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In how many ways can a commitee of 4 women and 5 men [#permalink]

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24 Aug 2012, 09:02

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In how many ways can a committee of 4 women and 5 men be chosen from 9 women and 7 men, if Mr.A refuses to serve on the committee if Ms.B is a member??

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In how many ways can a committee of 4 women and 5 men be chosen from 9 women and 7 men, if Mr.A refuses to serve on the committee if Ms.B is a member??

A. 1608 B. 1860 C. 1680 D. 1806 E. 1660

Total number of ways a committee of 4 women and 5 men can be chosen from 9 women and 7 men is \(C^4_9*C^5_7=2,626\).

The number of committees with Mr.A and Mr.B is \(C^1_1*C^1_1*C^3_8*C^4_6=840\).

The number of committees which do not have Mr.A and Mr.B (together) is 2,646-840=1,806.

Re: In how many ways can a commitee of 4 women and 5 men [#permalink]

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28 Aug 2012, 01:43

Bunuel wrote:

rajathpanta wrote:

In how many ways can a committee of 4 women and 5 men be chosen from 9 women and 7 men, if Mr.A refuses to serve on the committee if Ms.B is a member??

A. 1608 B. 1860 C. 1680 D. 1806 E. 1660

Total number of ways a committee of 4 women and 5 men can be chosen from 9 women and 7 men is \(C^4_9*C^5_7=2,626\).

The number of committees with Mr.A and Mr.B is \(C^1_1*C^1_1*C^3_8*C^4_6=840\).

The number of committees which do not have Mr.A and Mr.B (together) is 2,646-840=1,806.

Answer: D.

Thanks Bunuel for explanation, it is faster and clearer way (BTW i think there is typo its not 2,626). I have tried to think from different way, let say number of ways whithout Ms.B 9C5-8C3=70 is multiplied to number of commities with men which is 21 so overall 1470. Could you please clarify where i got wrong?
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bunuel.. question is saying if A will be the member then B will not.. but wat u did is that both will not be the member..

if we do ...8c3 *7c5 =?? in that b will not join but A will join..

if we do.. 9c4*6c4?? ..in this we l not have a will join??

wat actually m i asking cant we do with this way??

All committees are good but the committees which have Mr.A and Ms.B together. So, we found the number of all committees possible and subtracted the number of committees with Mr.A and Ms.B in them.

We could solve the question with direct approach as well, though it would be lengthier:

The number of committees with Ms.B but without Mr.A is \(C^3_8*C^5_6\);

The number of committees without Ms.B but with Mr.A is \(C^4_8*C^4_6\);

The number of committees without Ms.B and without Mr.A is \(C^4_8*C^5_6\);

Re: In how many ways can a commitee of 4 women and 5 men [#permalink]

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26 Aug 2014, 04:44

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In how many ways can a commitee of 4 women and 5 men [#permalink]

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10 Nov 2014, 08:08

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I think there's a slightly simpler way to do this:

The total number of committees that can be formed is \(C^9_4*C^7_5\). That's \(\frac{9*8*7*6}{4*3*2*1} * \frac{7*6}{2} = 63*2*21\).

The probability that BOTH A & B are selected are \(\frac{4}{9}*\frac{5}{7} = \frac{20}{63}\). Therefore, the probability that both of them are not selected = \(\frac{43}{63}\)

To find the total number of combinations without both A and B, we multiply \(63*2*21 * \frac{43}{63} = 42*43 = 1806.\)

Re: In how many ways can a commitee of 4 women and 5 men [#permalink]

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20 Oct 2016, 19:39

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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