Hakob wrote:

We should consider three scenarios:

1 prof - 7*2C10

2 prof - 2C7*1C10

3 fprof - 3C7

Add them all=560

what do you think?

YES. I took this one from my Permutations/Combinations guide available from this site. I think i will add many of hte problems you published here guys. Thanks.... Here is my formal explanation from back in the days.

Scenario 1. We get C(17, 3) = = 680 total possible committees. Now, the number of teams with students only is, using the same formula C(10,3) = = 120.

Now, 680-120=560.

Scenario 2. For a committee with 1 professor and 2 students, we will get 7xC(10,2) = = 45x7=315. (we multiply by 7 because For 2 professors and 1 student, we will get C(7, 2)x10 = 21x10=210, and for 3 professors, we will get C(7, 3)=35. Adding up the combinations we will get: 315+210+35=560.

Princeton Review suggests using Scenario 2 because it is supposedly simpler to understand, however, taking into consideration that you can make a mistake in the endless calculations and that it requires 3 complex operations in contrast to 2 in the first case, it has a weaker standing. In any case, both ways get the correct answer, but you need to choose the one that appeals more to youтАФthe one easier to use.