The rooms in the circular house plan shown below are to be painted using eight colors such that rooms with a common doorway must be different colors. In how many ways can this be done?
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The foregiong question is solved by exclusion-inclusion.The given answer is $$8^4-\binom{6}{1}8^3 +\binom{6}{2}8^2-\bigg[\bigg(\binom{6}{3}-2\bigg)8 +2 \times8^2\bigg]+\bigg[\binom{6}{4}-\binom{6}{5}+\binom{6}{6}\bigg]8$$
I agree with given answer except for the part $$\bigg[\bigg(\binom{6}{3}-2\bigg)8 +2 \times8^2\bigg]$$
because I think that there are more than $2$ selections which will cause $8^2$
Here my marked picture for my solution
I think that selection of $(e2,e4,e6),(e1,e2,e3),(e3,e4,e5),(e1,e5,e6)$ cause $8^2$, so the answer should have been $$8^4-\binom{6}{1}8^3 +\binom{6}{2}8^2-\bigg[\bigg(\binom{6}{3}-4\bigg)8 +4 \times8^2\bigg]+\bigg[\binom{6}{4}-\binom{6}{5}+\binom{6}{6}\bigg]8$$
Am I right or missing something ?
Each two rooms have a common door. Therefore each two rooms have different colors. Therefore $4$ distinct colors are used for the rooms. The ways to color $4$ rooms distinctly using $8$ colors are $\binom{8}{4}*4! = 1680$ which matches your answer. Your reasoning is correct.