I have $N>0$ coins that are thrown into $M>0$ boxes. Each box has the same chance to receive each coin. (i.e. the probability of a coin to land in one particular box is $1/M$)
How can I calculate the probability $p$, that each box contains at least one coin?
With $M > N$, $p = 0$
With $M = N$ $p$ would be $M!/M^N$
But with $N > M$ ?
Please give me some clues to solve this problem! Is it easier to just find $N = f(M)$ in order to have $p>90\%$?