Problem: A regular dodecahedron is a solid with 12 identical regular pentagonal faces. Imagine it oriented so that one face is directly on top and one is directly on the bottom. Each of these horizontal faces is surrounded by a ring of five faces that are slanted and adjacent to it—forming a top ring around the top face and a bottom ring around the bottom face. The remaining two faces connect the top and bottom rings, completing the solid. How many distinct paths are there from the top face to the bottom face, where each move goes to an adjacent face, no face is visited more than once, and once you enter the bottom ring, you are not allowed to return to the top ring? Solution: Grab a D12 if you have one or look at a picture and follow this solution. Let's pick paths greedily. There are $5$ ways to exit the top face, then $9$ ways to rotate around either direction of the first ring (or stay stationary). For each ending face, we have $2$ ways to get to the lower ring. Now, we have $9$ ways to rotate around or stay stationary and finally $1$ way to get to the last face. At each step, we perform a unique move that contributes to the constructed path, so there is no way we can follow this greedy approach with different choices and end up with the same path. Furthermore, it is clear every path can be written as a string of choices (this is to say yes, we do have a bijection between the paths we create and the desired set we wish to count). Hence, the answer is $5 \cdot 9 \cdot 2 \cdot 9 = 810$.