# 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.

• Wow, thanks a lot man! This explanation really helped. The small visualization helped me see exactly what's going on, thanks again! – ApprenticeOfMathematics Oct 22 '16 at 13:27

This is a classic error of overcounting, where identical selections are counted multiple times under different guises.

For this problem, suppose the teachers are named Abigail, Betty, Catherine and Donovan and one of the students is named Stanley. Then your incorrect method, which first selects three teachers and then two of the remaining people, counts the following identical selections as distinct: $$\{\text{A, B, D}\mid\text{C, S}\}$$ $$\{\text{A, C, D}\mid\text{B, S}\}$$ In particular, this incorrect method counts the three-teacher selections correctly ($\binom43\binom72=84$), but counts the four-teacher selections four times over ($\binom43\cdot1\cdot7=28$). By dividing the latter count by four, the correct result is obtained: $$84+\frac{28}4=91.$$