Wolfram-Alpha says that the chromatic polynomial for the path with $4$ vertices, $P_4$, is $x(x-1)^3.$ But putting $x = 4$ gives $108$, wheres the number of $4-$colorings should be $4!$, should it not? Am I misinterpreting the meaning of the chromatic polynomial?
2026-04-09 17:06:49.1775754409
Why does the chromatic polynomial for $P_4$ not output $4!$ for input $4$?
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$4!$ is the number of colourings using four colours on the complete graph on $4$ vertices. For the path, you have the opportunity to use the same colour on two different vertices in a number of different combinations, opening up for many more colourings. (No, you don't have to use all colours at your disposal, as that would, for instance, make it impossible to use $5$ colours.)