2022 AMC 12A 第 24 题

先试着解答 2022 AMC 12A 第 24 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2022 AMC 12A 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

24.

用数字 00,11,22,33,44 组成长度为 55 的字符串。有多少个这样的字符串满足:对每个 j∈{1,2,3,4}j\in\{1,2,3,4\},至少有 jj 个数字小于 jj?(例如,0221402214 满足条件,因为它含有至少 11 个小于 11 的数字,至少 22 个小于 22 的数字,至少 33 个小于 33 的数字,且至少 44 个小于 44 的数字。字符串 2340423404 不满足条件,因为它不含至少 22 个小于 22 的数字。)
How many strings of length 55 formed from the digits 0,0, 1,1, 2,2, 3,3, 44 are there such that for each j∈{1,2,3,4},j\in\{1,2,3,4\}, at least jj of the digits are less than j?j? (For example, 0221402214 satisfies the condition because it contains at least 11 digit less than 1,1, at least 22 digits less than 2,2, at least 33 digits less than 3,3, and at least 44 digits less than 4.4. The string 2340423404 does not satisfy the condition because it does not contain at least 22 digits less than 2.2.)
500500
625625
10891089
11991199
12961296
答案:E
知识点:有限制的排列分类讨论
小提示:
“至少有 jj 个数字小于 jj” 意味着第 jj 小的数字至多为 j−1j-1。
“At least jj digits are less than jj” means the jjth smallest digit is at most j−1j-1
大提示:
这些正是长度为 55 的停车函数;使用 66 个车位的圆形停车计数
These are parking functions of length 5;5; use the circular-parking count with 66 spaces
解答:
将五个数字排序为 d(1)≤d(2)≤⋯≤d(5)d_{(1)}\le d_{(2)}\le\cdots\le d_{(5)}。条件“至少有 jj 个数字小于 jj”等价于对 j=1,2,3,4j=1,2,3,4,有 d(j)≤j−1d_{(j)}\le j-1,即 d(1)=0, d(2)≤1d_{(1)}=0,\ d_{(2)}\le1, d(3)≤2, d(4)≤3\ d_{(3)}\le2,\ d_{(4)}\le3(而 d(5)≤4d_{(5)}\le4 自动成立)。
这些字符串恰好是长度为 55 的停车函数。为计数,将 66 个停车位排成圆形,让 55 辆有标号的汽车任意选择首选车位。每辆车向前行驶到第一个空位。在 656^5 个偏好字符串中,同时旋转所有偏好会使唯一的空位依次经过全部 66 个位置。因此,恰有 656=64=1296\frac{6^5}{6}=6^4=1296 个字符串使指定车位为空。将这个车位选作额外的第六个车位,恰好得到上述排序不等式。
因此,正确答案是 E。
Sort the five digits as d(1)≤d(2)≤⋯≤d(5).d_{(1)}\le d_{(2)}\le\cdots\le d_{(5)}. The requirement “at least jj digits less than jj” is equivalent to d(j)≤j−1d_{(j)}\le j-1 for j=1,2,3,4,j=1,2,3,4, i.e. d(1)=0, d(2)≤1,d_{(1)}=0,\ d_{(2)}\le1,  d(3)≤2, d(4)≤3\ d_{(3)}\le2,\ d_{(4)}\le3 (with d(5)≤4d_{(5)}\le4 automatic).
These strings are exactly the parking functions of length 5.5. To count them, arrange 66 parking spaces in a circle and let 55 labeled cars choose arbitrary preferred spaces. Each car moves forward to the first open space. Among the 656^5 preference strings, rotating all preferences cycles the unique empty space through all 66 positions. Therefore exactly 656=64=1296\frac{6^5}{6}=6^4=1296 strings leave a specified space empty. Choosing that space as the extra sixth space gives precisely the sorted inequalities above.
Thus, the correct answer is E.
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