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# If there are four distinct pairs of brothers and sisters

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Joined: 09 Sep 2013
Posts: 14238

Kudos [?]: 291 [0], given: 0

Re: If there are four distinct pairs of brothers and sisters [#permalink]

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14 Oct 2017, 20:22
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Kudos [?]: 291 [0], given: 0

Manager
Joined: 12 Nov 2016
Posts: 124

Kudos [?]: 12 [0], given: 71

Location: Kazakhstan
Concentration: Entrepreneurship, Finance
GMAT 1: 620 Q36 V39
GMAT 2: 650 Q47 V33
GPA: 3.2
Re: If there are four distinct pairs of brothers and sisters [#permalink]

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04 Nov 2017, 10:25
Bunuel wrote:
As committee shouldn't have siblings in it, then a pair can send only one "representative" to the committee. # of ways to choose which 3 pairs of brothers and sisters should send one "representative" to the committee is $$C^3_4$$ (choosing 3 pairs which will be granted the right to send one "representative" to the committee);

But each of these 3 pairs can send 2 persons to the committee either a brother or a sister: $$2*2*2=2^3$$;

So total # of ways is $$C^3_4*2^3=32$$.

What I fail to understand in this question and a similar one with 5 married couples, is a second part - why a number of alternatives for 3 pairs to send 2 persons equal to 2*2*2= 8
Here is a small example, let pairs be AB, CD, EF. Since 2 persons can't form the original pair of siblings then the alternatives are:
1)AC 2)AD 3) AE 4) AF 5) BC 6) BD 7) BE 8) BF 9) CE 10) CF 11) DE 12) DF
So, the number of alternatives is 12, not 8! What is it that I am missing?

Kudos [?]: 12 [0], given: 71

Re: If there are four distinct pairs of brothers and sisters   [#permalink] 04 Nov 2017, 10:25

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