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If there are four distinct pairs of brothers and sisters

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Joined: 12 Nov 2016
Posts: 140
Concentration: Entrepreneurship, Finance
GMAT 1: 620 Q36 V39
GMAT 2: 650 Q47 V33
Re: If there are four distinct pairs of brothers and sisters [#permalink]

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New post 04 Nov 2017, 11:25
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?
Re: If there are four distinct pairs of brothers and sisters   [#permalink] 04 Nov 2017, 11:25

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