Problem: As shown in the figure below, a regular dodecahedron (the polyhedron consisting of $12$ congruent regular pentagonal faces) floats in space with two horizontal faces. Note that there is a ring of five slanted faces adjacent to the top face, and a ring of five slanted faces adjacent to the bottom face. How many ways are there to move from the top face to the bottom face via a sequence of adjacent faces so that each face is visited at most once and moves are not permitted from the bottom ring to the top ring? [asy] import graph; unitsize(5cm); pair A = (0.082, 0.378); pair B = (0.091, 0.649); pair C = (0.249, 0.899); pair D = (0.479, 0.939); pair E = (0.758, 0.893); pair F = (0.862, 0.658); pair G = (0.924, 0.403); pair H = (0.747, 0.194); pair I = (0.526, 0.075); pair J = (0.251, 0.170); pair K = (0.568, 0.234); pair L = (0.262, 0.449); pair M = (0.373, 0.813); pair N = (0.731, 0.813); pair O = (0.851, 0.461); path[] f; f[0] = A--B--C--M--L--cycle; f[1] = C--D--E--N--M--cycle; f[2] = E--F--G--O--N--cycle; f[3] = G--H--I--K--O--cycle; f[4] = I--J--A--L--K--cycle; f[5] = K--L--M--N--O--cycle; draw(f[0]); axialshade(f[1], white, M, gray(0.5), (C+2*D)/3); draw(f[1]); filldraw(f[2], gray); filldraw(f[3], gray); axialshade(f[4], white, L, gray(0.7), J); draw(f[4]); draw(f[5]); [/asy] $\textbf{(A) } 125 \qquad \textbf{(B) } 250 \qquad \textbf{(C) } 405 \qquad \textbf{(D) } 640 \qquad \textbf{(E) } 810$

Solution: Like Solution 3, we can use casework, but in a different way. We can split the caseworks into the amount of moves we spend on each row (top and bottom, where both rows have 5 faces).
Case 1: 1 move on the top row;
There are 5 ways to make 1 move on the top row from the starting point (just moving on each of the five faces). Now we want to look at how many ways we can move towards the bottom. Notice how we have two faces on the bottom row that are adjacent to each of the top faces. If we spend one move on the bottom, there are 2 ways. If we spend two moves on the bottom, there are 4 ways (since order matters, we can flip the direction of these moves). If we spend three moves on the bottom, there are also 4 ways (similar reasoning). If we spend four moves on the bottom, there are 4 ways. And if we spend five moves on the bottom, there are again, 4 ways. In total there are $2 + 4 + 4 + 4 + 4 = 18$ ways to move from the bottom after the top moves are chosen. And since there are 5 ways to make 1 move on the top row, there are $5 \cdot 18 = 90$ ways.
Case 2: 2 moves on the top row;
From just counting, we can see that there are five ways to move once on the top, and after that, two options for the second move (either left or right), so there are $5 \cdot 2 = 10$ ways. And from using what we obtained in Case 1, there are $10 \cdot 18 = 180$ ways for this case.
Case 3: three moves on the top row;
There are, again, 10 ways to do this. And so, there are $10 \cdot 18 = 180$ ways for this case to happen.
Now, we can see that there will be 10 ways for the remaining cases of the first move for the top row (because 5 ways to choose the first, and then you can go either left or right, so $5 \cdot 2 = 10$ ).
So for Case 4 and Case 5, there will both also be $10 \cdot 18 = 180$ ways.
Finally, adding all these cases up, we obtain $90 + 180 + 180 + 180 + 180 = 810$ , or $\boxed{810}$ ~Misclicked

Answer: E 810