Is there a closed form for a sum $nPk +(n-1)Pk + (n-2)Pk + ... + kPk$? I would like to know if there is some closed form to solve for a sum in the form:
$nPk +(n-1)Pk + (n-2)Pk + ... + kPk$
For instance, if $n=7$ and $k=2$:
$7P2 + 6P2 + 5P2 + ... + 2P2$ = $\frac{7!}{5!} + \frac{6!}{4!} + \frac{5!}{3!}+ ...+ \frac{2!}{0!}$
 A: The sum you want is$$\sum_{i=0}^{n-k}\frac{(n-i)!}{(n-i-k)!}=k!\sum_{i=0}^{n-k}\frac{(n-i)!}{(n-i-k)!k!}=k!\sum_{i=0}^{n-k}\binom{n-i}{k}.$$
Here, note that $\sum_{i=0}^{n-k}\binom{n-i}{k}$ is the coefficient of $x^k$ in
$$(1+x)^n+(1+x)^{n-1}+\cdots+(1+x)^k$$
$$=(1+x)^k\cdot\frac{(1+x)^{n-k+1}-1}{x}=\frac{(1+x)^{n+1}}{x}-\frac{(1+x)^k}{x}.$$
Since there is no term $x^k$ in $\frac{(1+x)^k}{x}$, $\sum_{i=0}^{n-k}\binom{n-i}{k}$ is the coefficient of $x^{k+1}$ in $(1+x)^{n+1}$, which is $\binom{n+1}{k+1}$.
Hence, your sum is 
$$k!\binom{n+1}{k+1}=\frac{(n+1)\cdot n\cdot (n-1)\cdots (n-k+1)}{k+1}.$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
&\color{#66f}{\large%
\sum_{\ell\ =\ 0}^{n - k}{\pars{n - \ell}! \over \pars{n - \ell -k}!}}
=k!\sum_{\ell\ =\ 0}^{n - k}{\pars{n - \ell}! \over k!\pars{n - \ell -k}!}
=k!\sum_{\ell\ =\ 0}^{n - k}{n - \ell\choose k}
\end{align}

Note that
  $\ds{\sum_{\ell\ =\ 0}^{n - k}{n - \ell\choose k}
     =\sum_{\ell\ =\ 0}^{\color{#c00000}{\LARGE\infty}}{n - \ell\choose k}}$
  since $\ds{{n - \ell\choose k} = 0}$ when $\ds{\ell > n - k}$. Then, 

with $\ds{a > 2}$:
\begin{align}
&\color{#66f}{\large%
\sum_{\ell\ =\ 0}^{n - k}{\pars{n - \ell}! \over \pars{n - \ell -k}!}}
=k!\sum_{\ell\ =\ 0}^{\infty}\oint_{\verts{z}\ =\ a}
{\pars{1 + z}^{n - \ell} \over z^{k + 1}}\,{\dd z \over 2\pi\ic}
\\[5mm]&=k!\oint_{\verts{z}\ =\ a}{\pars{1 + z}^{n} \over z^{k + 1}}
\sum_{\ell\ =\ 0}^{\infty}\pars{1 \over 1 + z}^{\ell}\,{\dd z \over 2\pi\ic}
=k!\oint_{\verts{z}\ =\ a}{\pars{1 + z}^{n} \over z^{k + 1}}
{1 \over 1 - 1/\pars{1 + z}}\,{\dd z \over 2\pi\ic}
\\[5mm]&=k!\oint_{\verts{z}\ =\ a}{\pars{1 + z}^{n + 1} \over z^{k + 2}}
\,{\dd z \over 2\pi\ic}
=\color{#66f}{\Large k!\ {n + 1 \choose k + 1}}
\end{align}
A: Here is a combinatorial proof of the identity
$$ (k+1) \sum_{t=k}^n t^{\underline{k}} = (n+1)^\underline{k+1}. $$
The left-hand side counts the number of possibilities of choosing a number $s \leq n+1$ and $k$ distinct numbers less than $s$, and in addition a position $p \in \{1,\ldots,k+1\}$. If you put $s$ at position $p$ and the $k$ distinct numbers in the other position, you get $k+1$ distinct numbers at most $n+1$, and moreover this mapping is bijective.
Another simple proof uses induction. When $k = n$, the identity reads $(k+1) k! = (k+1)!$, which is clearly correct. For the inductive step, we need to prove that $n^\underline{k+1} + (k+1) n^\underline{k} = (n+1)^\underline{k+1}$. Indeed,
$$ n^\underline{k+1} + (k+1) n^\underline{k} = \frac{n!}{(n-k-1)!} + \frac{(k+1)n!}{(n-k)!} = \frac{(n-k)n! + (k+1)n!}{(n-k)!} = \frac{(n+1)n!}{(n-k)!} = \frac{(n+1)!}{(n-k)!} = (n+1)^\underline{k+1}. $$
