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# M 06 No. 10

Author Message
Manager
Joined: 11 Apr 2009
Posts: 160

Kudos [?]: 120 [0], given: 5

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05 Jun 2009, 14:06
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

Kudos [?]: 120 [0], given: 5

Math Expert
Joined: 02 Sep 2009
Posts: 42305

Kudos [?]: 133072 [1], given: 12403

Re: M 06 No. 10 [#permalink]

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17 Apr 2012, 08:35
1
KUDOS
Expert's post
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 $$C^4_{12}$$.

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

_________________

Kudos [?]: 133072 [1], given: 12403

Re: M 06 No. 10   [#permalink] 17 Apr 2012, 08:35
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# M 06 No. 10

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