# How many different ways can I arrange this committee?

A company has 20 employees, 12 males, and 8 females. Suppose we need to form a committee of 5 employees.

1. How many ways are there to form this committee if we need 3 males and 2 females? I believe this is a permutation problem and calculate it as one. 12 * 11 * 10 * 8 * 7 = 73,920

2. How many ways are there to form this committee if we need at least 4 females? Same as the above except we need four females so, 8 * 7 * 6 * 5 * 12 = 20,169

3. How many ways are there to form this committee if we need at least 2 males and at least 2 female? 12 * 11 * 8 * 7 * 16 (because that's how many people are left) = 118,272.

People got different answers, am I looking at this wrong?

• This seems like a homework problem; if so, it should have the self-study tag. See this help page – Peter Flom Feb 3 '14 at 10:38
• It's a practice midterm. Should I still put the self-study tag? – itsSLO Feb 3 '14 at 10:45
• Yes, you should. If you look at the link, the self-study tag is for "routine questions" - just of this sort. In fact, it was changed from "homework" some time ago, in order to deal with this sort of question. – Peter Flom Feb 3 '14 at 11:01

$${12 \choose 3} \times {8 \choose 2} = 220 \times 28$$
2) choosing 4 females AND a man, OR, all five are from females $${12 \choose 1} \times {8 \choose 4} + {8 \choose 5} = 840 + 56$$
3) either we have 3 men and two females, or 2 men and 3 females. $${12 \choose 3} \times {8 \choose 2} + {12 \choose 2} \times {8 \choose 3} = 6160 + 3696$$