Let a and b be coprime positive integers. Then the number a+b divides the binomial coefficient ${a+b \choose a}$. I know how to prove this combinatorially - for example after choosing an ordered set of a+b elements, there is a free action of a cyclic group of order a+b on the set of a-element subsets (and you can even choose a distinguished representative of each orbit if you so wish).

Passing to q-binomial coefficients, the quantum number $[a+b]_q$ divides the q-binomial coefficient $\begin{bmatrix}{a+b}\\ {a}\end{bmatrix}_q$.

My question today is whether this divisibility fact has a combinatorial proof or interpretation.

I remark that one can also ask whether or not $$\frac{1}{[a+b]_q}\begin{bmatrix}{a+b}\\ {a}\end{bmatrix}_q\in \mathbb{N}[q],$$ which would likely follow from any such combinatorial argument. When a+b=p is prime, I know this is the case by constructing algebraically a free $\mathbb{F}_p[d]/(d^p)$-module structure on the cohomology of the Grassmannian. I haven't done enough numerical experiments for non-prime values of a+b to be confident that this polynomial has non-negartive coefficients in general.