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22 Sep 2004, 00:03
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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.
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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Updated on: 22 Sep 2004, 06:38
Originally posted by hardworker_indian on 22 Sep 2004, 00:07.
Last edited by hardworker_indian on 22 Sep 2004, 06:38, edited 1 time in total.
Last edited by hardworker_indian on 22 Sep 2004, 06:38, edited 1 time in total.
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One answer by Paul (and this was my solution too):
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.
Any idea why Method #2 is wrong?
Paul
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
Any idea why Method #2 is wrong?
22 Sep 2004, 00:41
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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.
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.
