You have $N$ boxes and $M$ balls. The $M$ balls are randomly distributed into the $N$ boxes. What is the expected number of empty boxes?


I came up with this formula:


$\sum\_{i=0}^{N}i\binom{N}{i}\left(\frac{N-i}{N}\right)^{M}$


This seems to yield the right answer. However, it requires calculating large numbers, such as $\binom{N}{\frac{N}{2}}$. Is there a more direct way, perhaps using a probability distribution? It seems that neither the binomial nor the hypergeometric distributions fit the problem.



Probability Question
Let $X\_i$ be a random variable with value 1 when box $i$ is empty and 0 otherwise. Now
$P(X\_i=1)=(1-\frac{1}{N})^{M}$. And the expected number of empty boxes is just $\mathbb{E}(\sum X\_i)=N\mathbb{E}(X\_1)\approx \frac{N}{e^M}$


EDIT: gave the answer in terms of M,N instead of the numerical values given originally...


