Please check my work

Suppose a club has 10 men and 13 women. The club is going to form a committee of 8 people. How many possible 8 person committees have more women than men?

I first solved for the total combination by doing (23!/8!*15!) which came out to be 490314. Then I solved for just men. (10!/8!*2!) and got 40. Then I subtracted the 2 numbers and got 490274. Is this correct or am I way off?

The correct answer is a sum of four terms.
\(\displaystyle \sum\limits_{k = 5}^8 {{13 \choose k}{10 \choose 8-k} }\)

pka said:

The correct answer is a sum of four terms.
\(\displaystyle \sum\limits_{k = 5}^8 {{13 \choose k}{10 \choose 8-k} }\)

Click to expand...

OK, I see and I understand how you did that. I was so far off, it isn't funny! I guess I got that one wrong on my test. Bummer. Thanks for your help!