In his 1981 work, McKay demonstrates that the eigenvalue distribution of large regular graphs follow the law presented in his paper and which nowadays goes by his name.
At page 205 (the third page from the start of his paper) he counts the closed walks of $r$ steps on a regular graph starting by a point $\nu_0$, with the constraint that locally at $\nu_0$ (precisely in a ray of at least $\frac{r}{2}$) the graph is acyclic. This condition permit to map the problem to the study of the number of possible sequences $\pmb{\delta}=\delta_0\dots\delta_r$ (where is it is found that the walk could have only even steps, $r=2s$) where each $\delta_i$ is the distance of the vertex $\nu_i$ at the $i-$th step of the walk from the node $\nu_0$, (obviously $\delta_0=\delta_r=0, |\delta_{i+1}-\delta_{i}|=1$ and $\delta_i\geq0$). It turns out that the number of the sequences $\delta_1\dots\delta_r$ ($\delta_0$ is $0$ by definition and could be omitted in the study) with the given properties and $k$ element equal to $0$ (the walk returns $k$ times to $\nu_0$ after it started) is $$\binom{2s-k}{s}\cdot\frac{k}{2s-k}$$ The point is that I cannot derive this formula using combinatorics. I found that it could be related to Catalan numbers and Dyck words, but I cannot see how to impose the right constraints which rule out the unwanted path (the paper cite a reference for this result, but is a standard prob&stat book with a chapter on combinatorics, so I guess I am missing some point).