# 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.

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If your group is all four mathematicians, you will count it six times. Each unique pair could be the two you pick first and the other two will be the two you pick second. Similarly, all groups of three mathematicians and one physicist will be counted three times. Since there are 20 groups of 3+1, your overcount is $20*2+5=45=126-81$