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

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!

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

c is the correct one. i think.