Question

A committee must be formed with 3 teachers and 7 students. If there are 6 teachers tochoose from, and 10 students, how many different ways could the committee bemade?

Answer 1

In the selection, the order doesn't matter, then you could use a combination. The formula is:

$C(n,r)=\frac{n!}{r!(n-r)!}$

Where n is the number of things to choose from and r is the amount we choose.

Then if there are 6 teaches and the committee must be formed with 3 of then, the combinations are:

$\begin{gathered} C(6,3)=\frac{6!}{3!(6-3)!} \\ C(6,3)=\frac{6!}{3!(3)!} \\ C(6,3)=20 \end{gathered}$

And also there are 10 students and the committee must be formed by 7 of them, then the combinations are:

$\begin{gathered} C(10,7)=\frac{10!}{7!(10-7)!} \\ C(10,7)=\frac{10!}{7!(3)!} \\ C(10,7)=120 \end{gathered}$

And finally, the total number of ways the committee could be made is:

$C(6,3)\times C(10,7)=20\times120=2400$

Answer: 2400 ways