TITLE: Combinatorial formulas and interpretations
QUESTION [8 upvotes]: I found that
$$ \sum_{j=0}^{s}(n-s+j)!\binom{s}{j}(s-j)! =s!  \sum_{j=0}^{s} \frac{(n-s+j)!}{j!} = \frac{(n+1)!}{n+1-s}$$
I proved this formula with induction, but I was wondering if there is a (combinatorial?) interpretation that can explain it.
Moreover, I wanted to simplify in a similar way also the following:
$$ \sum_{j=0}^{s}(n-s+j)!\binom{s}{j}3^{s-j}(s-j)! = s! 3^s \sum_{j=0}^{s} \frac{(n-s+j)!}{j!3^j} = ?$$
is it possible?

REPLY [12 votes]: Here's another non-induction and non-combinatorial proof.
$$\begin{align}
\sum_{j=0}^{s}(n-s+j)!\binom{s}{j}(s-j)! &=s!  \sum_{j=0}^{s} \frac{(n-s+j)!}{j!} \cdot \color{lightgrey}{\frac{(n-s)!}{(n-s)!}}\\
&=s!(n-s)!\sum_{j=0}^s\binom {n-s+j}{n-s}\\
&=s!(n-s)!\binom {n+1}{n-s+1}
&&\tiny\text{using }\sum_{r=a}^b\binom {m+r}{m+a}=\binom {m+b+1}{m+a+1}\\
&=s! (n-s)!\frac {(n+1)!}{(n-s+1)!s!}\\
&=(n-s)!\frac{(n+1)!}{(n-s+1)!}\\
&=\frac{(n+1)!}{n-s+1}\qquad\blacksquare
\end{align}$$

Alternatively
$$\begin{align}
\sum_{j=0}^{s}(n-s+j)!\binom{s}{j}(s-j)! &=
n!\sum_{j=0}^s\frac{\binom sj}{\frac{n!}{(n-s+j)!(s-j)!}}\\
&=n!\sum_{j=0}^s\frac{\binom s{s-j}}{\binom n{s-j}}\\
&=\frac{n!}{\binom ns}\sum_{j=0}^s\frac{\binom ns\binom s{s-j}}{\binom n{s-j}}\\
&=\frac{n!}{\frac{n!}{s!(n-s)!}}\sum_{j=0}^s\frac{\binom n{s-j}\binom {n-s+j}{j}}{\binom n{s-j}}\\
&=s!(n-s)!\sum_{j=0}^s\binom {n-s+j}{n-s}\\
&=s!(n-s)!\binom{n+1}{n-s+1}\\
&=s!(n-s)! \frac{(n+1)!}{(n-s+1)!s!}\\
&=\frac{(n+1)!}{n-s+1}\qquad\blacksquare
\end{align}$$