Contiguous Triangle Coloration I somehow have this knack of imagining problems of my own, which I then myself find very hard to approach.  So there is no textbook to refer to, for the correct answer.
Here is a latest example of my ruminations, which goes into combinatorics
In how many ways can the triangle given below, consisting of 9 smaller triangles be colored, starting with 5 distinct colors, so that no two triangles that share an edge are of the same color?

I can sense that the problem can get more complicated as we venture into larger set of 16, 25, 36 and higher number of triangles in this arrangement having perfect square number of objects to fill. A generalization is most welcome, if it exists, but I would be glad even with a specific solution!!
 A: It’s actually a graph coloring problem: in how many ways can the $9$ vertices of the graph shown below be colored with $5$ colors so that adjacent vertices do not have the same color?
                       *
                       |
                       *
                      / \
                     *   *
                    /     \
               *---*---*---*---*

The three vertices of degree $3$ are the key to counting such colorings.
Suppose first that all three of them are the same color; then each of the remaining six vertices can be any of the other $4$ colors. There are $5$ choices for the color of the vertices of degree $3$ and $4^6$ choices for the colors of the other six vertices, so for this case we have a total of $5\cdot4^6$ possible colorings.
Now suppose that the three vertices of degree $3$ get three different colors; there are $5\cdot4\cdot3=60$ ways to assign $3$ of the $5$ colors to those three vertices. There are $4$ choices of color for each of the vertices of degree $1$ and $3$ choices for each of the vertices of degree $2$, so for this case we have a total of $60\cdot 4^3\cdot 3^3$ colorings.
Finally, suppose that the three vertices of degree $3$ are colored with two different colors. There are $5$ ways to choose the color that appears twice, $3$ ways to assign it to two of these vertices, and $4$ ways to choose a color for the third of these vertices, so there are $60$ ways to color those three vertices with two colors. Each of the vertices of degree $1$ and the vertex between the two vertices of the same color can be colored with any of $4$ colors, and each of the remaining two vertices can be colored with any of $3$ colors, so for this case we have a total of $60\cdot4^4\cdot 3^2$ colorings.
The grand total is then $20\,480+103\,680+138\,350=262\,400$.
