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
Combination
This topic has expert replies
Source: Beat The GMAT — Problem Solving |
GMAT/MBA Expert
-
Rahul@gurome
- GMAT Instructor
- Posts: 1179
- Joined: Sun Apr 11, 2010 9:07 pm
- Location: Milpitas, CA
- Thanked: 447 times
- Followed by:88 members
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
