Question

A committee must be formed with 3 teachers and 3 students. If there are 7 teachers tochoose from, and 8 students, how many different ways could the committee be made?

Answer 1

We have to choose 3 teachers from a group of 7 teachers, then, the amount of combinations is given by

$C^7_3$

We have to choose 3 students from a group of 8 students, then, the amount of combinations is given by

$C^8_3$

Since we have to choose a committee simultaneously formed by 3 teachers and 3 students, we want the intersection of those combinations, which is given by their product

$C^7_3\times C^8_3$

The formula for a combination is given by

$C^n_k=\frac{n!}{k!(n-k)!}$

Then, we have

$\begin{gathered} C^7_3\times C^8_3=\frac{7!}{3!(7-3)!}\times\frac{8!}{3!(8-3)!} \\ =\frac{7!}{3!4!}\times\frac{8!}{3!5!} \\ =\frac{7\cdot6\cdot5}{3!}\times\frac{8\cdot7\cdot6}{3!} \\ =\frac{7\cdot6\cdot5}{6}\times\frac{8\cdot7\cdot6}{6} \\ =7\times5\times8\times7 \\ =1960 \end{gathered}$