Counting partitions
List all partitions of \{1,2,3,4\} into two non-empty clusters and compare with S(4,2) from the Stirling formula. Compute S(10,3) from the formula.
Cluster labels are arbitrary, so swapping the two sides does not create a new partition. Enumerating this small case makes the division by label permutations concrete.
The partitions into two non-empty clusters are
\begin{aligned} &\{1\}\mid\{2,3,4\}, \quad\{2\}\mid\{1,3,4\}, \quad\{3\}\mid\{1,2,4\}, \quad\{4\}\mid\{1,2,3\},\\ &\{1,2\}\mid\{3,4\}, \quad\{1,3\}\mid\{2,4\}, \quad\{1,4\}\mid\{2,3\}. \end{aligned}
There are seven. The formula gives
S(4,2)=\frac12\Bigl[(-1)^1\binom{2}{1}1^4+(-1)^0\binom{2}{2}2^4\Bigr] =\frac12(-2+16)=7.
For ten observations and three clusters,
\begin{aligned} S(10,3) &=\frac{1}{6}\Bigl[\binom{3}{1}1^{10}-\binom{3}{2}2^{10}+\binom{3}{3}3^{10}\Bigr]\\ &=\frac{1}{6}(3-3072+59049)=\frac{55980}{6}=9330. \end{aligned}