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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