Given $d\in \mathbb{N}$, let $X\_d:= \{(\ell\_1, \ldots , \ell\_d): 0 \le \ell\_1 \le \ldots \le \ell\_d \le d\}\subset \mathbb{Z}^d$, and endow $X\_d$ with the (usual) partial order, namely, $x\le y$ if and only if $x\_j \le y\_j$ for all $j=1,\ldots , d$. Note that if $Y\subset X\_d$ is linearly ordered then $\# Y \le d^2 + 1$. Let $N\_d$ be the number of linearly ordered subsets of $X\_d$ of maximal length, that is, $N\_d :=\# \{Y\subset X\_d : Y \mbox{ linearly ordered, } \# Y = d^2 + 1\}$. 


My question: Is there any neat expression available for $N\_d$? If not, perhaps at least the asymptotics of $N\_d$ as $d\to \infty$ can be worked out? Any answers or pertinent references will be much appreciated.



Counting linearly ordered subsets of maximal length in partially ordered $d$-tuples of nonnegative integers
Denote $x!!=1!2!\cdots x!$. The number is
$$N\_d=\frac{(d-1)!!(d^2)!}{(2d-1)!!}\prod\_{1\leqslant i<j\leqslant n}(j-i).$$
It can be interpreted as the number of lattice paths from $(0,\ldots,0)$ to $(d,\ldots,d)$ in $\mathbb{Z}^d$ such that all points on the path satisfy $x\_1\leqslant x\_2\leqslant\cdots\leqslant x\_d$. More generally we might ask for the number of lattice paths from $(0,0,\ldots,0)$ to $(m\_1,\ldots,m\_d)\in\mathbb{Z}$ with $m\_1\leqslant m\_2\leqslant\cdots\leqslant m\_d$ such that in every step $x\_1\leqslant\cdots\leqslant x\_d$.
Using generating functions, these so-called **lattice permutations** are counted by Percy MacMahon in [Combinatory Analysis, 1915-1916](http://quod.lib.umich.edu/cgi/t/text/text-idx?c=umhistmath;idno=ABU9009.0001.001), chapter 5. 


An alternative proof is by Doron Zeilberger: [Andre's reflection proof generalized to the many-candidate ballot problem](http://www.sciencedirect.com/science/article/pii/0012365X83902005), Discrete Mathematics **44**(3), 1983, 325-326.


