Given a nonnegative integer $n$. Show that the Catalan number $C_n$ is the number of ways to arrange the integers $1, 2, \ldots, 2n$ as a standard Young tableau of rectangular shape with dimensions $2 \times n$.
For example, for $n = 3$, the standard Young tableaux of such shape are \begin{align*} \begin{array}{|c|c|c|} \hline 1&2&3\\ \hline 4&5&6\\ \hline \end{array} \quad \text{ and } \quad \begin{array}{|c|c|c|} \hline 1&2&4\\ \hline 3&5&6\\ \hline \end{array} \quad \text{ and } \quad \begin{array}{|c|c|c|} \hline 1&2&5\\ \hline 3&4&6\\ \hline \end{array} \\ \quad \text{ and } \quad \begin{array}{|c|c|c|} \hline 1&3&4\\ \hline 2&5&6\\ \hline \end{array} \quad \text{ and } \quad \begin{array}{|c|c|c|} \hline 1&3&5\\ \hline 2&4&6\\ \hline \end{array}. \end{align*}
Supposedly this can be derived using the Hook Length formula, but I can't see how. 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...
You don’t need a bijection if you’ve already shown that $C_n=\frac1{n+1}\binom{2n}n$. The following array shows the hook lengths of the cells in a $2\times n$ Young tableau:
$$\begin{array}{|c|c|c|} \hline n+1&n&n-1&\ldots&3&2\\ \hline n&n-1&n-2&\ldots&2&1\\ \hline \end{array}$$
The product of these hook lengths is clearly $(n+1)!n!$, so the hook-length formula says that there are
$$\frac{(2n)!}{(n+1)!n!}=\frac1{n+1}\cdot\frac{(2n)!}{n!\cdot n!}=\frac1{n+1}\binom{2n}n=C_n$$
standard Young tableaux of shape $2\times n$.
However, there is a natural bijection between such Young tableaux and up-down Dyck paths of length $2n$. If $s_1\ldots s_{2n}$ is a Dyck path of $n$ up-steps and $n$ down-steps, fill the top row of a $2\times n$ array with the indices $k$ such that $s_k$ is an up-step and the bottom row with the indices $k$ such that $s_k$ is a down-step. In each row arrange the entries in increasing order. Since each up-step in a Dyck path has a matching down-step that occurs later in the path, the resulting array is always a standard Young tableau, and it’s easy to see that each standard Young tableau can be produced in this way from a Dyck path.