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I can prove that number of partitions of a $k$-set into $m$ parts of sizes, from certain set $S$ that is subset of $\{0,1,2,\ldots\}$ is $$P_{m}(k,S)=\sum_{\sum_{s\in S}t_s=m,\sum_{s\in S}st_s=k}\prod_{s\in S-{0}}\frac{k!}{t_s!(s!)^{t_s}} $$and generating function considering parameter $k$ is $$\sum_{k=0}^{\infty}P_{m}(k,S)\frac{x^k}{k!}=\sum_{t_0=0}^{m-1}\frac{1}{(m-t_0)!}\left(\sum_{s\in S-{0}}\frac{x^s}{s!}\right)^{m-t_0}$$ Can someone find the generating function with two variables considering parameters $m$ and $k$?

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One wants to compute the two variables generating function $$ F_S(x,y)=\sum_{m=1}^{\infty}y^m\sum_{k=0}^{\infty}P_{m}(k,S)\frac{x^k}{k!}. $$ Note that, according to the second displayed formula in your post, $$ \sum_{k=0}^{\infty}P_{m}(k,S)\frac{x^k}{k!}=\sum_{i=1}^{m}\frac{1}{i!}u_S(x)^{i},\quad\mbox{with}\ u_S(x)=\sum_{s\in S-{0}}\frac{x^s}{s!}. $$ Hence, $$ F_S(x,y)=\sum_{i=1}^{\infty}\frac{1}{i!}u_S(x)^{i}\sum_{m=i}^{\infty}y^m=\sum_{i=1}^{\infty}\frac{1}{i!}u_S(x)^{i}\frac{y^i}{1-y}, $$ that is, $$ F_S(x,y)=\frac{\exp(yu_S(x))-1}{1-y}. $$

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