Question

A committee must be formed with 4 teachers and 5 students. If thereare 10 teachers to choose from, and 8 students, how many differentways could the committee be made?

Answer 1

We have to choose 4 teachers from 10 teachers.

So, the number of ways of doing this is

$10C_4=\frac{10!}{4!(10-4)!}=\frac{10!}{4!\times6!}=\frac{10\times9\times8\times7\times6!}{4\times3\times2\times6!}=10\times3\times7=210$

We have to choose 5 students from 8 students.

So, the number of ways of doing this is

$8C_5=\frac{8!}{5!(8-5)!}=\frac{8!}{5!\times3!}=\frac{8\times7\times6\times5!}{5!\times3\times2}=8\times7=56$

Now, we have to choose both teachers and students.

So the number of ways is

$10C_4\times8C_5=210\times56=11760$

The different number of ways the committee made is 11760 ways.