
Simple problem...
... I'm having trouble with it anyway, even though I think it might just be high school review, so I wasn't entirely sure where to put this topic. Anyway:
There are 15 items. Four are type A, five are type B, and six are type C. Selecting 3 randomly, what is the chance that exactly 2 will be type C?
There is no item replacement and it's unordered.
Since there are more questions using the same sort of thing later I was hoping to get a formula/method I could use with larger numbers instead of tracing through a diagram and multiplying every time.

Looking at the categories:
A: 4/15
B: 5/15
C: 6/15
Now, looking at these possible outcome combinations: C,C,N ; C,N,C ; or N,C,C. (N denoting "not C" in this case.)
This would be a simple binominal IF it were done with replacement:
$\displaystyle \binom{n}{k} p^kq^{nk}$
But since there is no replacement, each succeeding factor is conditional to the previous factor (i.e. you simply remove what you take out at each step):
$\displaystyle Pr(2 C's  3 Draws) = \binom{3}{2} (6/15)(5/14)(9/13)$
Where 9/13 is the "not C" part, i.e. 4/13 + 5/13