TITLE: Pairs of shortest paths
QUESTION [16 upvotes]: It is known that the binomial coefficient $2n \choose n$ is equal to number of shortest lattice paths from $(0,0)$ to $(n,n)$. The Catalan number $\frac{1}{n+1} {2n\choose n}$is equal to the number of shortest lattice paths that never go above the diagonal. Here, the diagonal may be viewed as a path from $(0,0)$ to $(n,n)$.
Is there a formula for the number of pairs $(P_{1},P_{2})$ where each $P_{i}$ is a shortest lattice path from $(0,0)$ to $(n,n)$ such that $P_{1}$ never goes above $P_{2}\ ?$ Here, "$P_{1}$ never goes above $P_{2}$" means that $P_{1}$ lies inside or on the boundary of the region determined by $P_{2}$, the $x$-axis, and the line $x=n$.

REPLY [7 votes]: If you ask the analogous question about p noncrossing paths in an m x n rectangle, you are essentially counting plane partitions (if you write in each box the number of paths it is above, then all rows and columns are weakly decreasing). There is then a beautiful formula by MacMahon for the number of these: it is the product of (i+j+k-1)/(i+j+k-2) over all 1<=i<=m, 1<=j<=n, 1<=k<=p.