Counting problem - fault in my reasoning. The problem is as follows:
The dean of science wants to select a committee consisting of mathematicians and physicists to discuss a new curriculum. There are $15$ mathematicians and $20$ physicists at the faculty; how many possible committees of $8$ members are there, if there must be more mathematicians than physicists (but at least one physicist) on the committee?
The given solution splits up the choices into the cases where there is $1$ physicist, $2$ physicists, and $3$ physicists. Then it's simply a matter of summing up $\binom{15}{5}\cdot\binom{20}{3}$ and  $\binom{15}{6}\cdot\binom{20}{2}$ and $\binom{15}{7}\cdot\binom{20}{1}$, giving $4503070$. 
This is what I tried to do:
first I pick a physicist, making sure I have at least one on the team. I have $20$ choices for this. Then I pick $5$ mathematicians, making sure there are more mathematicians than physicists on the team. I have $15\cdot14\cdot13\cdot12\cdot11$ choices for this. Then I pick two people from the $29$ people that are left because it doesn't matter whether they are mathematicians or physicists. I have $29\cdot28$ choices for this. After I multiply all those numbers, I need to divide by the number of ways I can permutate them, because order doesn't matter. I get $$\dfrac{20\cdot15\cdot14\cdot13\cdot12\cdot11\cdot29\cdot28}{8!} = 145145,$$ which is obviously not the right solution. Where did I go wrong?
Thanks in advance.

You don't have $8!$ possible permutations. For instance, the first person you chose can never be a mathematician and the second, third, fourth, fifth and sixth persons you chose can never be physicists.
