2021 AMC 10A Spring 第 20 题
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20.
将数列 重新排列,有多少种排列使得不存在连续三项递增,也不存在连续三项递减?
In how many ways can the sequence be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?
答案:D
视频讲解:
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文字解答:
若出现两个连续的“升”或两个连续的“降”,就会有连续三项递增或递减。
因此四个相邻比较符号必须交替,只可能是升降升降或降升降升。 对升降升降型,直接按中间值计数,或用五个不同数的标准交错排列数,可得 个。 。 , 。 , ,
把每个数 替换为 ,可与降升降升型一一对应,所以另有 个。
总共有 个有效排列。
所以正确答案是 D。
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 must be in position or position If it is in position let the entry in position be Its two neighbors must be distinct numbers less than which can be ordered in ways. Summing over gives permutations. By symmetry there are another when is in position for a total of with this comparison pattern.
Replacing every entry by gives a bijection to the down-up-down-up permutations, so there are another
The total number of valid rearrangements is
Thus, D is the correct answer.
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