2025 AMC 12A 第 23 题

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23.

若一个正整数不重复使用任何数字,没有数字 00,并且没有任何数字同时与两个更大的数字相邻,则称它为公平数。例如,1961962323, 和 1246312463 是公平数,但 15461546320320, 和 3432134321 不是。共有多少个公平正整数?

Call a positive integer fair if no digit is used more than once, it has no 00s, and no digit is adjacent to two greater digits. For example, 196,196, 23,23, and 1246312463 are fair, but 1546,1546, 320,320, and 3432134321 are not fair. How many fair positive integers are there?

511511

25842584

98419841

1771117711

1968219682

答案:C
知识点:有限制的排列二项式定理
难度评级:2340
解答:

数字互不相同,并且来自 {1,,9}\{1, \ldots, 9\},条件“没有数字同时与两个更大的数字相邻”等价于没有内部数字小于它的两个相邻数字。

对固定的 nn 个数字,从大到小插入来构造排列;每个新的较小数字必须放在当前排列的两个端点之一,因此有 2n12^{n-1} 个有效排列。

对所有非空数字子集求和, n=19(9n)2n1=12n=19(9n)2n=3912=196822=9841. \begin{gathered} \sum_{n=1}^{9}\binom{9}{n}2^{n-1} \\ = \frac{1}{2}\sum_{n=1}^{9}\binom{9}{n}2^{n} \\ = \frac{3^9 - 1}{2} = \frac{19682}{2} = 9841. \end{gathered}

因此,正确答案是 C

The digits are distinct and drawn from {1,,9},\{1, \ldots, 9\}, and "no digit adjacent to two greater digits" means no interior digit is smaller than both neighbors.

For a fixed set of nn digits, build the arrangement by inserting digits from largest to smallest; each new (smaller) digit must go to one of the two ends, giving 2n12^{n-1} valid arrangements.

Summing over all nonempty digit subsets, n=19(9n)2n1=12n=19(9n)2n=3912=196822=9841. \begin{gathered} \sum_{n=1}^{9}\binom{9}{n}2^{n-1} \\ = \frac{1}{2}\sum_{n=1}^{9}\binom{9}{n}2^{n} \\ = \frac{3^9 - 1}{2} = \frac{19682}{2} = 9841. \end{gathered}

Thus, the correct answer is C.

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