Find the chromatic polynomial of the following graph

Question

https://i.stack.imgur.com/Kyvbp.png

I am aware of the edge contraction method but after trying it out on this problem it seems like it will take too long.

So im going to try and solve this directly but im not very comfortable with this method, this is what I have so far.

1 has x options. 2 has (x-1) choices, 4 has (x-2) choices, 3 has (x-2) choices, 5 has (x-1) choices, 7 has (x-2) choices, and 6 has (x-2) choices, giving us

Pg(x) = x(x-1)^2(x-2)^4.

Is this correct?

Note "1" is the leftmost vertex, "2 and 3" are the adjacent vertices to "1", and so forth.

Answer 1

Try to prove the following:

Let denote $G$ a graph which is constructed "gluing" graphs $K$ and $H$ by a common vertex (like in your picture). We have that $P_G(x)=\frac{P_K(x)P_H(x)}{x}$

Answer 2

I think starting with the middle vertex is the best way to go, if only because of the symmetry of the problem.

Assume $x\geq 3$. In the middle vertex, you have $x$ choice of colour. Then the left top vertex and right top vertex each have $x-1$ choice. Then then left bottom and right bottom vertex each have $x-2$ choice. Then the leftmost and rightmost have $x-2$ choice. This leave us with $x(x-1)^{2}(x-2)^{4}$. Can easily check that this work for $x=1,2$ as well.

You get the same answer either way.