1

Find number of coloring of the circuit $C_7$ with 3 colors so that no $3$ adjacent vertices have identical colors.

I think that it will be just: $$3^7 - \underbrace{3}_{\mbox{color for 3 adjacent vertices}} \cdot \underbrace{3^4}_{\mbox{rest}} = 1944$$ but it looks so simple, especially that this exercise is connected with burnside lemma.

  • 1
    If you proceed clockwise around the circle, there are seven potential starting points for three consecutive vertices of the same color. – N. F. Taussig Aug 03 '19 at 19:51
  • Here is similar question. Read it! – callculus42 Aug 03 '19 at 19:54
  • By your logic, if the circuit were $C_4$ there would be $9$ colorings with three of a color in a row. In fact there are three all the same color, plus $4$ (for which vertex is different) times $3$ (for the color of the three vertices) times $2$ (for the color of the odd vertex) for a total of $27$ – Ross Millikan Aug 03 '19 at 20:01

0 Answers0