Question

令 $C = \{1, 2, 3, \ldots, 13\}$ 。令 $N$ 为最大的整数，使得存在一个有 $N$ 个元素的 $C$ 的子集，并且该子集不含五个连续整数。现在从 $C$ 中随机不放回地选出 $N$ 个整数。所选元素不包含五个连续整数的概率是多少？

Let $C = \{1, 2, 3, \ldots, 13\}.$ Let $N$ be the greatest integer such that there exists a subset of $C$ with $N$ elements that does not contain five consecutive integers. Suppose $N$ integers are chosen at random from $C$ without replacement. What is the probability that the chosen elements do not include five consecutive integers?

$\dfrac{3}{130}$

$\dfrac{3}{143}$

$\dfrac{5}{143}$

$\dfrac{1}{26}$

$\dfrac{5}{78}$

答案：D

知识点：组合基本概率极端原理

难度评级：1660

解答：

要避免五个连续整数，只需去掉两个元素（例如 $5$ 和 $10$ ），而去掉一个元素不可能同时击中两个不相交的区块 $\{1,\ldots,5\}$ 和 $\{9,\ldots,13\}.$ 因此 $N = 11.$

从 $13$ 个元素中选 $11$ 个，等价于去掉 $2,$ 个，共有 $\binom{13}{2} = 78$ 种。所选集合不含五个连续整数，当且仅当两个被去掉的元素合起来与每个窗口 $\{t, t+1, t+2, t+3, t+4\}$ 相交，其中 $t = 1, \ldots, 9.$

这迫使一个被去掉的元素在 $\{1,\ldots,5\},$ 中，另一个在 $\{9,\ldots,13\},$ 中，并且两者相距不超过 $5$ 。有效的去法为 $\{4,9\},$ $\{5,9\},$ 和 $\{5,10\},$ 共 $3$ 种。

概率为 $\dfrac{3}{78} = \dfrac{1}{26}.$

因此，正确答案是 D。

To avoid five consecutive integers, it suffices to remove two elements (for example $5$ and $10$ ), and no single removal can hit both disjoint blocks $\{1,\ldots,5\}$ and $\{9,\ldots,13\}.$ Thus $N = 11.$

Choosing $11$ of $13$ elements is the same as removing $2,$ which can be done in $\binom{13}{2} = 78$ ways. The chosen set avoids five consecutive integers exactly when the two removed elements together intersect every window $\{t, t+1, t+2, t+3, t+4\}$ for $t = 1, \ldots, 9.$

This forces one removed element in $\{1,\ldots,5\},$ the other in $\{9,\ldots,13\},$ and the two within $5$ of each other. The valid removals are $\{4,9\},$ $\{5,9\},$ and $\{5,10\},$ giving $3$ of them.

The probability is $\dfrac{3}{78} = \dfrac{1}{26}.$

Thus, the correct answer is D.

2025 AMC 12A 第 13 题

答案：D

难度评级：1660

解答：

其他年份的第 13 题