You have 40 $N$ boxes and 80 $M$ balls. The 80 $M$ balls are randomly distributed into the 40 $N$ boxes. What is the expected number of empty boxes?
I came up with this formula:
$\sum_{i=0}^{40}i\binom{40}{i}\left(\frac{40-i}{40}\right)^{80}$\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 $\binom{40}{20}$. \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.
