2021 AMC 10A Problem 20

Between consecutive terms the sequence either rises or falls. Two rises in a row would be three increasing terms, and two falls in a row three decreasing terms, so rises and falls must alternate. There are two shapes: up-down-up-down ($a_1 < a_2 > a_3 < a_4 > a_5$) and down-up-down-up.

Count the up-down shape. The value $5$ can only sit at a peak, position $2$ or $4$.

• $5$ in position $2$: the condition $a_1 < 5$ is automatic, so $a_1$ can be any of the other four values. The remaining three values fill positions $3, 4, 5$ with $a_3 < a_4 > a_5$: $a_4$ must be their maximum and $a_3, a_5$ can be in either order, $2$ ways. Total $4 \cdot 2 = 8$.
• $5$ in position $4$: the mirror image (read the sequence backwards), also $8$.

So there are $16$ up-down arrangements. Replacing each term $k$ by $6 - k$ flips every comparison, matching up-down arrangements with down-up ones, so there are $16$ of those too.

Total: $16 + 16 = 32$. The answer is $\boxed{\textbf{(D)}\ 32}$.

"No three consecutive increasing or decreasing" forces a strictly alternating pattern, and in an alternating pattern the largest element is always a peak. Placing the extreme element first collapses the count, and two symmetries (reversal, complement) mean only one small case needs to be counted by hand. These are the zigzag permutations; their count $2 \cdot 16$ comes from the Euler number $E_5 = 16$.

For the up-down shape, the peaks are $a_2$ and $a_4$, each larger than its neighbors. If the peaks are $\{5, 4\}$, the valleys $1, 2, 3$ go anywhere: $2 \cdot 3! = 12$ ways. If the peaks are $\{5, 3\}$, the valley $4$ must touch only the peak $5$, so it sits at an end next to $5$ and the other two valleys fill in: $4$ ways. No other peak set works. That gives $16$, then double for the down-up shape.

• Forgetting the down-up pattern (answering 16), or placing 5 in a valley position, where it cannot be smaller than a neighbor.
• Counting the two shapes for each placement of $5$ and then also doubling, which overcounts to $64$.