This question is a continuation of this one.
Here are nine squares, connected by lines.
Each square must be colored, and two squares connected by a line must be colored differently.
Question 1. What is the minimum number of colors required?
Question 2. How many different colorings are possible with the minimum number of colors?
Let's name the squares (nodes in a graph) as follows: $T$ is the top node, $L1, \dots, L4$ denote the four nodes on the left column (from top to bottom), and $R1, \dots, R4$ similarly for the right column.
The answer is
3 colors.
Reasoning:
First, observe that $T, L3, L4$ form a triangle, so we'll need at least three colors.
Then, excluding the node $T$, I noticed that the remaining graph of 8 nodes is a bipartite graph in disguise, so the 8 nodes can be colored with just two colors: color $L1, L3, R1, R3$ red, and $L2, L4, R2, R4$ blue.
Finally, we can assign a third color to the remaining $T$ (say, green), completing the proof that the graph can be colored with exactly 3 colors.
The answer is
the graph can be colored using 3 colors in 42 different ways.
Reasoning:
We can color the triangle $T, L3, L4$ using three colors in $3! = 6$ different ways. Then $R3$ is forced by the triangle $T, L4, R3$, and then $R4$ by $T, R3, R4$.
Now assume $T$ is green, $L3, R3$ are red, and $L4, R4$ are blue. $L2$ and $R2$ can be either green or blue.
If at least one of $L2$ and $R2$ is blue, both $L1$ and $R1$ are forced to be red (3 possibilities).
If both $L2$ and $R2$ are green, all three neighbors of both $L1$ and $R1$ are green, giving two choices for each (4 possibilities).
In total, the graph can be colored using three colors in $3! \times (3+4) = 42$ different ways.
