What is the polynomial asimptotic bound of $ (a_n)_{n=10}^\infty$, where $a_n$ represents the number of ways to distribute $10$ identical balls into $n$ non-identical boxes, such that there are no two nonempty boxes with the same number of balls?
There are, of course, ${{10+n-1}\choose {n-1}} = \Theta (n^{10})$ ways to distribute $10$ identical balls into $n$ labeled boxes. I guess that $a_n=\Theta(n^4)$, since when $n$ is large, almost all boxes remain empty; so the "closest to empty" configuration is the one in which there exist four boxes with $1,2,3$ and $4$ balls respectively. Showing that this is a lower bound is easy, but I am not sure how to show that it is an upper bound (if it is indeed the correct answer).
The answer can be written as a sum over the partitions of $10$ into distinct positive integers; these can be counted by hand and they are given by
$$4 + 3 + 2 + 1, 5 + 3 + 2, 5 + 4 + 1, 6 + 3 + 1, 6 + 4, 7 + 2 + 1, 7 + 3, 8 + 2, 9 + 1, 10$$
so there are $10$ of them (and this can be verified on the OEIS). Given such a partition we can consider the number of ways to distribute its parts among $n$ labeled boxes and of course the answer is ${n \choose k}$ where $k$ is the number of parts. So the exact answer is
$$a_n = {n \choose 4} + 4 {n \choose 3} + 4 {n \choose 2} + {n \choose 1}$$
with asymptotic $a_n \sim \frac{n^4}{4!}$ as expected.