Combinatorics question??

Question

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

Can anyone explain how to solve this? Thanks :)

Answer 1

In the following, C(n,k) is the choose function C(n,k) = n!/ (k! (n-k)!)  which is the number of ways to choose k elements for a set of n elements.

Now for your committee, there are 3 possibilities:

1) the stubborn man and stubborn woman are both not serving

2) the stubborn man is serving and not the stubborn woman

3) the stubborn woman is serving and not the stubborn man

# possibilities in each case

case 1: choose 3 women out of the 7 non stubborn women and 3 men from the 5 non stubborn men

C(7,3)*C(5,3)

case 2: choose 3 women out of the 7 non-stubborn, the stubborn man and 2 other men out of the 5 other

C(7,3)*C(5,2)

case 3:  choose the stubborn woman, 2 more from the 7 remaining, and 3 from the 5 non-stubborn men.

C(7,2)*C(5,3)



So the total is 35*10 + 35 * 10 + 21 * 10 =910 possibilities

Another way to see it, is there are C(8,3)*C(6,3) total possibilities for the committee if you didn't consider that man and woman not wanting to work together.  Of those, C(7,2)*C(5,2) have both the man and the woman.  So you have

C(8,3)*C(6,3) - C(7,2)*C(5,2) = 56*20 - 21*10 = 910 possibilities for your final committee.