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In how many ways can a commitee of 4 women and 5 men

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

A. 1608
B. 1860
C. 1680
D. 1806
E. 1660
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D
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Bunuel
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Re: In how many ways can a commitee of 4 women and 5 men [#permalink] In how many ways can a commitee of 4 women and 5 men   24 Aug 2012, 10:16
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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.
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In how many ways can a commitee of 4 women and 5 men [#permalink] In how many ways can a commitee of 4 women and 5 men   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 \(63*2*21 * \frac{43}{63} = 42*43 = 1806.\)

Answer: D.
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Re: In how many ways can a commitee of 4 women and 5 men [#permalink] In how many ways can a commitee of 4 women and 5 men   28 Aug 2012, 01:43
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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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Re: In how many ways can a commitee of 4 women and 5 men [#permalink] In how many ways can a commitee of 4 women and 5 men   28 Aug 2012, 11:22
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??
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Re: In how many ways can a commitee of 4 women and 5 men [#permalink] In how many ways can a commitee of 4 women and 5 men   29 Aug 2012, 01:46
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sanjoo wrote:
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 \(C^4_8*C^5_6\);

Total: \(C^3_8*C^5_6+C^4_8*C^4_6+C^4_8*C^5_6=1,806\).

Hope it's clear.
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Re: In how many ways can a commitee of 4 women and 5 men [#permalink] In how many ways can a commitee of 4 women and 5 men   17 Jun 2019, 17:57
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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


We can use the formula:

Total number of ways to select the committee = number of ways when B is with A + number of ways when B is not with A

The total number of ways to select the committee = 9C4 x 7C5:

9C4 x 7C5 = [(9 x 8 x 7 x 6)/(4 x 3 x 2)] x [(7 x 6 x 5 x 4 x 3)/(5 x 4 x 3 x 2)] = 126 x 21 = 2646

The number of ways when B is with A = 8C3 x 6C4:

8C3 x 6C4 = [(8 x 7 x 6)/(3 x 2)] x [(6 x 5 x 4 x 3)/(4 x 3 x 2)] = 56 x 15 = 840

Therefore, the number of ways when B is not with A is:

2646 - 840 = 1806

Answer: D
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Re: In how many ways can a commitee of 4 women and 5 men [#permalink] In how many ways can a commitee of 4 women and 5 men   17 Jun 2019, 19:49
The total number of ways in which a committee of 4 women and 5 men can be chosen from 9 women and 7 men is 9C4∗7C5 = 2,626.

The number of committees with Mr.A and Mr.B (together) is 1C1∗1C1∗8C3∗6C44 = 840.

Therefore, The number of committees which do not have Mr.A and Mr.B (together) is 2,646-840=1,806.
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Re: In how many ways can a commitee of 4 women and 5 men [#permalink] In how many ways can a commitee of 4 women and 5 men   19 Oct 2023, 22:08
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