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# How many ways can 14 men be partitioned into 6 committes

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Intern
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How many ways can 14 men be partitioned into 6 committes [#permalink]  12 Dec 2004, 10:47
How many ways can 14 men be partitioned into 6 committes where 2 of the committes contain 3 men and others 2?

Manager
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144,144

I could explain, but please do tell me if my answer is anywhere near the OA.
Intern
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No. Your answer is not right and I want to give others a chance before I reveal it. I hope you understand.
Manager
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I was not asking you to "reveal" the OA.
Intern
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Let me give a try.
Is it 14C3 + 11C3 + 9C2 + 7C2 + 5C2 + 3C2 = 607?

Thanks!
Intern
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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
_________________

Let's crack GMAT

Senior Manager
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I got,

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

Last edited by Dookie on 13 Dec 2004, 10:25, edited 1 time in total.
Director
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14C3.11C3.8C2.6C2.4C2.2C2/(2!.4!) for me
Intern
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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?
Director
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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!)
Intern
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merci beaucoup monsieur!!
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