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Permutation/Combination Problem

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Permutation/Combination Problem

by dzelkas » Sun May 25, 2008 11:47 am
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?


a) 8
b) 24
c) 32
d) 56
e) 80


correct answer is C. Can someone please help explain the solution?
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by aatech » Sun May 25, 2008 7:58 pm
https://www.beatthegmat.com/committee-num-t11273.html
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by VP_Jim » Sun May 25, 2008 9:08 pm
I was thinking that this question sounds familiar... yes, it's pretty much the exact same set up as the question aatech linked to. My explanation for that one still stands here: just think about it logically. In my opinion, equations tend to get students in trouble.
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by mandy12 » Sun May 25, 2008 11:55 pm
One approach to solve this question is -

total no of ways of selecting 3 people from 8 = 8C3 = 56

Now if the siblings are present in the combination then 2 places are booked and only one place is left to be filled....

1*1*6 for each sibling pair..We have 4 pairs ...so total number of combinations in which sibling pair will be present is -

4*1*1*6 = 24

Hence the number of combinations in which the sibling pair will not be presnet = 56 - 24 = 32
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by getneonow » Mon May 26, 2008 12:37 am
2* 4C3 ( 3 boys or gals committe) + 2 * 4C2 * 2C1 ( 2 boys and one gal or 2 gals and one boy) = 32


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