TITLE: How many ways are there to choose a group of $5$ people from $10$ men and $12$ women if at least $2$ are women?
QUESTION [5 upvotes]: Question: There are $10$ men and $12$ women. How many ways are there to choose a group of $5$ people where at least $2$ are women?
The correct answer is $${12 \choose 2} * {10 \choose 3} + {12 \choose 3} * {10 \choose 2} + {12 \choose 4} * {10 \choose 1} + {12 \choose 5} * {10 \choose 0}$$ which is pretty straight-forward and correct. 
But say we chose $2$ women first, and chose the rest $3$ from the remaining people. So the combination becomes ${12 \choose 2} * {20 \choose 3}$. But it doesn't match with the first answer. What am I doing wrong in the second process?

REPLY [3 votes]: The problem with your second approach is that you are over counting. 
For instance, if the group you pick is one with three women, you would count the same group 3 times. 
To see this let the people you pick be $A,B,C,D,E$
where $A,B,C$ are the three women. You could have $A,B$ be the women you first choose and then have $C,D,E$ by the people you choose from the rest. Or A,C could be the first women and $B,D,E$ be the rest. Or $B,C$ be the first women and $A,D,E$ be the rest. 
In the first answer this group is correctly only counted once.