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Family X consists 2 pairs of brothers, Family Y consists of

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Family X consists 2 pairs of brothers, Family Y consists of [#permalink] New post 10 May 2005, 13:57
:oops:

Family X consists 2 pairs of brothers, Family Y consists of 3 sisters and Family Z consists of 1 brother / 1 sister. All of them divide themselves into 3 groups to play a game (the group cant be empty). In how many ways can they play :?:
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Re: Permutation / Combination [#permalink] New post 10 May 2005, 16:30
Total number of people = 4 + 3 + 2 = 9.
Therefore number of ways of distributing them in 3 groups, with no group empty = 3^6.
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 [#permalink] New post 10 May 2005, 19:20
with no group empty, i got 8
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Re: Permutation / Combination [#permalink] New post 11 May 2005, 14:59
kapslock wrote:
Total number of people = 4 + 3 + 2 = 9.
Therefore number of ways of distributing them in 3 groups, with no group empty = 3^6.


What is your reasoning. :!: ( the OA is 9!/ (3!)^4 )
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Re: Permutation / Combination   [#permalink] 11 May 2005, 14:59
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Family X consists 2 pairs of brothers, Family Y consists of

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