Question

From a group of 6 women and 8 men a committee consisting of 3 men and 4 women is to be formed. How many different committees are possible if
(a) 2 of the men refuse to serve together?
(b) 2 of the women refuse to serve together?
(c) 1 man and 1 woman refuse to serve together?

Answer 1

I already have the answer for part c, it's 630. Help with a and b please?

Answer 2

For (a) Let name the men as \(A_1,A_2,\cdots A_8 \) and \( A_1 \) and \( A_2\) don't want to serve together. Now you will have two cases.

(1) Taking either of \( A_1 \) or \( A_2\) from those two men and rest two from the remaining 6 men. \[ \binom 2 1 \times \binom 6 2 \times \binom 8 4 \]
(2) Taking the three men from 6 men.
\[ \binom 6 3 \times \binom 8 4 \]
Your answer is sum of these two cases.

For (b) the idea is practically same.

Answer 3

wazzup

Answer 4

Thank you so much :) :)

Answer 5

Glad to help :)