Consider a graph with chromatic polynomial $P(x)$ joined to a clique of order $k$ in two distinct points (joining here just means interesection of points). Then, what is the chromatic polynomial of the new graph obtained by this joining?


Specifically, I think it is of the form $\frac{P(x)\cdot x\_{(k)}\cdot (x-2)}{x(x-1)^2}$, where $x\_{(k)}$ denotes the falling factorial. This is the form that was obtained when I experimented with some low order cliques. Is this true in general? If the two graphs share one single point or a clique in between them, then the resulting chromatic polynomial of the graph is well known. But, there is no formula in the above case. Though the deletion-contraction formula would give the appropriate polynomial, but is there a known formula beforehand? Thanks in advance. 



Chromatic Polynomial when two disjoint graphs are joined at $2$ distinct points
As Gordon Royle remarked, it may depend on the choice of vertices to attach to the clique.


Let $G$ be your original graph, and $a$ and $b$ your two points.  

If If $x \ge k$, an $x$-colouring of your new graph can be obtained by taking an $x$-colouring of $G$ in which $a$ and $b$ have different colours, and then colouring the remaining $k-2$ vertices of the $k$-clique using the $x-2$ colours not used for $a$ and $b$. Thus the number of such colourings is $(x-2)!/(x-k)!$ times the number $\widetilde{P}(x)$ of $x$-colourings of $G$ in which $a$ and $b$ have different colours.  

If $a$ and $b$ are adjacent in $G$, then $\widetilde{P} = P$. If they are not adjacent, then $\widetilde{P}(x) = P(x)- P\_{G\_{a=b}}(x)$, where $P\_{G\_{a=b}}(x)$ is the chromatic polynomial of the graph $G\_{a=b}$ obtained from $G$ by identifying $a$ with $b$. 


