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Does there exist a nice closed form formula for the sum $$\sum_{k=0}^m P(m,k)x^k$$ where $P(m,k)=C(m,k)*k!$, $C(m,k)$ being the "m choose k" number. Formula given by Maple 11 is complicated. I thought there may be an easier one. Of course $$\sum_{k=0}^m C(m,k)x^k=(1+x)^m.$$

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up vote 2 down vote accepted

We can write \begin{align} \sum_{k=0}^m\binom mkk!x^k&=\sum_{k=0}^m\frac{m!}{(m-k)!}x^k\\ &=m!\sum_{j=0}^m\frac{x^{m-j}}{j!}\\ &=m!x^mS_m\left(\frac 1x\right), \end{align} where $S_n(t):=\sum_{j=0}^n\frac{t^j}{j!}$. But the $S_n$ are hard to simplify.

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Your variables in your definition of $S_n$ seem to be mixed up... – Ben Millwood May 19 '12 at 17:46
We could use the integral representation $$ S_m(t) = \frac{e^t}{m!} \int_t^\infty u^m e^{-u}\,du = \frac{e^t}{m!}\Gamma(m+1,t) $$ to get $$ \sum_{k=0}^m\binom mkk!x^k = e^{1/x} x^m \Gamma(m+1,1/x), $$ where $\Gamma(s,t)$ is the incomplete gamma function. – Antonio Vargas May 19 '12 at 20:14

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