# How many ways can I choose a committee of $3$ people from $7$ males and $2$ females.

Among $$9$$ people there are $$7$$ females and $$2$$ males. You want to make a committee of $$3$$ people. What is the probability the committee has at least two females?

I wanted to go about this problem $$2$$ ways to check my answer...however I'm getting different answers and I'm not sure which to trust.

So ideally there should be $${9\choose3}$$ different total combinations of committees which is equal to $$84$$. However if I break this into cases I don't get the same answer.

The first case would be a committee of $$3$$ females: $${7\choose3} =35$$, then $$2$$ females and $$1$$ male: $${7\choose2}{2\choose1}=42$$ and finally $$1$$ female and $$2$$ males $${7\choose1}{2\choose2}= 14$$ but this sums to $$91$$. So which total amount of arrangements is correct, and why aren't I getting the same result?

• $\binom 22=1$ not $2$. Your approach is sound otherwise. – lulu Sep 23 '18 at 13:48
• silly mistake! thank you! – Lil Sep 23 '18 at 14:04
• Oh, I make worse mistakes than that all the time. Points to you for checking the sum. – lulu Sep 23 '18 at 14:07
• The word arrange is misleading since it implies order. Using the word select or choose would be more appropriate in this context. – N. F. Taussig Sep 23 '18 at 14:13