2021 AMC 10A Spring Problem 20

Below is the video solution and professionally curated solution for Problem 20 of the 2021 AMC 10A Spring, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 10A Spring solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:arrangements with restrictionsbijection

Difficulty rating: 1950

20.

In how many ways can the sequence 1,2,3,4,51,2,3,4,5 be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?

1010

1818

2424

3232

4444

Video solution:
Solution video thumbnail
Play video

Click to load, then click again to play

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, a direct count by the middle value, or equivalently the standard alternating-permutation count for five distinct numbers, gives 1616 permutations. Replacing every entry xx by 6x6-x gives a bijection from these to the down-up-down-up permutations, so there are another 1616.

The total number of valid rearrangements is 16+16=3216+16=32.

Thus, D is the correct answer.

Problem 20 in Other Years