<s>[INST] 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!
 [/INST] From a set of 5 women you always take two women.
From a set of two men  you take one or none of them
Then you take 2 or 3 from a set of 5 men. So: 
\begin{align}
P=\binom{5}{2}\binom{2}{1}\binom{5}{2} +\binom{5}{2}\binom{2}{0}\binom{5}{3} \\
P=\underbrace{\binom{5}{2}}_\text{Two women}(\underbrace{\binom{2}{1}\binom{5}{2} +\binom{2}{0}\binom{5}{3}}_\text{Three men})\\
P=\binom{5}{2}(\frac {2!}{1!1!}\frac {5!}{2!3!}  +\frac {2!}{2!0!} \frac {5!}{3!2!})\\
P=\binom{5}{2}(2*10+1*10)\\
P=\frac {5!}{2!3!}(30)\\
P=10*30=300
 \end{align}
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