From a group of 4 professors and 6 students, a supervisory committee

From a group of 4 professors and 6 students, a supervisory committee of 3 members is to be assembled. If the committee must include at least one professor, how many ways can the committee be formed?

A. 36
B. 60
C. 72
D. 80
E. 100


{The number of committees with at least one professor} =

= {The total number of possible committees} minus {The number of committees with no professors} (i.e., committees made up solely of students).

Using this approach, calculate \(C^3_{10}\) (the total ways to select 3 out of 10) and subtract \(C^3_6\) (the ways to choose 3 individuals exclusively from the 6 students, hence no professors):

\(C^3_{10}-C^3_6=\frac{10!}{7!*3!}-\frac{6!}{3!*3!}=120-20=100\).

Alternatively, we can use a direct method:

{The number of committees with at least one professor} =

= {The committees with 1 professor and 2 students} plus {The committees with 2 professors and 1 student} plus {The committees with 3 professors and no students}:

This gives: \(C^1_4*C^2_6+C^2_4*C^1_6+C^3_4=100\).

Answer: E.

But you already know this. Your question is different.

And, I think I understand what you mean: when saying that "288 and 100 are very different numbers," you may imply that we cannot divide 288 by any factorial to compensate for duplications, as we almost always do when the order doesn't matter and when we need to get rid of the same selections.

But not this time: because here we may have THREE DIFFERENT cases (1p2s; 2p1s or 3p0s), with a different factorial correction in each. For example, the case when we have 1 professor and 2 students needs a different factorial correction than the case when we have 2 professors and 1 student. Hence, you cannot divide 288 by one factorial (say 2! or 3!) to fix it. So, 288 has duplications, but it cannot be compensated just by dividing it by any factorial.

Hope it's clear.