# Choosing People For A Committee With Limitations

From a group of 8 women and 6 men, a committee consisting of 3 men and 3 women is to be formed.
How many different committees are possible if
(c) 1 man and 1 woman refuse to serve together?

committee without the "problematic" woman ${6\choose 3}\cdot{7\choose 3}$
committee without the "problematic" man ${5\choose 3}\cdot{8\choose 3}$
Now it seems that I counted twice committes without the "problematic" people so overall it is ${6\choose 3}\cdot{7\choose 3} + {5\choose 3}\cdot{8\choose 3}-{5\choose 3}\cdot{7\choose 3}=910$.

Is there a way to calculate the committees without "over-counting"?

• Note that neither of the "problematic" people refuse to serve on their own. Instead, count the total number of committees and subtract the ones that include both of the people who don't wish to work with each other. Correcting for this overcounting is called "inclusion-exclusion" and is completely standard. – Eric Tressler May 10 '15 at 18:21

There are ${8 \choose 3} \cdot {6 \choose 3}$ committees in total, of which ${7 \choose 2} \cdot {5 \choose 2}$ contain the problematic pair.
This gives ${8 \choose 3} \cdot {6 \choose 3}-{7 \choose 2} \cdot {5 \choose 2}$ committees without the problematic pair.
• why ${7\choose 2}\cdot {5\choose 2}$ count the problematic pair? – gbox May 10 '15 at 18:25