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# There are four distinct pairs of brothers and sisters. In

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Intern
Joined: 12 Jun 2006
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There are four distinct pairs of brothers and sisters. In [#permalink]  12 Jun 2006, 23:04
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There are four distinct pairs of brothers and sisters. In how many ways can a 3 person committee be selected so as NOT to have siblings in it?

A. 24
B. 32
C. 48
D. 60
E. 80

Can sombody explain....
Senior Manager
Joined: 09 Mar 2006
Posts: 445
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B it is

First, you have to select the pairs whose members will be elected to committee. That is 4C3 = 4 distinct ways.
Now for each pair there are 2 ways to select a person.

So 4*2*2*2 = 32
Intern
Joined: 12 Jun 2006
Posts: 49
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But I thought it was 4 pairs.... with not ot have siblings in it.....

I LOST IT!!!
SVP
Joined: 30 Mar 2006
Posts: 1739
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Kudos [?]: 46 [0], given: 0

Number of ways 3 people can be chosen from 8 = 8C3 = 56

Number of ways siblings appear in the commitee = 24
Hence 56-24 = 32 ways
Intern
Joined: 12 Jun 2006
Posts: 49
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Kudos [?]: 1 [0], given: 0

Thanks a ton!!!
But is there a particular way that we can tackle Combination questions, 'cos each question sounds different???
Manager
Joined: 10 May 2006
Posts: 186
Location: USA
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Total, there are 8C3 ways to choose a group of 3 = 56 ways

To calculate the # of ways to choose a group WITH siblings:

Assume that there is a sibling in the group - 4 pairs of siblings = 4 choices
For each pair of sibling that's chosen, there are 6 choices for the third member of the group.

Therefore, there are 4x6 = 24 groups with sibling pairs.

Number of groups without sibling pairs = 56-24 = 32.

I don't think there's a particular way to tackling comb/permutation questions. The key is to really understand the concepts behind these questions so you can manipulate them when you need to...this is esp important since these questions come in all shapes/forms.
Manager
Joined: 10 May 2006
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Location: USA
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http://www.gmatclub.com/phpbb/viewtopic.php?t=14706

Check out this link started by Honghu....look through it in depth if you have time to go over it before your exam.
VP
Joined: 02 Jun 2006
Posts: 1267
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Kudos [?]: 42 [0], given: 0

The first position can be picked in 8, the second in 6 and third in 4. As the order is not important the three choices in any order are the same.

Therefore, # of ways = 8x6x4/3x2 =32
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