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There are four distinct pairs of brothers and sisters. In

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There are four distinct pairs of brothers and sisters. In [#permalink] New post 09 Oct 2008, 23:13
There are four distinct pairs of brothers and sisters. In how many ways can a committee of 3 be formed and NOT have siblings in it?

Ans 32
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Re: siblings [#permalink] New post 09 Oct 2008, 23:42
Four possible cases.

Case 1: select 3 members from 4 brothers.....4c3 = 4
Case 2: select 3 members from 4 sisters.....4c3 = 4
case 3: select 1 member from brothers and 2 members from the 3 sisters (excluding the sibling) = 4c1*3c2 = 4*3 = 12
case 4: select 1 member from sisters and 2 members from the 3 brothers (excluding the sibling) = 4c1*3c2 = 4*3 = 12

Hence, total number = 12 + 12 + 4 + 4 = 32.
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Re: siblings [#permalink] New post 10 Oct 2008, 01:17
Total members is 8 (4 pairs of brother & sister).
committe of 3 can be chossed 8C3 = 56

Now suppose 3 member committe consist of 1 pair than 3rd member can be choosed from remaing 6 members 6C1 = 6
Now there are 4 pairs, so no of committe in which siblings are there = 4x6=24

Ans 56 - 24 = 32
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Re: siblings [#permalink] New post 10 Oct 2008, 02:26
Now suppose 3 member committe consist of 1 pair than 3rd member can be choosed from remaing 6 members 6C1 = 6
Now there are 4 pairs, so no of committe in which siblings are there = 4x6=24

Good work Amit. I was stumped earlier but your answer makes sense. But is it really correct ?
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Re: siblings [#permalink] New post 10 Oct 2008, 06:01
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32=>(8*6*4)/(2*3)
Solution:
There are 8 persons, we can take anyone out of 8, in order to pick next person for the committee we have only 6 people, otherwise we can accidentally take brother or sister of one that we have already chosen, and for the third place we can choose from only 4 persons. And we divide by 3! In order to get rid of replications.
Re: siblings   [#permalink] 10 Oct 2008, 06:01
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