This answer follows from a bijection of De Médicis and Viennot (1994, Adv. Appl. Math.) Let $\mathcal{M}_n$ denote the set of perfect matchings of $[2n]$, i.e. the set of partitions of $[2n] := \{1,2,\ldots,2n\}$ into pairs. Let $M \in \mathcal{M}_n$. For $p = \{a,b\}, q = \{c,d\} \in M$ with $a<b$, $c<d$, and $a<c$, we say that $p$ and $q$ *cross* if $a < c < b< d$ and we say they *nest* if $a<c<d<b$. Finally, we say they are *aligned* if they neither cross nor nest, i.e., $a<b<c<d$. Define:

$\mathrm{ne}(M):= |\{\{p,q\}\subset M\colon \textrm{$p$ and $q$ nest}\}|;$
 
$\mathrm{cr}(M):= |\{\{p,q\}\subset M\colon \textrm{$p$ and $q$ cross}\}|;$ 

$\mathrm{al}(M):= |\{\{p,q\}\subset M\colon \textrm{$p$ and $q$ are aligned}\}|.$ 


Then $\sum_{M \in \mathcal{M}_n}x^{\mathrm{ne}(M)}y^{\mathrm{cr}(M)}=\sum_{M \in \mathcal{M}_n}x^{\mathrm{cr}(M)}y^{\mathrm{ne}(M)}$. However, crossings and alignments (or nestings and alignments) are *not* equidistributed: $\sum_{M \in \mathcal{M}_n}x^{\mathrm{al}(M)}y^{\mathrm{cr}(M)} \neq \sum_{M \in \mathcal{M}_n}x^{\mathrm{cr}(M)}y^{\mathrm{al}(M)}$.