A committee of 6 is chosen from 8 men & 5 women so as to contain atleast 2M & 3 W. How many different committees could be formed if two of the men refuse to work together.
Ans: 635
Ans: 635
Solution.
The committee can either have
(a) 2 M and 4 W
(b) 3 M and 3 W
Let the two men who refuse to work together be M1 and M2.
So if (a) is the case then there are three possibilities.
Firstly, the committee has M1, no M2, 1 other man and 4 women. So the possible number of ways this can happen is 6C1 * 5C4 = 30 ways.
Secondly, the committee has M2, no M1, 1 other man and 4 women. The possible number of ways this can happen is 6C1 * 5C4 = 30
Thirdly, the committee has neither M1 and nor M2, 2 other men and 4 women. The possible number of ways this can happen is 6C2 * 5C4 = 75.
If (b) is the case, again there are three possibilities.
Firstly the committee has M1, no M2, 2 other men and 3 women. So the possible number of ways this can happen is 6C2 * 5C3 = 150.
Secondly the committee has M2, no M1, 2 other men and 3 women. The possible number of ways this can happen is 6C2 * 5C3 = 150.
Thirdly the committee has neither M1, nor M2, 3 other men and 3 women.
So the possible number of ways this can happen is 6C3 * 5C3 = 200.
Adding up all the above possibilities we have that the number of committees that can be formed is 30 + 30 + 75 + 150 + 150 + 200 = 635
