Find chromatic polynomial of a graph $K_{n,m}$ 
Find chromatic polynomial of a graph $K_{n,m}$

My solution:
 Let's define the lower and upper factorials by examples:


*

*lower factorial: $$t^{\underline{3}}=t(t-1)(t-2)$$

*upper factorial: $$t^{\overline{3}}=t(t+1)(t+2)$$
Knowing this my polynomial is:
$$W(t)=(t-{\underline{\text{min}(n,m)}})^2 (t-\text{min}(a,b))^{|n-m|}$$
Can you check my reasoning?
 A: *

*The chromatic polynomial $\chi_G(t)$ of a graph $G=(V,E)$ can always be written as $$\chi_G(t)=\sum_k a_k(t)_k,\qquad(t)_k=t(t-1)\ldots(t-k+1),$$ where $a_k$ is the number of partitions of $V$ into $k$ nonempty independent sets.

*Generalizing the construction of $K_{n,m}$, the graph join of two graphs is the graph obtained by joining (with an edge) each vertex of the first graph with each vertex of the second graph. Formally, the graph join of $G_1=(V_1,E_1)$ and $G_2=(V_2,E_2)$, $V_1\cap V_2=\emptyset$, is the graph $G_1\diamond G_2=(V,E)$ with $V=V_1\cup V_2$ and $E=E_1\cup E_2\cup(V_1\diamond V_2)$, where $V_1\diamond V_2=\big\{\{v_1,v_2\} : v_1\in V_1,v_2\in V_2\big\}$ is "unordered" $V_1\times V_2$.

*If $\chi_{G_1}(t)=\sum\limits_k a_k(t)_k$ and $\chi_{G_2}(t)=\sum\limits_k b_k(t)_k$, then $$\chi_{G_1\diamond G_2}(t)=\sum_k c_k(t)_k,\qquad c_k=\sum_j a_j b_{k-j}.$$ This can be seen from the interpretation of $a_k,b_k,c_k$ given in 1.

*Now $K_{n,m}=\overline{K}_n\diamond\overline{K}_m$ where $\overline{K}_n$ is the graph with $n$ vertices and no edges. As $\chi_{\overline{K}_n}(t)=t^n$, we need expressions connecting $t^n$ and $(t)_n$; this is where Stirling numbers appear. The outcome is $$\chi_{K_{n,m}}(t)=\sum_{k=0}^{m}\begin{Bmatrix}m\\k\end{Bmatrix}(t)_k(t-k)^n.$$
Here is an example to check the formula.
