Show that $\ S(2n, n) \ ≥ n!$ for all $ n ≥ 1$.

Question

I have been given the following practice question about Stirling numbers of the second kind:

Show that $\ S(2n, n)\ ≥ n!$ for all $ n ≥ 1$.

I don't know where to start with this and any help would be appreciated!

Answer 1

Some hints for one possible solution:

• Since $2n/n=2$, some of the partitions of $\{1,\cdots,2n\}$ into $n$-parts are specfically partitions into size-two subsets that look like $\{a,b\}$.
• A factorial $n!$ counts the number of permutations of $\{1,\cdots,n\}$
• The sets $\{1,\cdots,n\}$ and $\{n+1,\cdots,2n\}$ have the same size...
• A function $f:X\to Y$ can be treated as a collection of ordered pairs in $X\times Y$