It is known that $\sum_{k = 0}^{n } {n \choose k}(k!) = \lfloor e \cdot n! \rfloor $ But is it known what $\sum_{p = 0}^{n} \sum_{q = 0}^{n - p} {n \choose p}{{n - p} \choose q} p! \cdot q! \cdot (n-p-q)!$ is equal to?
It is known that $\sum_{k = 0}^{n } {n \choose k}(k!) = \lfloor e \cdot n! \rfloor $ But is it known what $\sum_{p = 0}^{n} \sum_{q = 0}^{n - p} {n \choose p}{{n - p} \choose q} p! \cdot q! \cdot (n-p-q)!$ is equal to?
