# 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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## 2 Answers

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 $20*2+5=45=126-81$

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Sorry, but I couldn't see why I'm counting the group of four mathematicians six times. Could you clarify a little more, please? –  Ian Mateus Jul 2 '12 at 13:55
@IanMateus: Let them be a,b,c, and d. In step 1, you could choose ab and then in step 2 you could choose cd. Or in step 1 you could choose ac and in step 2 choose bd. Or .... Your approach counts these as different cases. The official solution avoids this by making sure no case is counted in different ways. That is the point of picking the number of mathematicians first, and then picking them separately –  Ross Millikan Jul 2 '12 at 13:58
Everything makes sense now :D many thanks! –  Ian Mateus Jul 2 '12 at 14:00
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You have counted the number of ways to chose two mathematicians leading the group, plus two regular members which can be mathematicians or physicians.

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