# Expression for summation involving factorial

It is known that $\sum_{k = 0}^{n} {n \choose k} = 2^n$ and $\sum_{k = 0}^{n} {n \choose k} (!k)= n!$. But is it known what $\sum_{k = 0}^{n } {n \choose k}(k!)$ is equal to?

• $\sum_{k=0}^n C_n^k k!=\int_0^\infty C_n^k x^k e^{-x}dx=\int_0^\infty (1+x)^n e^{-x}dx$ expressible via the incomplete Gamma function – Peter Kravchuk Jun 6 '16 at 12:13
• Useful summary here: oeis.org/wiki/Subfactorial. – Todd Trimble Jun 6 '16 at 14:30

See https://oeis.org/A000522 . It is known (and not too hard to prove) that it is $\lfloor e \cdot n! \rfloor$ for $n \geq 1$.
• Note that the sum expands into $n!(\frac{1}{0!}+\dots+\frac{1}{n!})$; the sum in parentheses converges to $e$ from below, and can easily be shown to be in $(e-\frac{1}{n!},e)$. – Klaus Draeger Jun 6 '16 at 16:17