Suppose I have a Standard Young Tableaux with dimensions $2$ x $n$. The Catalan numbers, $C_n$ count the number of ways to arrange the numbers $\left\{1, ..., 2n\right\}$. This can be derived using the Hook Length formula, but I can't see why? Any ideas?
Since the Catalan numbers show up in a lot of places, I'm wondering if a Bijection to another Catalan application can be made...
Here is a bijection to Dyck paths (or to well-formed bracketings):
Take a SYT of shape $2\times n$ (so it contains the numbers $\{1,\ldots,2n\}$, and we aim to form a word of $n$ up-steps (opening brackets) and $n$ down-steps (closing brackets). Well, there is an obvious way of doing so: make the $i$-th symbol for $1 \leq i \leq 2n$ an up-step if $i$ appears in the top row of the STY and a down-step if $i$ appears in the bottom row.
Then, indeed, the property of being a STY simply says that any prefix of the constructed word of $2n$ symbols contains at least as many up-steps as down-steps. This is exactly the property of being a Dyck path of length $2n$, or, equivalently, a well-formed backeting of $2n$ opening and closing brackets.
I haven't actually checked, but this bijection gotta be in Stanley's list, and you certainly find it in various other places.
Indeed, there is an $m$-dimensional version that gives a bijection between rectangular SYT and multidimensional Dyck paths: Let $e_i$ ($1 \leq i \leq m$) be the $i$-th standard basis vector in $\mathbb{R}^m$. One can then send a SYT of shape $m \times n$ to the sequence of length $mn$ by making the $i$-th symbol for $1 \leq i \leq mn$ equal to $e_j$ if $i$ appears in the SYT in row $j$.
This map sends SYT of shape $m \times n$ to multi-dimensional Dyck paths given by all paths from $(0,\ldots,0) \in \mathbb{R}^m$ to $n\cdot(1,\ldots,1) \in \mathbb{R}^m$ which stays inside the wedge $x_1 \geq x_2 \geq \ldots \geq x_m$. These are then counted by the multi-dimensional Catalan numbers, given by https://oeis.org/A060854, and you can find this correspondence for example in this paper of mine with Paco Santos and Volkmar Welker on page 3.
The hooks lengths across the first row are $n+1,n,n-1, \ldots 3, 2$, and for the second row we get $n,n-1, \ldots 2,1$. So, the hook length formula gives the number of such tableaux as $\frac{(2n)!}{(n+1)!n!} = \frac{1}{n+1} \frac{(2n)!}{n!n!} = \frac{1}{n+1} \binom{2n}{n}$. I don't know a bijective proof off the top of my head, but I agree with T. Amdeberhan in the comments - I am willing to bet Stanley knows one.
