$$S(n,k)=\sum_{i=1}^k \frac{i^n(-1)^{k-i}}{(k-i)!(i-1)!}$$ Calculate $S(5,3)$ by using this formula and by listing all possible partitions of $\{1,2,3,4,5\}$ into $3$ blocks. Are they the same?
By the formula I got $S(5,3)=90$ Now how am I going to list this number of partitions? And next it says correct the formula.
By simplifying the formula \begin{eqnarray*} S(n,k) =\frac{1}{k!} \sum_{i=0}^{k} (-1)^{k-i} \binom{k}{i} i^n \end{eqnarray*} you have probably bumped $n$ down by one. The value $S(6,3)=90$ is correct, try looking at the partitions of $[6]$ into $3$ blocks.