Ways to stack 65 different disks in 3 piles with constraints. 
How many ways are there to stack 65 different disks in 3 piles if pile 1 but have at least 15 disks and pile 3 must be non-empty.

Attempt:
1) Ways to arrange all the disks in a horizontal line: $65!$
2) Then I have to place separator characters that would mark where pile 1,2 and 3 start. 
Originally, I have $66 $ places to put these 2 separators, but I have to discount the first 15 as the first pile must have at least $15$ disks, and I have to discount the last one as the last pile can't be empty. So I end up with: $\binom {50}{2}$ ways to place these separators.
But now I'm lacking the case where pile 2 is empty. In this case everything is the same except I only have one separator: $65!\cdot \binom{50}1$.
So the answer would be
$$65! \left (\binom{50}2+\binom{50}1\right)$$
Is my reasoning correct?

What you’ve done is fine. Pascal’s identity says that your answer is equal to $65!\binom{51}2$, so you might wonder whether it’s possible to get that form directly. It is, but it takes a little cleverness. Throw away the first $15$ disks, as you did, leaving $50$. There are now $51$ places to put your two separators, so you can place them in $\binom{51}2$ different ways. After they’ve been placed, move the righthand separator one disk to the left, and you’ll have your three piles: the middle pile can now be empty, and the third pile is guaranteed to have at least one disk.
