I was reading this article on "The Search for Simple Symmetric Venn Diagrams" by Frank Ruskey, Carla D. Savage, and Stan Wagon and on the first page page they prove that $n$ is prime if an $n$-Venn diagram has an $n$-fold rotational symmetry. From the introductory paragraph:
In any symmetric $n$-Venn diagram the fixed point of the rotations, the center of the diagram, must lie in the unique region of rank $n$. The unbounded outer region has rank $0$. Regions of rank $0 < r < n$ must be distributed symmetrically and thus their number, ${n\choose r}$, must be divisible by $n$. This property holds exactly when $n$ is prime.
I am confused about how they came to conclude that ${n\choose r}$ must be divisible by $n$ when an $n$-Venn diagram has $n$-fold rotational symmetry.
Since there is $n$-fold rotational symmetry, for each region of rank $r$ there must $n-1$ other regions of rank $r$ located symmetrically around the center (provided $0<r<n$). Hence the number of regions of rank $r$ must be a multiple of $n$ for each $0<r<n$. Call this multiple $k$. However, since this is a $n$-Venn diagram, the number of regions of rank $r$ is $n\choose r$. This means that by a double counting argument we have $nk={n\choose r}$ for some integer $k$, or more simply put, $n$ divides $n\choose r$.