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The Adams family consists of 2 pairs of brothers. The

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Intern
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The Adams family consists of 2 pairs of brothers. The [#permalink] New post 14 Aug 2007, 12:03
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The Adams family consists of 2 pairs of brothers. The Simpsons family consists of 3 sisters and the Flinstones family has a brother and a sister.

All of these people divide themselves into 3 groups to play the Ouija board. In how many ways can they play?

Answer: 9!/(3!)^4

Can anyone explain how we come up with the answer?

Thanks,
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 [#permalink] New post 14 Aug 2007, 12:18
Looking at it, it is actually a very simple problem. Now u need to understand that the you can form 3 groups and all the 3 groups are identical.

Selecting the first group of 3 is done in 9C3 ways. Second group is 6C3 ways. and the third group is 3C3 ways. Due to the identicality of the groups, you have finally divide by 3!.

So 9C3*6C3*3C3/3! = 9!/(3!)^4
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 [#permalink] New post 14 Aug 2007, 12:44
Can you please explain division by 3! - identicality of the groups?

thanks,

dahcrap wrote:
Looking at it, it is actually a very simple problem. Now u need to understand that the you can form 3 groups and all the 3 groups are identical.

Selecting the first group of 3 is done in 9C3 ways. Second group is 6C3 ways. and the third group is 3C3 ways. Due to the identicality of the groups, you have finally divide by 3!.

So 9C3*6C3*3C3/3! = 9!/(3!)^4
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 [#permalink] New post 14 Aug 2007, 18:23
Quote:
Now u need to understand that the you can form 3 groups and all the 3 groups are identical.


Can we assume on the GMAT that they're talking about 3 identical groups? Couldn't it be split up 7-1-1 or 5-2-2 or any other combination?

EDIT: yeah, I'm also wondering why it wouldn't be simply 9!
  [#permalink] 14 Aug 2007, 18:23
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