How many ways can hexagonal tiles can be arranged in a triangle?
Let us assume that there is an equilateral triangle of side $s$. The triangle is subdivided by edges on $b$ segments, so that there are $b^2$ smaller equilateral triangles inside, in a triangular fragment of hexagonal lattice.
There are hexagonal tiles of side $a=s/b$.
How can we count the number ($L$) of different arrangement of the hexagonal tiles inside the triangle, if there is a condition that the tiles must touch by vertices or by sides and fit into triangle?
One of the solutions is as in uniform tiling hexadeltile $3.6^2$. The other one is a fragment of regular tiling - homeycomb ($6^3$). But how many total ways ($L$) there are?
The problem is also in finding the number of cases with the same number ($C$) of the hexagonal tiles within the given subdivision.
For example:
if we divide the triangle on $b=5$
the total number of different arrangements $L= 4$.
There are: $1$ case ($L_1$) with $3$ tiles ($C_1=3$) and 3 cases ($L_3$) with $2$ tiles ($C_2=3$).
I have tried to solve this with some combinatorics, but the best results I could achieve is to draw all the cases (from $b = 3$ to $b=8$) and count them... Help.
