A question is already asked here 9 years ago. But unfortunately, it is not answered (nor confirmed as a typo). The question is from the classic texbook "An introduction to probability theory" by William Feller. The question want us to prove that
$x_m(r,n)$, the probability of finding $m$ or more cells empty is
$$ \binom{n}{m}\sum_{\nu=0}^{n-m}(-1)^{\nu}\binom{n-m}{\nu}\left(1-\frac{m+\nu}{n}\right)^{r}\frac{m}{m+v} $$
Any help ?
The general inclusion/exclusion formula for the probability that $m$ or more of the events $A_1, \dots , A_n$ occur simultaneously is $$P_m = \sum_{\nu = 0}^{n-m} (-1)^{\nu} \binom{m+\nu-1}{m-1} S_{m+\nu}$$ where $S_1 = \sum_i P(A_i)$, $S_2 = \sum_{ij} P(A_i A_j)$, $S_3 = \sum_{ijk} P(A_i A_j A_k)$ etc.
(Reference: Equation 5.2 in section IV.5 of An Introduction to Probability Theory and Its Applications, Third Edition by William Feller)
In the case of $r$ balls in $n$ cells with $A_i$ being the event that cell $i$ is empty, we have $$S_{\nu} = \binom{n}{\nu} \left( 1 - \frac{\nu}{n} \right)^r$$ for $0 \le \nu \le n$, so $$x_m(r,n) = \sum_{\nu = 0}^{n-m} (-1)^{\nu} \binom{m+\nu-1}{m-1} \binom{n}{m+\nu} \left( 1 - \frac{m+\nu}{n} \right)^r$$ Now apply the identity $$\binom{m+\nu-1}{m-1} \binom{n}{m+\nu} = \binom{n}{m} \binom{n-m}{\nu} \frac{m}{m+\nu}$$ and we have $$x_m(r,n) = \binom{n}{m} \sum_{\nu=0}^{n-m} (-1)^{\nu} \binom{n-m}{\nu} \left( 1 - \frac{m+\nu}{n} \right)^r \frac{m}{m+\nu}$$
