# Is there a combinatorial proof that the Catalan number $C_n$ satisfies $(n+1)C_n={2n \choose n}$?

I saw this question and thought that may be it is possible to prove that the $$n^{\text{th}}$$ Catalan number $$C_n$$ equals $$\frac{1}{n+1}{2n\choose n}$$ by taking a set $$A$$ of size $$n+1$$ and another set $$B$$ of size $$C_n$$ such that there exists a $$1$$-$$1$$ correspondence $$A\times B\to T$$, where $$T$$ is the set of subsets of $$\{1,2,\ldots,2n\}$$ of size $$n$$. I made my attempt but failed, but I am curious if there is a known bijection $$A\times B\to T$$ for some $$A$$, $$B$$.

I know that there are combinatorial proofs that show $$C_n={2n\choose n}-{2n\choose{n-1}}$$, but I want a specific proof that shows $$(n+1)C_n={2n\choose n}$$. Below is my attempt.

Write $$[k]=\{1,2,\ldots,k\}$$. Furthermore, $$\binom{X}{k}$$ is the set of all subsets of cardinality $$k$$ of a given set $$X$$.

Let there be $$2n$$ people, named, $$1$$, $$2$$, $$\ldots$$, $$2n$$. The $$2n$$ people are seated around a round table in the counterclockwise order. Let $$\mathcal{P}$$ denote the set of all pairings $$\big\{\{x_1,y_1\},\{x_2,y_2\},\ldots,\{x_n,y_n\}\big\}$$ of $$[2n]$$ in such a way that, when $$x_i$$ shakes hand with $$y_i$$ simultaneously for every $$i\in[n]$$, there are no crossing arms. Wlog, we assume that $$x_i for each $$i\in[n]$$ and that $$x_1.

Define $$f:[n+1]\times\mathcal{P}\to\binom{[2n]}{n}$$ as follows: $$f\Big(k,\big\{\{x_1,y_1\},\{x_2,y_2\},\ldots,\{x_n,y_n\}\big\}\Big)=\{y_1,y_2,\ldots,y_{k-1},x_k,x_{k+1},\ldots,x_n\}$$ for each $$k\in[n+1]$$ and $$\big\{\{x_1,y_1\},\{x_2,y_2\},\ldots,\{x_n,y_n\}\big\}\in\mathcal{P}$$. Well, this is where my idea fails. I was hoping that $$f$$ will be a bijection, but it isn't even injective.