${p}_{m}(r+1,n)={p}_{m}(r,n)\frac{n\u0101\x88\x92m}{n}+{p}_{m+1}(r,n)\frac{m+1}{n}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}(1)$

* The results which I can't use must be

${E}_{m}(r,n)={\textstyle (}\genfrac{}{}{0ex}{}{n}{m}{\textstyle )}A(r,n\u0101\x88\x92m)={\textstyle (}\genfrac{}{}{0ex}{}{n}{m}{\textstyle )}\underset{\mathrm{\u012a\xbd}=0}{\overset{n\u0101\x88\x92m}{\u0101\x88\x91}}(\u0101\x88\x921{)}^{\mathrm{\u012a\xbd}}{\textstyle (}\genfrac{}{}{0ex}{}{n\u0101\x88\x92m}{\mathrm{\u012a\xbd}}{\textstyle )}(n\u0101\x88\x92m\u0101\x88\x92\mathrm{\u012a\xbd}{)}^{r}$

and by association

$A(r,n+1)=\underset{k=1}{\overset{r}{\u0101\x88\x91}}{\textstyle (}\genfrac{}{}{0ex}{}{r}{k}{\textstyle )}A(r\u0101\x88\x92k,n)$

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.