Why rhombus tilings are just piling boxes with decending restrictions

Suppose $H$ is a hexagon with sidelength $a\times b\times c$, and it's innner angles are all $\frac{2\pi}{3}$. Then it's well known that the rhombus tilings of $H$ is in one-to-one correspondece with plane partitions of shape $a\times b$ and largest number not exceeding $c$.

That is, if we draw a 3D graph of any tiling of $H$, it will look like piling unit boxes at the corner of the wall, but with decending conditions. This fact is quite obvious, but I still want a rigorous proof. My question is, can we give a real bijection between these two objects, rather than just saying "draw a graph and it's obvious that ..." ?

• Great question! An idea: there is certainly an injection from the set of piles to the set of rhombus tilings by taking a projection onto a plane (ignoring the details on how to define it rigorously, but that's not the interesting part). So it suffices to count both sizes and prove that they're equal. OEIS might help. – punctured dusk Nov 13 '16 at 21:19
• As mentioned by Dijkstra in a note that can be found here, the following article, aiming to prove that any tiling has an equal number of rhombi of each orientation, asserts this correspondence without proof: G. David and C. Tomei, The problem of the calissons, Amer. Math. Monthly. 96 (1989), 429–431. (Dijkstra then gives a rigorous proof of that assertion, but does not touch upon the correspondence.) – punctured dusk Nov 13 '16 at 21:38
• An explicit formula for the number of rhombus tilings: OEIS: oeis.org/A008793 or arxiv.org/pdf/math/9909038v1.pdf (page 1, (1.1) case $a=b=c$). Counting cube pilings translates as counting matrices with entries in $\mathbb N_{\leq n}$ whose rows and columns are weakly decreasing. (The 1-dimensional analogue is a lot easier.) – punctured dusk Nov 13 '16 at 22:03
• @punctureddusk Unfortunately the proof that paper refers to uses the correspondence, as is mentioned in the line after the equation. – Abhimanyu Pallavi Sudhir Dec 12 '19 at 10:47