2021 AMC 10A Problem 20

A permutation is valid exactly when the four comparison signs between consecutive terms alternate. Thus the signs must be either up-down-up-down or down-up-down-up. For the up-down-up-down pattern, the largest entry $5$ must be in position $2$ or position $4.$ If it is in position $2,$ let the entry in position $4$ be $r.$ Its two neighbors must be distinct numbers less than $r,$ which can be ordered in $(r-1)(r-2)$ ways. Summing over $r=1,2,3,4$ gives $$0+0+2+6=8$$ permutations. By symmetry there are another $8$ when $5$ is in position $4,$ for a total of $16$ with this comparison pattern. Replacing every entry $x$ by $6-x$ gives a bijection to the down-up-down-up permutations, so there are another $16.$ The total number of valid rearrangements is $16+16=32.$ Thus, D is the correct answer.