TITLE: A divisibility of q-binomial coefficients combinatorially
QUESTION [11 upvotes]: 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.

REPLY [4 votes]: Supposing $\gcd(a,b)=1$ throughout this answer. Haiman was able to prove that 
$$\frac{1}{[a+b]_q}\begin{bmatrix}{a+b}\\ {a}\end{bmatrix}_q\in \mathbb{N}[q]$$
by interpreting it as the Hilbert series of some quotient of a polynomial ring. See propositions 2.5.(2-4) in his paper "Conjectures On The Quotient Ring By Diagonal Invariants".
Regarding combinatorial proofs: It is known in rational Catalan land that the quantity $\frac{1}{a+b}\binom{a+b}{a}$ counts  paths from $(0,0)$ to $(a,b)$ with steps $(1,0),(0,1)$ that don't go below the $y=bx/a$ line
simultaneous $a,b$ core partitions 
and there are explicit statistics on both of these sets which conjecturally give $\frac{1}{[a+b]_q}\begin{bmatrix}{a+b}\\ {a}\end{bmatrix}_q$ (see conjectures 2 and 3 in this paper) but to my knowledge they are still open.