This question is motivated by this link. The statement is as follows. (Edit: Even if there are already two great answers, I would love to have a couple more answers. Especially, I would like to see another, hopefully more combinatorial, proof of the bonus question below.)
Question. Let $x$ and $y$ be noncommuting variables such that $x^2=y^2=1$. Multiplication is associative. Prove that, for each positive integer $n$, there are exactly $\displaystyle\binom{2n}{n}$ expressions of length $2n$ in $x$ and $y$ that are equal to $1$. For example, when $n=1$, there are $2$ such expressions: $xx$ and $yy$. When $n=2$, there are $6$ such expressions: $xxxx$, $xxyy$, $xyyx$, $yxxy$, $yyxx$, and $yyyy$.
For a technical clarification, consider the free product $G:=C_2*C_2$, where $C_2$ is the cyclic group of order $2$. Then, $G$ has the following presentation: $G=\langle x,y\,|\,x^2=y^2=1\rangle$. We want to find the number of strings of length $2n$ formed by $x$ and $y$ that can be reduced to $1$.
I would like to see how to prove this statement using a combinatorial argument, such as constructing a bijection, finding a generating function, etc. However, any proof that is different from the proofs below is welcome. (If you can visit the referred link and give it a combinatorial proof, that would be most appreciated.)
Bonus. Let $s$ be a reduced word in $x$ and $y$ (that is, it can no longer be reduced using the rules $x^2=y^2=1$). If $s$ has length $k$, then show that, for any integer $n\geq 0$, there are exactly $\displaystyle\binom{n+2k}{n}$ words in $x$ and $y$ of length $k+2n$ that can be reduced to $s$.
Elementary Proof
We work in $R:=\mathbb{Z}[x,y]$. Note that $$x+y=x+xxy=x(1+xy)$$ and $$(x+y)^2=\big(x(1+xy)\big)^2=x(1+xy)\,x(1+xy)\,.$$ Because $$(1+xy)x=x+xyx=x(1+yx)\,,$$ we have $$(x+y)^2=xx(1+yx)(1+xy)=yx(1+xy)^2=(xy)^{-1}(1+xy)^2\,.$$ Therefore, $$(x+y)^{2n}=\Big((xy)^{-1}(1+xy)^2\Big)^n=(xy)^{-n}(1+xy)^{2n}\,.$$ Thus, there are $\displaystyle\binom{2n}{n}$ expressions of length $2n$ that are equal to $1$.
Algebraic Proof
Here is another approach borrowing the idea from Julian Rosen. Let $R$ denote the unital $\mathbb{Z}$-algebra generated by $x$ and $y$ (i.e. $R=\mathbb{Z}[G]$). Then, the $\mathbb{Z}$-algebra homomorphism $$\varphi:R\to\text{Mat}_{2\times2}\big(\mathbb{Z}[t,t^{-1}]\big)$$ defined by sending $$x\mapsto\begin{bmatrix}0&1\\1&0\end{bmatrix}\text{ and }y\mapsto\begin{bmatrix}0&t^{-1}\\t&0\end{bmatrix}$$ is injective. We can easily see that $$\varphi\big((x+y)^2\big)=t^{-1}(1+t)^2\,I\,,$$ where $I$ is the $2$-by-$2$ identity matrix. Hence, $$\varphi\big((x+y)^{2n}\big)=t^{-n}(1+t)^{2n}\,I\,,$$ and the assertion follows immediately.
Geometric Proof
Using the notations from the geometric proof in my answer here, recall that $x=\sigma_\alpha$ and $y=\sigma_\beta$. Thus, $$(x+y)^2=2+\sigma_\alpha\sigma_\beta+\sigma_\beta\sigma_\alpha=2+\rho_{2\alpha-2\beta}+\rho_{2\beta-2\alpha}\,.$$ (It can also be proven that $$\sigma_{\theta_1}+\sigma_{\theta_2}=2\,\cos(\theta_1-\theta_2)\,\sigma_{\frac{\theta_1+\theta_2}{2}}$$ for all $\theta_1,\theta_2\in\mathbb{R}$.) Note that $$\rho_{+\theta}+\rho_{-\theta}=2\,\cos(\theta)$$ for all $\theta\in\mathbb{R}$. Hence, $$(x+y)^2=2\,\big(1+\cos(2\alpha-2\beta)\big)=2^2\,\big(\cos(\alpha-\beta)\big)^2\,.$$ Then, the number of expressions of length $2n$ that are equal to $1$ is given by $$\begin{align}\frac{1}{(2\pi)^2}\,\int_0^{2\pi}\,\int_0^{2\pi}\,(x+y)^{2n}\,\text{d}\beta\,\text{d}\alpha&=\frac{1}{(2\pi)^2}\,\int_0^{2\pi}\,\int_0^{2\pi}\,2^{2n}\,\big(\cos(\alpha-\beta)\big)^{2n}\,\text{d}\beta\,\text{d}\alpha \\&=\frac{1}{(2\pi)^2}\,2^{2n}\,\frac{\pi}{2^{2n-1}}\,\binom{2n}{n}\,(2\pi)=\binom{2n}{n}\,. \end{align}$$
