Question

How many different committees can be formed from 8 teachers and 31 students if the committee consists of 3 teachers and 3 ​students?

Answer 1

I'm out at the moment, so I'll explain asap.
There are 8C3 * 31C3 ways in selecting 3 teachers and 3 students.

Since order doesn't matter, we use combinatoric, nCr, instead of permutation, nPr.

Now, we merely need to select 3 teachers from 8 teachers (objects) and 31 students (objects).

Since, we want them simultaneously, we need to multiply the two together.

In essence, there are 8C3 × 31C3 ways

Answer 2

There are many students, however, because there has to be three teachers in each committee, there can only be 2 full committees.
This is because 8 divided by three equals 2 committees, with 2 left over. Even though you have some left over, you cannot create another full committee. Hope that this helped you!