find chromatic polynomial of $K_{3,3}$ - full connected $3-3$ bipartite
my solution
I have considered $9$ cases:
Left side is colored using $1$ color AND right using $1$ color
Left side is colored using $1$ color AND right using $2$ colors
Left side is colored using $1$ color AND right using $3$ colors
Left side is colored using $2$ colors AND right using $1$ color
Left side is colored using $2$ colors AND right using $2$ colors
Left side is colored using $2$ colors AND right using $3$ colors
Left side is colored using $3$ color AND right using $1$ color
Left side is colored using $3$ color AND right using $2$ colors
Left side is colored using $3$ color AND right using $3$ colors
I got such result:
$$ f(t) = (t-2) (t-1) ((t-5) (t-4) (t-3)+3 (t-4) (t-3)+(t-3)) t+3 (t-1) ((t-4) (t-3) (t-2)+3 (t-3) (t-2)+(t-2)) t+((t-3) (t-2) (t-1)+3 (t-2) (t-1)+(t-1)) t $$
that gives me $$f(t) = -31 t + 78 t^2 - 75 t^3 + 36 t^4 - 9 t^5 + t^6 $$
some first values: $$f(1) = 0 \\ f(2) = 2 \\ f(3) = 42 \\ f(4) = 420 $$
It can be boiled down to 3 cases only:
Case 1: All vertices on the left have a different color, thus all vertices on the right can choose from $t-3$ colors $$t(t-1)(t-2)(t-3)^3$$ Case 2: Choose two vertices on the left to share a color $\binom{3}{2}=3$, the remaining vertex on the left will have a different color, hence all the vertices on the right side can choose from $t-2$ colors $$3t(t-1)(t-2)^3$$ Case 3: All vertices on the left share the same color $$t(t-1)^3$$
We get then $$t(t-1)(t-2)(t-3)^3 + 3t(t-1)(t-2)^3 + t(t-1)^3 = t^6 -9t^5 + 36t^4 -75t^3 + 78t^2 -31t$$