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.
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Difficulty rating: 1950
20.
In how many ways can the sequence be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?
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, a direct count by the middle value, or equivalently the standard alternating-permutation count for five distinct numbers, gives permutations. Replacing every entry by gives a bijection from these to the down-up-down-up permutations, so there are another .
The total number of valid rearrangements is .
Thus, D is the correct answer.
Problem 20 in Other Years
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