Powerful Limit

Note that $\begin{array}{rcl} \displaystyle f(n) & = & \displaystyle \sum_{r=0}^n \binom{n}{r} r^r (2-r)^{n-r} \; = \; \sum_{r=0}^n \sum_{s=0}^{n-r} \binom{n}{r}\binom{n-r}{s}2^s (-1)^{n-r-s}r^{n-s} \\ & = & \displaystyle \sum_{0 \le r+s \le n} \frac{n!}{r! s! (n-r-s)!} (-1)^{n-r-s}2^s r^{n-s} \; = \; \sum_{s=0}^n \sum_{r=0}^{n-s} \binom{n}{s}\binom{n-s}{r} (-1)^{n-s-r}2^sr^{n-s} \\ & = & \displaystyle \sum_{s=0}^n \binom{n}{s}2^s \sum_{r=0}^{n-s} \binom{n-s}{r} (-1)^{n-s-r}r^{n-s} \end{array}$ But, using the Inclusion-Exclusion Principle to count the number of bijections from a set with $N$ elements to itself, $\sum_{r=0}^N \binom{N}{r} (-1)^{N-r}r^N \; = \; N! \hspace{1cm} N \ge 0 \;,$ and hence $\begin{array}{rcl} \displaystyle f(n) & = & \displaystyle \sum_{s=0}^n \binom{n}{s} 2^s (n-s)! \\ \displaystyle \frac{f(n)}{n!} & = & \displaystyle \sum_{s=0}^n \frac{2^s}{s!} \end{array}$ and hence $\lim_{n \to \infty} \frac{f(n)}{n!} \; = \; e^2$ making the answer $\lfloor 1000e^2\rfloor = \boxed{7389}$ .