2020 AMC 10A Problem 19

First understand which faces touch which. Each face in the top ring touches the top face, its two neighbors in the ring, and two faces of the bottom ring. The bottom ring is the same, flipped over.

A route cannot go back to the top face (no repeats), and it cannot climb from the bottom ring to the top ring. So every route looks like this:

top face $\to$ some faces of the top ring $\to$ some faces of the bottom ring $\to$ bottom face.

Count each stage.

1. Enter the top ring. Any of its $5$ faces: $5$ ways.
2. Walk along the top ring. Label the ring faces $1, 2, 3, 4, 5$ in a circle and say you entered at face $1$. You may:
- stop right away at face $1$: $1$ way;
- go clockwise $1$, $2$, $3$ or $4$ steps: $4$ ways;
- go counterclockwise $1$, $2$, $3$ or $4$ steps: $4$ ways.

You cannot turn back, because that would repeat a face. So there are $1 + 4 + 4 = 9$ ways, and each one ends at a definite face.
3. Step down. The face where you stopped touches $2$ faces of the bottom ring: $2$ ways.
4. Walk along the bottom ring. Same reasoning as step 2: $9$ ways.
5. Step onto the bottom face. Every bottom-ring face touches it: $1$ way.

Multiply the stages:
$$5 \times 9 \times 2 \times 9 \times 1 = 810.$$

The answer is $\boxed{\textbf{(E)}\ 810}$.

The rule against going back up forces every route to pass through the layers in order. Then the choices at each stage do not depend on earlier choices (only on where you are), so the total is a product. On a ring of $5$ faces, a walk that never repeats a face can stop immediately or go $1$ to $4$ steps in either direction, which gives $9$ walks.