[Math] Committee combinatorics


A committee of 6 people is to be chosen from a group consisting of 7 men and 8 women. If the committee must consist of at least 3 women and at least 2 men, how many different committees are possible?

The answer given in the text is $\binom{8}{3} \binom{7}{3} + \binom{8}{4} \binom{7}{2} = 3430$. I understand this line of thinking–first count all the ways we can choose 3 of the 8 women and 3 of the 7 men. Then, add to this all of the we can choose 4 of the 8 women and 2 of the 7 men.

I thought about this question a bit differently, however, and am failing to see where my line of thinking is flawed. This is how I performed the calculation:

$$\binom{8}{3} \binom{7}{2} 10 = 11760$$

In other words, we choose 3 women and then 2 men. Then, we are left with 10 other choices, the gender of whom is irrelevant. What am I doing wrong?

Best Answer

You are counting things too many times: If the sixth member is a female, then you get a committee of $4$ women and $2$ men, but exactly this committee was chosen $3$ other times. If the sixth member is a man then this $3$ women / $3$ men committee was chosen $2$ other times.

Note that you can thus fix your reasoning to compute: $$\binom{8}{3}\binom{7}{2}\left(\frac{5}{4}+\frac{5}{3}\right)=3430.$$