Another problem on combinatorics. This time I'm asking for a hint and if it is possible a general strategy when dealing with this kind of problems.
Show without using the preceding results * that the probability $p_{m}(r,n)=n^{-1}E_{m}(r,n)$ of finding exactly $m$ cells empty satisfies
$$p_{m}(r+1,n)=p_{m}(r,n)\frac{n-m}{n}+p_{m+1}(r,n)\frac{m+1}{n} \qquad \qquad (1)$$
* The results which I can't use must be
$$E_{m}(r,n)=\binom{n}{m}A(r,n-m)=\binom{n}{m}\sum_{\nu=0}^{n-m}(-1)^{\nu}\binom{n-m}{\nu}(n-m-\nu)^{r}$$ and by association $$A(r,n+1)=\sum_{k=1}^{r}\binom{r}{k}A(r-k,n)$$
By the way, $A(r,n)$ is equal to $n!S(r,n)$ where $S(r,n)$ are the Stirling numbers of the second kind.
I'm not sure how to proceed since the 'preceding result' I can't use is basically the definition for $E_{m}(r,n)$. Even if I use such definition in the left hand side of equation $(1)$, I end up with the first term $p_{m}(r,n)\frac{n-m}{m}$ and a nasty second term $\binom{n}{m}\sum_{s=0}^{n-m}(-1)^{s}\binom{n-m}{s}(1-\frac{m+s}{n})^{r}(-\frac{s}{n})$which doesn't resemble the equation $(1)$.