Given two $k$-subsets $I$ and $J$ of $\{1 \dots n\}$, denote by $\min(J)$ the minimal element in $J$ and by $\max(I)$ the maximal element in $I$, we write $I \prec J$ if $\max(I)<\min(J)$. The sets $I$ and $J$ are called weakly separated if at least one of the following two conditions holds:
(1) $J-I$ can be partitioned into a disjoint union $J-I = J' \sqcup J''$ so that $J' \prec I-J \prec J''$;
(2) $I-J$ can be partitioned into a disjoint union $I-J = I' \sqcup I''$ so that $I' \prec J-I \prec I''$.
This condition is introduced in Leclerc and Zelevinsky's paper Quasi-commuting families of quantum minors.
Let $N$ be the number of all weakly separated pairs $I, J$ where $|I|=|J|=k$ and $I, J \subset {1, \ldots, n}.
I have counted certain special cases. In the case of $k=2$, $n$ arbitrary, $N$ is the number of diagonal dissections of an $n$-gon into two regions. In the case of $k=3, n=6$, $N=54$. In the case of $k=3, n=7$, $N=224$. In the case of $k=3,n=8$, $N=652$. In the case of $k=3,n=9$, $N=1557$.
Are there some method in combinatorics to count the number $N$? Thank you very much.
Edit: I forgot to say that I assume that $I \ne J$, $I$ and $J$ are not of the form $a, a+1, \ldots, a_k$ (mod $n$). I counted the pair $I, J$ as a set. So I counted $(I, J)$, $(J, I)$ as one pair (not two pairs). I am sorry about the confusion.
