jhenezhubby01ff

2022-10-02

Prove that number of partitions of a $k$-set into $m$ parts of sizes, from certain set $S$ that is subset of $\left\{0,1,2,\dots \right\}$ is
${P}_{m}\left(k,S\right)=\sum _{\sum _{s\in S}{t}_{s}=m,\sum _{s\in S}s{t}_{s}=k}\prod _{s\in S-0}\frac{k!}{{t}_{s}!\left(s!{\right)}^{{t}_{s}}}$
and generating function considering parameter $k$ is
$\sum _{k=0}^{\mathrm{\infty }}{P}_{m}\left(k,S\right)\frac{{x}^{k}}{k!}=\sum _{{t}_{0}=0}^{m-1}\frac{1}{\left(m-{t}_{0}\right)!}{\left(\sum _{s\in S-0}\frac{{x}^{s}}{s!}\right)}^{m-{t}_{0}}$
Can someone find the generating function with two variables considering parameters $m$ and $k$?

graulhavav9

One wants to compute the two variables generating function
${F}_{S}\left(x,y\right)=\sum _{m=1}^{\mathrm{\infty }}{y}^{m}\sum _{k=0}^{\mathrm{\infty }}{P}_{m}\left(k,S\right)\frac{{x}^{k}}{k!}.$
Note that, according to the second displayed formula in your post,

Hence,
${F}_{S}\left(x,y\right)=\sum _{i=1}^{\mathrm{\infty }}\frac{1}{i!}{u}_{S}\left(x{\right)}^{i}\sum _{m=i}^{\mathrm{\infty }}{y}^{m}=\sum _{i=1}^{\mathrm{\infty }}\frac{1}{i!}{u}_{S}\left(x{\right)}^{i}\frac{{y}^{i}}{1-y},$
that is,
${F}_{S}\left(x,y\right)=\frac{\mathrm{exp}\left(y{u}_{S}\left(x\right)\right)-1}{1-y}.$

Do you have a similar question?