2021 AMC 10A Spring Problem 20

Attempt Problem 20 of the 2021 AMC 10A Spring below, then check your answer against the video solution and professionally curated solution 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.

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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

Answer: D
Concepts:arrangements with restrictionsbijection
Difficulty rating: 1950
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 55 must be in position 22 or position 4.4. If it is in position 2,2, let the entry in position 44 be r.r. Its two neighbors must be distinct numbers less than r,r, which can be ordered in (r1)(r2)(r-1)(r-2) ways. Summing over r=1,2,3,4r=1,2,3,4 gives 0+0+2+6=80+0+2+6=8 permutations. By symmetry there are another 88 when 55 is in position 4,4, for a total of 1616 with this comparison pattern.

Replacing every entry xx by 6x6-x gives a bijection to the down-up-down-up permutations, so there are another 16.16.

The total number of valid rearrangements is 16+16=32.16+16=32.

Thus, D is the correct answer.

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