Stirling numbers with $k=n-2$

Question

Is there a more general method of calculating:

$$ \genfrac\{\}{0pt}{}{n}{n-2} $$

Like for :$$ \genfrac\{\}{0pt}{}{n}{n-1} $$ we can use $nC_2 $

Answer 1

Dividing $n$ objects into $n-2$ sets can be done in two ways. Either you can have two sets of size 2, or one set of size 3. Hence

$$\genfrac\{\}{0pt}{}{n}{n-2} = \frac{1}{2}{n\choose 2}{n-2\choose 2} + {n\choose 3} $$ The $\frac{1}{2}$ is necessary because the two doubletons are indistinguishable, so we are overcounting.

We similarly have $$\genfrac\{\}{0pt}{}{n}{n-3} = \frac{1}{6}{n\choose 2}{n-2\choose 2}{n-4\choose 2} + {n\choose 3}{n-3\choose 2} + {n\choose 4} $$