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* C^8_3`

The formula for a combination is given by

`C^n_k=(n!)/(k!(n-k)!)`

Then, we have

`\begin{gathered} C^7_3* C^8_3=(7!)/(3!(7-3)!)*(8!)/(3!(8-3)!) \\ =(7!)/(3!4!)*(8!)/(3!5!) \\ =(7\cdot6\cdot5)/(3!)*(8\cdot7\cdot6)/(3!) \\ =(7\cdot6\cdot5)/(6)*(8\cdot7\cdot6)/(6) \\ =7*5*8*7 \\ =1960 \end{gathered}`