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Re: How many ways can 14 men be partitioned into 6 committes [#permalink]
I was not asking you to "reveal" the OA.
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Re: How many ways can 14 men be partitioned into 6 committes [#permalink]
Let me give a try.
Is it 14C3 + 11C3 + 9C2 + 7C2 + 5C2 + 3C2 = 607?

Thanks!
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Re: How many ways can 14 men be partitioned into 6 committes [#permalink]
rohitprabhu wrote:
Let me give a try.
Is it 14C3 + 11C3 + 9C2 + 7C2 + 5C2 + 3C2 = 607?

Thanks!



I think this appraoch is correct, but it should be

14C3 + 11C3 + 8C2 + 6C2 + 4C2 + 1 = Whatever
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Re: How many ways can 14 men be partitioned into 6 committes [#permalink]
I got,

(14C3 * 11C3 * 8C2 * 6C2 * 4C2 * 2C2) / 6!

Originally posted by Dookie on 13 Dec 2004, 05:49.
Last edited by Dookie on 13 Dec 2004, 11:25, edited 1 time in total.
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Re: How many ways can 14 men be partitioned into 6 committes [#permalink]
14C3.11C3.8C2.6C2.4C2.2C2/(2!.4!) for me
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Re: How many ways can 14 men be partitioned into 6 committes [#permalink]
Here's the answer given

14! 1
-------------------- x ---------
3!*3!*2!*2!*2!*2!* 2!*4!

which works out to 3153150

I understand the first part but I don't get the second part.

twixt: can you explain?
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Re: How many ways can 14 men be partitioned into 6 committes [#permalink]
Shumi,

Groups are not distinct here so you have to divide your comb number by the number of groups !. In this case these groups are not equally sized so you have to consider them as different entities : first you have 2 groups of 3 (so divide by 2!) and then 4 groups of 2 (so divide by 4!)
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Re: How many ways can 14 men be partitioned into 6 committes [#permalink]
merci beaucoup monsieur!!



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