Calculating a chromatic polynomial that contains several identical subgraphs. So I learned about the theorem that states that if a graph $G$ has two subgraphs $X$ and $Y$ such that $X \cup Y = G$ and $X\cap  Y=K_n$, where $K_n$ is an $n$-clique, then the chromatic polynomial of $G$ is given by $$cr(G)=\frac{cr(X)cr(Y)}{cr(K_N)}$$Can this theorem be generalized so that instead of $K_n$ we have a graph $A$ and instead of $X$ and $Y$ we have a family of graphs $\mathcal X=\{X\subseteq G:\forall_{Y\subseteq G\setminus\{X\}} X\cap Y=A\}$, and the chromatic polynomial of G is given by $$cr(G)=\frac{1}{cr(A)}\prod_{X\in \mathcal X}cr(X)$$
 A: There's some restricted cases when it works, e.g. if $X$ and $Y$ are both trees, but not in general.
Suppose a $k$-coloring of $X$ is possible, and we list all $\mathrm{cr}_k(X)$ distinct $k$-colorings of $X$.  By permuting the $k$ colors, we find that each possible $k$-coloring of its $n$-clique occurs exactly
$$\frac{\mathrm{cr}_k(X)}{k(k-1)\cdots(k-n+1)}=\frac{\mathrm{cr}_k(X)}{\mathrm{cr}_k(K_n)}$$
times.
This special property does not hold for any subgraph $A$ other than $K_n$: if an $n$-vertex graph has a non-edge, then there are ways to color it with $n$ colors, and there are ways to color it with $n-1$ colors.  So if we again list of $k$-colorings of $X$ (where $k \geq n$), some $k$-colorings of its subgraph $A$ will occur fewer times than others.
As a concrete counterexample, let's glue $C_6$ and $C_7$ along a $5$-vertex path $P_5$, and count $3$-colorings (the chromatic polynomials of these graphs are well known; see Wikipedia).



*

*$P_5$ has $48$ distinct $3$-colorings.

*$C_6$ has $66$ distinct $3$-colorings.

*$C_7$ has $126$ distinct $3$-colorings.


But we find $66 \times 126/48$ is not even an integer, so $\mathrm{cr}(C_6)\mathrm{cr}(C_7)/\mathrm{cr}(P_4)$ is not the chromatic polynomial of any graph (since we get a non-integer when we evaluate it at $3$).
(Note: there's no squared in the denominator in the formula.)
