determine chromatic polynomial i need to determine the chromatic polynomial of the following graph:

i know that the $P(k) = k(k − 1)(k − 2)· · ·(k − n + 1)$, so every vertex has one color. so i assume that total is  10 different colors.
but i do not know how to find the polynomial? do i need to change the graph?
every help, appreciate.
thanks 
 A: The formula you propose is only true for complete graphs, so it does not hold here.
Decompose your graph $G$ into three graphs $G_1,G_2,G_3$ where


*

*$G_1$ is the cycle of length $4$ on the left (vertices $\{i,g,h,j\}$), with chromatic polynomial
$$
P_{G_1}(k)=(k-1)^4+(-1)^4(k-1)
$$

*$G_2$ is the tree in the middle (vertices $\{g,f,e\}$), with chromatic polynomial
$$
P_{G_2}(k)=k(k-1)^2
$$

*$G_3$ is the cycle of length $5$ on the right (vertices $\{a,b,c,d,e\}$), with chromatic polynomial
$$
P_{G_3}(k)=(k-1)^5+(-1)^5(k-1)
$$


Then, use the fact that if two graphs $G$ and $H$ share a single vertex, the chromatic polynomial of $G \cup H$ is given by
$$
P_{G\cup H}= \frac{P_G(k)P_H(k)}{k}
$$
If I am not mistaken, you should get 
$$
\boxed{
P_{G=G_1\cup G_2 \cup G_3} = \frac{(k-1)^{11}+(k-1)^{8}-(k-1)^{7}-(k-1)^{4}}{k}
}
$$
To check if your answer is sensible, remember that the chromatic number of $G$ is the smallest integer that is not a root of the polynomial. Here, real roots are $1$ and $2$. It is straightforward to color the graph with $3$ colors, and since $G_3$ is a pentagon, at least three colors are necessary to color the graph. Therefore, this is in accordance with our chromatic polynomial.
