From a group of 4 men and 5 women, how many committees of size 3 are possible with 2 men and 1 woman if 2 of the men are feuding and refuse to serve on the committee together?
The indirect method of solving this problem is to do the following :
Total number of possible cases = 5C1 * 4C2 = 30
No of ways feuding men serve together = 2C2 * 5C1 = 5
Answer = 30 - 5 = 25
However, i wanted to solve this problem using the direct method. I tried to split the 4 men into 2 groups, one group was the feuding group of 2 men, and the other 2 were the passive ones. Then i chose 1 man from the feuding group and 1 man from the passive group. May i know what is wrong with this logic?
5C1 * 2C1 * 2C1 = 20