Back to: Chromatic Polynomial

Case 1.1: Calculation of \(P_{1.1^{\circ}}(k)\)

Observe that the vertex \(F\) cannot have any of the colors \(1\), \(3\), \(4\), while the vertex \(E\) cannot have any of the colors \(1\) and \(3\).

There are \(k-3\) colors available for the vertex \(F\). For each of the choices, there are \(k-3\) color available for the vertex \(E\) because the vertex \(E\) cannot have the colors \(1\), \(3\), and the color that is assigned to the vertex \(F\).

Hence we obtain \[P_{1.1^{\circ}}(k)=(k-3)^2.\]