Suppose that $n$ indistinguishable balls are to be arranged in N distinguishable boxes so that each distinguishable arrangement is equally likely. If n is greater than or equal to N, what is that probability that no box will be empty?
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The first question has $\left(\!{N\choose n}\!\right)$ arrangements. If we insist that each box is nonempty, we first put one ball in each, and there are $\left(\!{N\choose n-N}\!\right)$ arrangements of the remainder. The ratio is your desired answer.
Edit: This notation denotes counting multisets, i.e. $\left(\!{a\choose b}\!\right)={a+b-1\choose b}$

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