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# M02#05

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Senior Manager
Status: Do and Die!!
Joined: 15 Sep 2010
Posts: 312

Kudos [?]: 603 [0], given: 193

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03 Jan 2011, 10:44
If there are four distinct pairs of brothers and sisters, then in how many ways can a committee of 3 be formed and NOT have siblings in it?

A. 8
B. 24
C. 32
D. 56
E. 192

question-1 8*6*4 => can we Make permutations like this ?
question-2 why its is divided by 6. we have already made permutation as 8*6*4 and not like 8*7*6*5 etc

Thanks
_________________

I'm the Dumbest of All !!

Kudos [?]: 603 [0], given: 193

Math Expert
Joined: 02 Sep 2009
Posts: 42352

Kudos [?]: 133300 [0], given: 12445

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03 Jan 2011, 11:02
shrive555 wrote:
If there are four distinct pairs of brothers and sisters, then in how many ways can a committee of 3 be formed and NOT have siblings in it?

8
24
32
56
192

question-1 8*6*4 => can we Make permutations like this ?
question-2 why its is divided by 6. we have already made permutation as 8*6*4 and not like 8*7*6*5 etc

Thanks

$$C^3_4*2^3=32$$;

$$C^3_4$$ - # of ways to select the sibling pair, which will be "granted" the right to give member for committee;
$$2^3$$ - each selected sibling pair can give either brother or sister for membership: 2*2*2=2^3.

As for your question: you should divide 8C1x6C1x4C1 by 3! to get rid of duplications. With 8C1x6C1x4C1 you can have group ABC as well as group BCA, which is basically the same group.

This issue is discussed here: ps-combinations-94068.html
_________________

Kudos [?]: 133300 [0], given: 12445

Re: M02#05   [#permalink] 03 Jan 2011, 11:02
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# M02#05

Moderator: Bunuel

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