# How many committees of $2$ women and $3$ men can be formed if two of the men refuse to serve on the committe together?

From a group of 5 women and 7 men, how many different committees consisting of 2 women and 3 men can be formed? What if 2 of the men are feuding and refuse to serve on the committee together?

This is an example in the book that has already been solved in the text. I understand how to answer the first question. However, it goes on to say that

"Because a total of $\binom{2}{2}\binom{5}{1} = 5$ out of the $\binom{7}{3} = 35$ possible groups of 3 men contain both of the feuding men, it follows that there are $35-5=30$ groups that do not contain both of the feuding men. Because there are still $\binom{5}{2}=10$ ways to choose the 2 women, there are $30*10=300$ possible committees in this case"

I'm just not sure where they got the fact that $\binom{2}{2}\binom{5}{1} = 5$ out of the $\binom{7}{3} = 35$ possible groups of 3 men contain both of the feuding men.

Thanks!

• If your committee contains the 2 feuding men, then you have to select 1 more man out of the 5 remaining men (since there are 3 men on the committee and 7 men altogether), and there are $\binom{5}{1}=5$ ways to do this. – user84413 Sep 13 '17 at 20:33