\frac{C^4_6*2^4}{C^4_{12}}=\frac{16}{33}

Answer: D.
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Re: M 06 No. 10   [#permalink] 17 Apr 2012, 07:35
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