Stirling number proof, proving that $s(n, n-2) = 2{n\choose3} + 3{n\choose4}$

Question

Let $s(n,k)$ denote the unsigned Stirling numbers of the first kind. Prove that $$s(n, n-2) = {n\choose3} + 3{n\choose4} $$

-I had a similar question for the $s(n, n-1)$ case, but unsure of how to solve for this? Any help is appreciated.

Answer 1

Note: The formula in the question is wrong: the righthand side should be $2\binom{n}3+3\binom{n}4$.

HINT: Suppose that a permutation of $[n]=\{1,\ldots,n\}$ has $n-2$ cycles. There are two possibilities.

• It might have a $3$-cycle; how many such permutations are there?
• If it does not have a $3$-cycle, what must it look like? How many of those permutations are there?