Let $n$ be a nonnegative integer. Determine the number of ways that one can choose $(n+1)^2$ sets $S_{i,j}\subseteq\{1,2,\ldots,2n\}$, for integers $i,j$ with $0\leq i,j\leq n$, such that:
$\bullet$ for all $0\leq i,j\leq n$, the set $S_{i,j}$ has $i+j$ elements; and
$\bullet$ $S_{i,j}\subseteq S_{k,l}$ whenever $0\leq i\leq k\leq n$ and $0\leq j\leq l\leq n$.
I have a coherent solution involving a lemma, but I was wondering if there was a shorter way. Thanks for any help!
Note that there are $(2n)!$ ways to choose $S_{1, 0}, S_{2, 0}... S_{n, 0}, S_{n, 1}, S_{n, 2}... S_{n, n}$, because there are $2n$ ways to choose which number $S_{1, 0}$ is, $2n-1$ ways to choose which number to append to make $S_{2, 0}$, $2n-2$ ways to choose which number to append to make $S_{3, 0}$.... After that, note that $S_{n-1, 1}$ contains the $n-1$ in $S_{n-1. 0}$ and $1$ other element chosen from the $2$ elements in $S_{n, 1}$ not in $S_{n-1, 0}$ so there are $2$ ways for $S_{n-1, 1}$. By the same logic there are 2 ways for $S_{n-1, 2}$ as well so $2^n$ total ways for all $S_{n-1, j}$, so doing the same thing $n-1$ more times yields a final answer of $(2n)!\cdot 2^{n^2}$.
Hope this helps- it's a bit shorter.