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"?
Best Answer
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.