We follow [28] here. We first change the way we look at the problem; we imagine now to have the stones in a line, and to assign to each of them a number between and - this number is the box the stone falls in. There are possible distributions, each with probability . We seek the probability of finding all cells occupied.

Let be the event that cell number is empty
(
). In this situation all stones are placed in the remaining
cells, and this can be done in different ways. Similarly,
there are arrangements, leaving two preassigned cells empty,
etc. Accordingly and with
:

We use now an important theorem proved in [28, pg 99]: