Wrong solution in commission problem. In one congress there are 15 physics and 15 math teachers. How many committees of 8 teachers can be formed with at least 4 math teachers and at least 2 physics teachers?
I know how to solve this problem. However. I can't explain why the following solution is incorrect
1) 
Commission with 6 mathematicians and 2 physicists:
possibilities to select 4 mathematicians * possibilities to select 2 physicists * possibilities to select the 2 remaining mathematicians from the 11 left: 
${14 \choose 4}  {15 \choose  2} {11 \choose 2}$
2) 
Commission with 4 mathematicians and 4 physicists:
possibilities to select 4 mathematicians * possibilities to select 2 physicists * possibilities to select the 2 remaining physicists from the 13 left: 
${15 \choose 4}  {15 \choose  2} {13 \choose 2}$
3)
Commission with 5 mathematicians and 3 physicists:
possibilities to select 4 mathematicians * possibilities to select 2 physicists * possibilities to select the 1 remaining mathematicians from the 11 left * possibilities to select the 1 remaining physicists from the 13 left: 
${15 \choose 4}  {15 \choose  2} {11 \choose 1} {13 \choose 1}$
The answer would then be 1) + 2) + 3).
Why is this solution wrong?

You're double counting. For instance, suppose we label the math teachers $M_1$ through $M_{15}$. In the 6 mathematicians case, you're treating choosing $M_1$ as the "4" as different from choosing $M_1$ as one of the "2 remaining mathematicians from the 11 left". But there's no difference between those two cases. For the 6 mathematicians case, you should just have ${15 \choose 6} {15 \choose  2}$.
Consider a simpler case. How many ways are there to choose three people with from 2 math teachers and 2 physics teachers, if you need at least one of each? For the case where we end up with two math teachers, the logic you present would say "there are ${2 \choose 1}=2$ ways to pick which math teacher we're using to satisfy the 'at least one math teacher' requirement, and ${1 \choose 1}=1$ ways to choose a math teacher from the remaining 1 math teacher". But there's no difference between choosing one math teacher versus the other to be the "at least one".
