2021 AMC 10A Spring 第 20 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

20.

将数列 1,2,3,4,51,2,3,4,5 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 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

答案:D
知识点:有限制的排列双射
难度评级:1950
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 1616 个。 55 22 442244 rrrr(r1)(r2)(r-1)(r-2) r=1,2,3,4r=1,2,3,4 88 55 440+0+2+6=80+0+2+6=8

把每个数 xx 替换为 6x6-x,可与降升降升型一一对应,所以另有 1616 个。

总共有 16+16=3216+16=32 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 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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