$s(n, n-3)={n \choose 2}{n \choose 4}$
I'm not sure how to prove this. I know there are 3 cases.
1st case has a 4-cycle: $|s(4, 1)|{n \choose 4}$
2nd case has a 3-cycle and a 2-cycle: $|s(5,2)|{n \choose 5}$
3rd case has three 2-cycles: $|s(6,3)| {n \choose 6}$
Are these correct? I don't see this equal to the identity. Is it possible to provide combinatorial proof?
Thanks
Good start,You are correct there are $3$ cases. ... but a little more care is required to get the correct factors of each type.
We trying to count permutations of $[n]$ with $n-3$ cycles (in the cycle decomposition.)
Case $1$: A $4$ cycle and $n-4$ fixed points \begin{eqnarray*} (\alpha \beta \gamma \delta) (*) \cdots (*). \end{eqnarray*} The first $4$ elements can be chosen in $\binom{n}{4}$ and they can be arranged in $\color{blue}{6}$ ways.
Case $2$: A $3$ cycle, a $2$ cycle and $n-5$ fixed points \begin{eqnarray*} (\alpha \beta \gamma)( \delta \epsilon) (*) \cdots (*). \end{eqnarray*} The first $5$ elements can be chosen in $\binom{n}{5}$ and they can be arranged in $\color{blue}{20}$ ways.
Case $3$: Three $2$ cycles and $n-6$ fixed points \begin{eqnarray*} (\alpha \beta)( \gamma \delta)( \epsilon \zeta) (*) \cdots (*). \end{eqnarray*} The first $6$ elements can be chosen in $\binom{n}{6}$ and they can be arranged in $\color{blue}{15}$ ways.
So that's the combinatorial description of the elements. I will leave you to do the algebra ... \begin{eqnarray*} {n \brack n-3} =\color{blue}{6} \binom{n}{4} + \color{blue}{20} \binom{n}{5} +\color{blue}{15} \binom{n}{6} = \cdots. \end{eqnarray*}