Forming a committee of 5 with at least 3 teachers I am working on the following problem 

In how many ways can a committee of 5 be formed from a group of 11 people consisting of 4 teachers and 7 students if the committee must include at least 3 teachers

So I know that the answer is ${4 \choose 3}{7 \choose 2} + {4 \choose 4}{7 \choose 1} = 91$ and I can follow the reasoning behind this answer, but I still do not understand why my original approach to solving the problem is incorrect. 
My thought was to first choose 3 teachers and then add the extra teacher left to the group of students and pick two. In other words
$${4 \choose 3}{8 \choose 2}$$
I thought that both cases of groups with 3 and 4 teachers would be covered by including the teacher in the group of students when choosing two more people to complete the committee, but I seem to overestimate the correct answer (ie the above is equal to 112), but I don't understand why.
Question: What is wrong with my approach to this problem?

Your approach counts some possibilities twice.
If you have a committee of $4$ teachers and $1$ student, it will be counted $4$ times in your formula, because you are differentiating between the possible ways of selecting $3$ teachers and then another one. For example, for a commitee consisting of $4$ professors $A,B,C,D$ and a student $E$ you are counting


*

*$\{A,B,C\}+\{D,E\}$

*$\{A,B,D\}+\{C,E\}$

*$\{A,C,D\}+\{B,E\}$

*$\{B,C,D\}+\{A,E\}$


as four different options.
Notice by the way that you have $21$ too much, which is exactly $3$ times the seven ways to choose only $1$ student and $4$ teachers.
