The hat guessing number of a graph.
Suppose that each vertex of a graph corresponds to a player, wearing a hat whose color is arbitrarily chosen from a set of $q$ possible colors. Each player can see the hat colors of his neighbors, but not his own hat color. All of the players are asked to guess their own hat colors simultaneously, according to a predetermined guessing strategy and the hat colors they see, where no communication between them is allowed. The hat guessing number $HG(G)$ of a graph $G$ is the largest integer $q$ such that there exists a guessing strategy guaranteeing at least one correct guess for any hat assignment of $q$ possible colors.
Because it suffices that a single player guesses correctly, the hat guessing number of a graph is the maximum of the hat guessing numbers of its connected components.

The Billey-Jockusch-Stanley bijection to 321-avoiding permutations.

The graph of inversions of a permutation.
For a permutation of $\{1,\dots,n\}$, this is the graph with vertices $\{1,\dots,n\}$, where $(i,j)$ is an edge if and only if it is an inversion of the permutation.

The zeta map on Dyck paths.
The zeta map $\zeta$ is a bijection on Dyck paths of semilength $n$.
It was defined in [1, Theorem 1], see also [2, Theorem 3.15] and sends the bistatistic (area, dinv) to the bistatistic (bounce, area). It is defined by sending a Dyck path $D$ with corresponding area sequence $a=(a_1,\ldots,a_n)$ to a Dyck path as follows:

• First, build an intermediate Dyck path consisting of $d_1$ north steps, followed by $d_1$ east steps, followed by $d_2$ north steps and $d_2$ east steps, and so on, where $d_i$ is the number of $i-1$'s within the sequence $a$.
  For example, given $a=(0,1,2,2,2,3,1,2)$, we build the path
  $$NE\ NNEE\ NNNNEEEE\ NE.$$
• Next, the rectangles between two consecutive peaks are filled. Observe that such the rectangle between the $k$th and the $(k+1)$st peak must be filled by $d_k$ east steps and $d_{k+1}$ north steps. In the above example, the rectangle between the second and the third peak must be filled by $2$ east and $4$ north steps, the $2$ being the number of $1$'s in $a$, and $4$ being the number of $2$'s. To fill such a rectangle, scan through the sequence a from left to right, and add east or north steps whenever you see a $k-1$ or $k$, respectively. So to fill the $2\times 4$ rectangle, we look for $1$'s and $2$'s in the sequence and see $122212$, so this rectangle gets filled with $ENNNEN$.
  The complete path we obtain in thus
  $$NENNENNNENEEENEE.$$