Simplification of a Finite Sum

Question

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.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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.

Answer 2

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).