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# I got it right but that was just a wild guess

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Manager
Joined: 27 Mar 2008
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I got it right but that was just a wild guess [#permalink]  09 Aug 2008, 17:56
I got it right but that was just a wild guess.
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Manager
Joined: 15 Jul 2008
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Re: Counting and Permutation and Probablity [#permalink]  09 Aug 2008, 18:18
How many ways can you choose the first member of the committee ? 8
How many ways can you choose the second member ? 6 (you are not allowed to choose the spouse of the first)
How many ways can you choose the third member ? 4 (spouses of the first two are not allowed)

so 8x6x4. But in this you have included the order of arrangements. 3x2x1 possible arrangements of these 3 members.

Director
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Re: Counting and Permutation and Probablity [#permalink]  10 Aug 2008, 21:51
dishant007 wrote:
I got it right but that was just a wild guess.

Total ways of choosing 3 people = 8 choose 3 = 56
Total # of groups with a married couple = 4*6 = 24

Total # of groups w/o a married couple = 56-24 = 32
Manager
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Re: Counting and Permutation and Probablity [#permalink]  11 Aug 2008, 12:46
zoinnk wrote:
dishant007 wrote:
I got it right but that was just a wild guess.

Total ways of choosing 3 people = 8 choose 3 = 56
Total # of groups with a married couple = 4*6 = 24

Total # of groups w/o a married couple = 56-24 = 32

Where does the 6 come from?
SVP
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Re: Counting and Permutation and Probablity [#permalink]  11 Aug 2008, 12:59
dishant007 wrote:
I got it right but that was just a wild guess.

No of ways form commite of 3 out of 8 members = 8C3 = 56

M1 M2 M3 M4 W1 W2 W3 W4

No of way for commite of 3 out of 8 with couples to gether
= Chose one man out of Four * Chose second member (man's spouse) * Chose third member (other than couple)
= 4C1 *1C1* 6C1
= 4 * 6 =24

Ans = 56-24=36
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Re: Counting and Permutation and Probablity   [#permalink] 11 Aug 2008, 12:59
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