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There are four distinct pair of brothers and sisteres. In

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There are four distinct pair of brothers and sisteres. In [#permalink] New post 10 Nov 2006, 04:25
There are four distinct pair of brothers and sisteres. In how many ways can a committee of three be formed and not have siblings in it?

8
24
32
56
80

Pls explain how u arrived at the answer also. :)
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 [#permalink] New post 10 Nov 2006, 05:08
first choose which three (out of 4) pairs, have a representative in the committee (4 ways to do that... you can think of it as which pair doesn't have....)
now each pair can send, independently, 1 out of 2 possible representative.
so we have 2*2*2 = 8 possible committees once the three pairs are chosen.

since choosing the three pairs, and choosing the committee out of the given three pairs can be done independently... jut multiply it altogether to get 32
which is the right answer.

alternatively (i.e. what i would do on test day) just count.
for pairs 1,2,3: we have aaa,aab,aba,abb,baa,bab,bba,bbb
for pairs 1,2,4: aaa,aab,aba,abb,baa,bab,bba,bbb
now you can see that for each choices of pair you have 8 options. you have 4 choices of pairs and get 32 options altogether.

amit.
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 [#permalink] New post 10 Nov 2006, 05:11
B....

Simply pick all opportunities:

AA
BB
CC
DD

ABC=3!
ACD=3!
ABD=3!
BCD=3!

Total 24....
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 [#permalink] New post 10 Nov 2006, 11:30
8C3 -4*6C,1

Total possibilities of 8 people taken by 3 minus the pairs of siblings, 4 pairs *6 other people

56-24 =32
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 [#permalink] New post 10 Nov 2006, 17:51
ugo_castelo wrote:
8C3 -4*6C,1

Total possibilities of 8 people taken by 3 minus the pairs of siblings, 4 pairs *6 other people

56-24 =32


32 too.

The first member can be chosen 8 ways, the second 6 ways and the third 4 ways. Then the total has to be divided by 3! as the order doesn't matter.
(8*6*4)/(3*2*1) = 32
  [#permalink] 10 Nov 2006, 17:51
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