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If there are four distinct pairs of brothers and sisters

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

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New post 11 Jun 2018, 02:57
seekmba 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?

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


# of ways to from a committee of 3 with no siblings on it = (8 * 6 * 4)/3! = 32

The First person in the committee can be chosen in 8 ways.
The Second person in the committee can be chosen in 6 ways, since we will remove the sibling of first person selected & then choose.
The Third person in the committee can be chosen in 4 ways, since we will remove the sibling of second person selected, as well & then choose.

Now we are asked for the # of ways to form a committee & not the arrangement of the persons, hence order does not matter. To account for this we divide by 3!, which is the # of possible arrangements of 3 persons.

Answer C.

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

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New post 15 Jul 2019, 23:16
I know the answer is 32. But what's the difference between the 2 approaches?

Case 1: selected any 3 pairs of bro-sis and then one from each pair:
4C3 X 2C1 X 2C1 X 2C1

Case 2: Selected 1 person out of each pair and then 3 out of those people
2C1 X 2C1 X 2C1 X 2C1 X 4C3

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Re: If there are four distinct pairs of brothers and sisters   [#permalink] 15 Jul 2019, 23:16

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