Simplification of a Finite Sum For $m, n \in \mathbb{N}$, can  $$\sum_{k=0}^{m} (m-k)! { m \choose m-k } { n \choose n - (m-k) } ( 1 - p)^{k} \cdot p^{m-k} \cdot \frac{ e^{-\lambda} \lambda^{n-(m-k)} } { (n - (m-k) )! }  $$ be simplified?
This is related to a previous question I asked regarding the number of $r$-permutations of a multiset $ S = \{\!\!\{ m \cdot 0, 1, ..., n \}\!\!\}$  [1]. Here, each permutation is weighted by a product of a Bernoulli and a Poisson distribution - I am trying to work out the normalising constant for a joint distribution.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
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\begin{align}
&\sum_{k = 0}^{m}\pars{m - k}!{m \choose m - k}{n \choose n - \bracks{m - k}}
\pars{1 - p}^{k}\, p^{m - k}\,{\expo{-\lambda}\lambda^{n - \pars{m -k}}
\over  \bracks{n - \pars{m - k}}!}
\\[5mm] = &\
\sum_{k = 0}^{m}k!{m \choose k}{n \choose n - k}
\pars{1 - p}^{m - k}\, p^{k}\,{\expo{-\lambda}\lambda^{n - k}
\over  \pars{n - k}!}
\\[5mm] = &\
\pars{1 - p}^{m}\expo{-\lambda}\lambda^{n}
\sum_{k = 0}^{m}{k! \over \pars{n - k}!}{m \choose k}{n \choose n - k}
\bracks{p \over \pars{1 - p}\lambda}^{k}
\\[5mm] = &\
\pars{1 - p}^{m}\expo{-\lambda}\lambda^{n} 
\sum_{k = 0}^{m}{m!\, n! \over \pars{n - k}!\pars{m - k}!\pars{n - k}!}
{\bracks{p/\pars{1 - p}\lambda}^{\, k} \over k!}
\\[5mm] = &\
\pars{1 - p}^{m}\expo{-\lambda}\lambda^{n} 
\sum_{k = 0}^{m}{\bracks{\prod_{i = 1}^{k}\pars{m - i + 1}}
\bracks{\prod_{j = 1}^{k}\pars{n - j + 1}} \over
\prod_{\ell = 1}^{k}\pars{n - \ell}}
{\bracks{p/\pars{1 - p}\lambda}^{\, k} \over k!}
\\[5mm] = &\
\pars{1 - p}^{m}\expo{-\lambda}\lambda^{n} 
\sum_{k = 0}^{m}{\bracks{\pars{-1}^{k}\prod_{i = 1}^{k}\pars{i - m - 1}}
\bracks{\pars{-1}^{k}\prod_{j = 1}^{k}\pars{j - n - 1}} \over
\pars{-1}^{k}\prod_{\ell = 1}^{k}\pars{\ell - n}}
{\bracks{p/\pars{1 - p}\lambda}^{\, k} \over k!}
\\[5mm] = &\
\pars{1 - p}^{m}\expo{-\lambda}\lambda^{n} 
\sum_{k = 0}^{m}{\pars{-m}_{k}\pars{-n}_{k} \over \pars{1 - n}_{k}}
{\bracks{-p/\pars{1 - p}\lambda}^{\, k} \over k!}
\\[5mm] = &\
\bbx{\pars{1 - p}^{m}\expo{-\lambda}\lambda^{n}\
\mbox{}_{2}\mrm{F}_{1}\pars{-m,-n;1 - n;-\,{p \over \pars{1 - p}\lambda}}}
\end{align}

See
  The Hypergeometric Series.

A: Not sure that there's a "nice/compact" form available.  Using Mathematica one obtains the following:
Sum[(m - k)! Binomial[m, m - k] Binomial[n, n - m + k] (1 - p)^k p^(m - k)
  Exp[-\[Lambda]] \[Lambda]^(n - m + k)/(n - m + k)!, {k, 0, m}]

$$\frac{e^{-\lambda } n! p^m \lambda ^{n-m} \, _1F_2\left(-m;-m+n+1,-m+n+1;\frac{(p-1) \lambda }{p}\right)}{((n-m)!)^2}$$
where $\, _1F_2$ is the generalized hypergeometric function (1F2).
