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There was a problem discussed in an another post by

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Director
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There was a problem discussed in an another post by [#permalink] New post 21 Sep 2004, 23:03
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There was a problem discussed in an another post by jjomallls that I wanted to bring up for discussion. The question was:

3) A club has a membership of 9 men and 12 women. A 5-member committee is to be selected to attend a conference. How many committes with no more than three women be formed?

PS: People who want to try this, please do not see the next post.
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 [#permalink] New post 21 Sep 2004, 23:07
One answer by Paul (and this was my solution too):
Paul wrote:
12C3*9C2 + 12C2*9C3 + 12C1*9C4 + 12C0*9C5 = 7920+5544+1512+126 = 15102 possible ways
Possible ways of forming the 5 person committee given that there must be 3 women or less:
I) 3W+2M
II) 2W+3M
III) 1W+4M
IV) 0W+5M
The sum of these possibilities will give you the total possible favorable outcomes. Since there are 12 women from which to select the women in the committee and 9 men from which to select the men in the committee, we have the following:
12C3 ways of selecting 3 women * 9C2 ways of selecting other 2 men
12C2 ways of selecting 2 women * 9C3 ways of selecting other 3 men
12C1 ways of selecting 1 woman * 9C4 ways of selecting other 4 men
12C0 ways of selecting 0 woman * 9C5 ways of selecting other 5 men
The answer is the sum of those combination outcomes and as shown in first line for question 3.


There was an interstingly another method too by Paul. I initially thought this would be the fastest method to solve, but the approach produced a different result.

Paul wrote:
21C5 - 9C4*12C1 - 9C5 = 20349 - 1512 -126 = 18711.


Any idea why Method #2 is wrong?

Last edited by hardworker_indian on 22 Sep 2004, 05:38, edited 1 time in total.
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 [#permalink] New post 21 Sep 2004, 23:41
Ok, did not see the other post..
My approach would be

9C5(all men committee)+12C1*9C4 +12C2*9C3+12C3*9C2 =whatever should be the answer.

No more than three women could mean no women or 1 women or 2 women or 3 women in the committee.
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 [#permalink] New post 21 Sep 2004, 23:47
I calculated the numbers..its coming to 19638.
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 [#permalink] New post 22 Sep 2004, 07:05
Hi hardworker,

the method #2 must be formulated in this way:

21C5 - 9C1*[b]12C4 - 9C0*[/b]12C5

the result should be same compared with method #1


cheers
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 [#permalink] New post 22 Sep 2004, 07:09
opss.... here the correction of method #2 again

21C5 - 9C1*12C4 - 9C0*12C5


cheers
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 [#permalink] New post 22 Sep 2004, 07:25
GMATPIPO wrote:
opss.... here the correction of method #2 again
21C5 - 9C1*12C4 - 9C0*12C5
cheers


Got it. Silly mistake - it escaped my eye.
I was wondering it has got to work both ways.
  [#permalink] 22 Sep 2004, 07:25
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