We throwing $m$ balls to $n$ cells.... We throwing $m$ balls to $n$ cells randomly...
At each cell can be more then one ball, or (of course) it can still empty.
What is the expectation of the empty cells?
I'd like to get any help!
Thank you!  
 A: The number of balls per cell has a multinomial distribution. But for any one cell, it is binomial. The probability a ball lands in that one cell is $1/n$ and outside it is $(1-1/n). $ The probability that that particular cell remains empty after all balls are thrown is $(1-1/n)^m. $ Now define the Indicator variable $I_k=1$ if cell number $k$ remains empty and $=0$ if cell $k$ is not empty. Then the expected number of empty cells is
$$E\sum_{k=1}^n I_k = \sum_{k=1}^nP(\text{cell k is empty})= n(1-1/n)^m   $$
A: Using stars and bars there are $\binom{m+n-1}{n-1}$ ways to distribute the balls allowing for empty cells.
Again by stars and bars there are $\binom{n}{i}\binom{m+n-i-1}{n-i-1}$ ways to leave exactly $i$ cells empty. Since there are $\binom{n}{i}$ ways to select the empty cells and $\binom{m+n-i-1}{n-i-1}$ ways to divide $m$ balls in $n-i$ cells.
Thus the expected number of empty cells is:
$$\sum_{i=1}^{n-1}\dfrac{\binom{n}{i}\binom{m+n-i-1}{n-i-1}i}{\binom{m+n-1}{n-1}}$$

Two years later I realize this is wrong, as for example $(1,1)$ does not have the same probability as $(2,0)$. The answer by Mr.Spot is correct, and more easily computable.
A: I see that this problem has been answered, but I thought I'd share my approach anyway.
I find it's most helpful to look at this problem from the perspective of a cell. "If I am a cell, what is the probability that I will remain empty after all $m$ balls are thrown?" Well, each time a ball is thrown, there is a $\frac{n-1}{n}$ chance that it will not land in me. So the probability that I remain empty after all the balls are thrown is $\left(\frac{n-1}{n}\right)^m$, since this event must occur $m$ times. So now we have a formula for the probability that an individual cell will be empty after all of the balls are thrown; to get the expected number of empty cells, we simply multiply this probability by the number of cells, i.e. $n\left(\frac{n-1}{n}\right)^m$.
