smc

This solution essentially explains other ways of thinking about the cases stated in Solution 2. Case 1: $\left(\genfrac{}{}{0ex}{}{6}{5}\right)\cdot 5!$ 5 colors need to be chosen from the group of 6. Those 5 colors have 5! distinct arrangements on the pentagon's vertices. Case 2: $\left(\genfrac{}{}{0ex}{}{6}{4}\right)\cdot 4\cdot 5\cdot 3!$ 4 colors need to be chosen from the group of 6. Out of those 4 colors, one needs to be chosen to form a pair of 2 identical colors. Then, for arranging this layout onto the pentagon, one of the five sides (same thing as pair of adjacent vertices) of the pentagon needs to be established for the pair. The remaining 3 unique colors can be arranged 3! different ways on the remaining 3 vertices. Case 3: $\left(\genfrac{}{}{0ex}{}{6}{3}\right)\cdot \left(\genfrac{}{}{0ex}{}{3}{2}\right)\cdot 5\cdot 2$ 3 colors need to be chosen from the group of 6. Out of those 3 colors, 2 need to be chosen to be the pairs. Then, for arranging this layout onto the pentagon, start out by thinking about the 1 color that is not part of a pair, as it makes things easier. It can be any one of the 5 vertices of the pentagon. The remaining 2 pairs of colors can only be arranged 2 ways on the remaining 4 vertices. Solving each case and adding them up gets you 3120. $\boxed{3120}$