Histogram of duplication in n choose k Imagine having 17 balls to distribute to 4 people. One algorithm for distributing these balls is to give each ball to one out of the four randomly. This means, in an extreme case, it is possible for 1 person to receive all 17 balls, while the other 3 get none. More likely, however, would be some fairly even distribution, with every person getting at least one ball.
This means it is possible to assign probabilities to the cases in which 0, 1, 2 or 3 people gets no balls.
In general, you have n to distribute among k and probabilities can be given to all k cases in which one or several people get no balls. The number n need not be greater than k.
How does one compute this generally?

The stars and bars technique can be used here. There are $\binom{n+k-1}{k-1}$ ways to distribute the balls. Suppose that $1\le m\le k$. There are $\binom{k}m$ ways to choose a subset of the $k$ to receive balls, and there are $\binom{n-1}{m-1}$ ways to distribute the balls to that subset in such a way that everyone receives at least one ball. Thus, the probability that exactly $m$ get balls is
$$\frac{\binom{k}m\binom{n-1}{k-1}}{\binom{n+k-1}{k-1}}\;.$$
