A "rhomblet" is formed by joining the sides of two equilateral triangles with side length $1$.
How many ways can you tile a regular hexagon of side length $3$ using rhomblets?
One possible tiling is shown below.
Solution: This is identical to putting cubes in a 3x3x3 volume. Viewing the logo as if we are looking down on it, all cubes must be pushed as far right and back as possible, as far left and back as possible, and as far down as possible. Imagine slicing the 3x3x3 into three 3x3x1s. These are the bottom section, the middle section, and the top one. We determine how many possibilities are there for each of the 3 sections. You can make a bijection between a legal section and a path from the leftmost corner of the section to the rightmost. Each path corresponds to all the blocks being placed beyond it and closer to the back. There are $\binom{6}{3} = 20$ ways of doing this. There are now $20^3 = 8000$ possibilities to check. I know of no good way to proceed from here, so I wrote it in Python to check each of the 8000 cases. I am fairly sure of my answer's correctness.
I don't know if there is a non-bash way of doing this.
Using the MacMahon formula$^*$ for the number of plane partitions (stacks of boxes which are pushed all the way to the bottom, back and right) which fit in an $r\times s\times t$ box, the answer is $$ \prod_{i=1}^3\prod_{j=1}^3\frac{i+j+3-1}{i+j-1}=\frac41\cdot\frac52\cdot\frac52\cdot\frac63\cdot\frac63\cdot\frac63\cdot\frac74\cdot\frac74\cdot\frac85=980.$$ I do not know how to derive this formula.
$^*$See the second to last equation in that section, $N_1(r,s,t)=\dots$