After a little thought, I figured it would be:
C(15,0) + C(15,1) + C(15,2) + C(15,3) + ... + C(15,15)
Granted, some of these would be unlikely, e.g., C(15,0), but this expression would at least figure the upper bound. Then I was told that the total could be found with 2 to the 15th power. I never knew that. If true, that is cool. I don't know if that is true, because I have not worked out the proof. I assume I would not be smart enough to do so.
I did try out a few cases to see if it worked for them.
c(2,0) + c(2,1) + c(2,2) = 1 + 2 + 1 = 4. This is equal to 2^2.
Also c(3,0) + c(3,1) + c(3,2) + c(3,3) = 1 + 3 + 3 + 1 = 8. This also happens to be 2^3.
It's looking pretty good. As I look at this, I'm seeing Pascal's Triangle. So there is another thing I wasn't previously aware of. I'm learning.
Anyway, most combinations of ways to group those 15 people was 2^15 = 32,768.
