2021 AMC 10A Problem 20

1. Between each pair of neighbours the sequence goes either up or down. Two ups in a row would make three increasing terms, and two downs in a row three decreasing terms. So ups and downs must alternate.
2. That leaves two patterns: up, down, up, down (like $2, 5, 1, 4, 3$) and down, up, down, up.
3. Count the up-down pattern first: $a_1 < a_2 > a_3 < a_4 > a_5$. Positions $2$ and $4$ are "peaks" (bigger than both neighbours) and positions $1, 3, 5$ are "valleys".
4. The biggest number, $5$, cannot be in a valley, because a valley is smaller than a neighbour. So $5$ is at position $2$ or position $4$.
5. $5$ at position $2$. The condition $a_1 < 5$ holds automatically, so $a_1$ can be any of the other four numbers: $4$ choices. The three numbers left fill positions $3, 4, 5$ with $a_3 < a_4 > a_5$, so the biggest of the three goes in the middle and the other two can go either way round: $2$ choices. That is $4 \times 2 = 8$ ways.
For example, with $a_1 = 3$ the numbers left are $1, 2, 4$, giving $3, 5, 1, 4, 2$ and $3, 5, 2, 4, 1$.
6. $5$ at position $4$. Reading a sequence backwards turns this case into the last one, so it also gives $8$ ways.
7. So there are $16$ up-down sequences.
8. For the down-up pattern, replace every number $k$ by $6 - k$ (so $1 \leftrightarrow 5$, $2 \leftrightarrow 4$, and $3$ stays). This turns every up into a down, so each up-down sequence becomes a down-up one: $3, 5, 1, 4, 2$ becomes $3, 1, 5, 2, 4$. There are $16$ down-up sequences too.
9. Total: $16 + 16 = 32$.

The answer is $\boxed{\textbf{(D)}\ 32}$.

"No three in a row going the same way" forces the sequence to zigzag, and in a zigzag the biggest number must be a peak. Placing the biggest number first makes the rest of the count short, and two flips (reading backwards, and replacing $k$ by $6 - k$) mean only one small case has to be counted by hand.

In the up-down pattern, count by which two numbers are the peaks (positions $2$ and $4$).
- Peaks $5$ and $4$: they can be in either order, and the valleys $1, 2, 3$ can go anywhere, since every valley is smaller than every peak. That is $2 \times 6 = 12$ ways.
- Peaks $5$ and $3$: the valley $4$ is bigger than $3$, so it may only touch $5$. It must sit at the end next to $5$, and then $1$ and $2$ fill the other two valleys in $2$ ways. With $5$ in either peak spot, that is $2 \times 2 = 4$ ways.
- A peak of $2$ or less is impossible, since it would need two smaller neighbours.

That gives $12 + 4 = 16$, and doubling for the down-up pattern gives $32$.