simplify summation with binomial coefficients

Question

Is there a way to simplify this summation?:

$$\sum ^n _{k=0} \left(n-k\right)!\binom{n}{k}^2$$

Answer 1

Alright, some of the comments have pointed out that it seems pretty difficult to find a definite formula but we can simplify it down to something much nicer to handle. We have, $$\sum ^n _{k=0} \left(n-k\right)!\binom{n}{k}^2$$ Now, multiplying and dividing $k!$ and $n!$we get, $$n! \sum^n _{k=0} \frac{1}{k!} \frac{\left(n-k\right)! \cdot k!}{n!} \cdot\binom{n}{k}^2 $$ This is equivalent to $$n! \sum^n _{k=0} \frac{\binom{n}{k}^2}{\binom{n}{k}k!}$$ Which is $$n! \sum^n _{k=0} \frac{\binom n k}{k!}$$

Answer 2

The simplest form seems to be $n! L_n(-1)$, where $L_n(x)$ is a Laguerre polynomial.