Combinatorics thinking I saw this question in a book I've been reading: in a group of four mathematicians and five physicians, how many groups of four people can be created if at least two people are mathematicians?
The solution is obtained by ${4 \choose 2}{5 \choose 2} + {4 \choose 3}{5 \choose 1} + {4 \choose 4}{5 \choose 0} = 81$. But I thought of the present (wrong) solution:
Step 1. Choose two mathematicians. It gives ${4 \choose 2}$ different ways of choosing.
Step 2. Choose two people from the seven people left. It gives ${7 \choose 2}$ ways of choosing.
Step 3. Multiply. ${4 \choose 2}{7 \choose 2} = 126 = {9 \choose 4}$. It is equivalent of choosing four people in one step. Clearly wrong.
I really don't know what's wrong in my "solution". It seems like I am counting some cases twice, but I wasn't able to find the error. What am I missing here?
PS. It is not homework.

You have counted the number of ways to chose two mathematicians leading the group, plus two regular members which can be mathematicians or physicians.
