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

  • $\begingroup$ please use $n \choose r $ to use $n \choose r$ $\endgroup$ Aug 13 '20 at 8:23
  • $\begingroup$ You can also use $\binom{n}{k}$ to produce $\binom{n}{k}$. $\endgroup$ Aug 13 '20 at 8:25
  • $\begingroup$ Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. $\endgroup$ Aug 13 '20 at 9:19

You are missing the case where you may choose both men from the non-feuding group. This adds $5$ more possibilities. Your expression counts only those cases where one man is selected from each of the feuding and non-feuding groups.


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