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

Question

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

Bunuel · Accepted Answer

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.

GSBae · Answer

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.

sanjoo · Answer

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??

Bunuel · Answer

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.

ScottTargetTestPrep · Answer

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 =   x   = 126 x 21 = 2646

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

8C3 x 6C4 =   x   = 56 x 15 = 840

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

2646 - 840 = 1806

Answer: D

punyaslok · Answer

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.

Riyaagoell · Answer

Shouldnt the ans for the question be : 1 women is fixed so we have to choose (9-1)8 women from remaining 3 =8C3
1 man refused to be on commitee so (7-1)6 from remaining 5=6C5
8C3*6C5=336

Bunuel · Answer

Your reasoning counts only committees in which Ms. B is included and Mr. A is excluded. But the following committees are all allowed:

• Ms. B is on the committee and Mr. A is not
• Mr. A is on the committee and Ms. B is not
• Neither Ms. B nor Mr. A is on the committee

The only forbidden case is when  both  Ms. B and Mr. A are on the committee together.

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

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

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