2021 AMC 10A Spring Problem 20

Question

In how many ways can the sequence

1,

2,

3,

4,

5

be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?

10

18

24

32

44

Answer: D

Concepts:arrangements with restrictions bijection

Difficulty rating: 1950

Small hint:

A valid permutation must have comparison signs that alternate

Big hint:

Count the up-down-up-down permutations and use symmetry for the reverse pattern

Video solution:

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Written solution:

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.