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

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

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12 Dec 2004, 11:47
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

How many ways can 14 men be partitioned into 6 committes where 2 of the committes contain 3 men and others 2?

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12 Dec 2004, 12:24
144,144

I could explain, but please do tell me if my answer is anywhere near the OA.
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12 Dec 2004, 12:36
No. Your answer is not right and I want to give others a chance before I reveal it. I hope you understand.
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12 Dec 2004, 14:37
I was not asking you to "reveal" the OA.
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12 Dec 2004, 20:09
Let me give a try.
Is it 14C3 + 11C3 + 9C2 + 7C2 + 5C2 + 3C2 = 607?

Thanks!
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13 Dec 2004, 05:10
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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Senior Manager
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13 Dec 2004, 05:49
I got,

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

Last edited by Dookie on 13 Dec 2004, 11:25, edited 1 time in total.
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13 Dec 2004, 09:17
14C3.11C3.8C2.6C2.4C2.2C2/(2!.4!) for me
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13 Dec 2004, 09:31

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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14 Dec 2004, 04:03
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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14 Dec 2004, 13:57
merci beaucoup monsieur!!
14 Dec 2004, 13:57
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