I'm trying to evaluate the sum given below
$$ S(x,y) = \frac{1}{k!}\sum_{j=0}^k \binom{k}{j} (-x)^j y^{k-j} j! $$
where $x,y > 0$. Is there any clever closed form way of writing this sum? If not, maybe there is a clever way of writing it as a product? I was considering products of $(y-jx)$ factors perhaps $\pm (y-x)^k$ but could not get this work. Any ideas?
Since ${k \choose j} = {k \choose k-j} = \frac{k!}{j!(k-j)!}$, upon making the substitution $j \leftrightarrow k-j$ you can write your sum as
$$ S(x,y) = (-x)^k \sum_{j=0}^k \frac{1}{j!} \left( - \frac{y}{x} \right)^j = (-x)^k S(1, -y/x) \; .$$
Now we recognize $S(u) := S(1, u) = \sum_{j=0}^k \frac{u^j}{j!}$ as the $k$-th partial sum of the MacLaurin expansion of $e^{u}$, which admits a closed form in terms of the (incomplete) Gamma function, as expressed in this MSE post. This gives some perspective on the other possible 'closed forms' one might expects.