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 $N(G_{\mathfrak{S}_{n+1}})=C_n$?
Best Answer
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.
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.