Given a multi-variable polynomial $F$, denote the number of monomials by $N(F)$. Take for instance, \begin{align\*}N(x(x+y)+(x+y)^2-(x-y)^2)=N(x^2+5xy)&=2 \qquad \text{and} \\
N((x+z)(x+y)^2)=N(x^3 + 2x^2y + x^2z + xy^2 + 2xyz + y^2z)&=6.\end{align\*}
Denote the symmetric group on $n$ letters $\{1,2,\dots,n\}$ by $\mathfrak{S}\_n$. Define the action of $\mathfrak{S}\_n$ on a function $F(x\_1,\dots,x\_n)$ in a natural way: given $w\in\mathfrak{S}\_n$, then $w\cdot F(x\_1,\dots,x\_n)=F(x\_{w(1)},\dots,x\_{w(n)})$. Introduce the (multi-variable) rational functions
$$G(\mathbf{x},\mathbf{z})=\prod\_{k=1}^n\frac{x\_1z\_1+x\_2z\_2+\cdots+x\_kz\_k}{x\_k-x\_{k+1}}.$$


Assuming that the symmetric groups act only on the $x$-variables, let's compute the polynomial
$$G\_{\mathfrak{S}\_{n+1}}=\sum\_{w\in\frak{S}\_{n+1}}w\cdot G.$$
I would like to ask:



> 
> **QUESTION.** Let $C\_n=\frac1{n+1}\binom{2n}n$ be the Catalan numbers. Is there a combinatorial proof that $N(G\_{\mathfrak{S}\_{n+1}})=C\_n$?
> 
> 
> 



Counting monomials and the Catalan numbers
Any monomial $P:=\prod x\_i^{c\_i}$ of degree $\sum c\_i=n$ maps to a non-zero constant after symmetrization
$$
P\to \Phi(P):=G\_{\mathfrak{S}\_{n+1}}\frac{P}{(x\_1-x\_2)(x\_2-x\_3)\ldots (x\_n-x\_{n+1})}.
$$
Indeed, $\Phi(P)$ is a constant by a degree consideration, to prove that this constant is non-zero you may use, for example, Theorem 2 [here](https://arxiv.org/pdf/1512.07136.pdf).


Thus any monomial $\prod (x\_iz\_i)^{c\_i}$ maps to a non-zero constant times $\prod z\_i^{c\_i}$.


Thus, you simply count the number of monomials which may arise, when you multiply the linear forms $\prod\_{k=1}^n(x\_1z\_1+x\_2z\_2+\cdots+x\_kz\_k)$. The only condition is that $c\_1+\ldots+c\_i\geqslant i$ for all $i$. Such sequences are indeed enumerated by Catalan numbers, a bijection with lattice paths below diagonal is straightforward.


