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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] New post 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?

Could you explain your approach please?
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 [#permalink] New post 12 Dec 2004, 11:24
144,144 :?

I could explain, but please do tell me if my answer is anywhere near the OA.
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 [#permalink] New post 12 Dec 2004, 11:36
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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 [#permalink] New post 12 Dec 2004, 13:37
I was not asking you to "reveal" the OA.
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 [#permalink] New post 12 Dec 2004, 19:09
Let me give a try.
Is it 14C3 + 11C3 + 9C2 + 7C2 + 5C2 + 3C2 = 607?

Thanks!
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 [#permalink] New post 13 Dec 2004, 04:10
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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 [#permalink] New post 13 Dec 2004, 04:49
I got,

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

Last edited by Dookie on 13 Dec 2004, 10:25, edited 1 time in total.
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 [#permalink] New post 13 Dec 2004, 08:17
14C3.11C3.8C2.6C2.4C2.2C2/(2!.4!) for me
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 [#permalink] New post 13 Dec 2004, 08:31
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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 [#permalink] New post 14 Dec 2004, 03:03
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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 [#permalink] New post 14 Dec 2004, 12:57
merci beaucoup monsieur!!
  [#permalink] 14 Dec 2004, 12:57
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How many ways can 14 men be partitioned into 6 committes

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