AMC12 2025 A

AMC12 2025 A · Q13

AMC12 2025 A · Q13. It mainly tests Basic counting (rules of product/sum), Pigeonhole principle.

Let $C = \{1, 2, 3, \dots, 13\}$. 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?

令 $C = \{1, 2, 3, \dots, 13\}$。令 $N$ 为最大整数，使得存在 $C$ 的一个 $N$ 元子集不包含五个连续整数。從 $C$ 中不放回地随机选择 $N$ 个整数。所选元素不包含五个连续整数的概率是多少？

(A) \frac{3}{130} \frac{3}{130}

(B) \frac{3}{143} \frac{3}{143}

(D) \frac{1}{26} \frac{1}{26}

(E) \frac{5}{78} \frac{5}{78}

Answer

Correct choice: (D)

正确答案：（D）

Solution

We first find what $N$ is by figuring out how many numbers we need to take out of the set so that the set does not contain $5$ consecutive integers. Since $N$ must be maximized, we must minimize what numbers are removed, and we quickly find that taking two numbers out works. Consider taking out $5$ and $10$. You are left with $\{1,2,3,4,6,7,8,9,11,12,13\}$, which does not have a string of $5$ consecutive integers. There are only $3$ ways to take out two integers such that the resulting set meets our condition ($5$ and $10$, $5$ and $9$, or $4$ and $9$), and ${\binom{13}{2}}=78$ total ways to choose such integers. Therefore, the probability is $\boxed{\dfrac{1}{26}}$. Minor edits ~aashrithm29

首先通过找出需要从集合中取出多少个数使得剩余集合不包含 $5$ 个连续整数来求 $N$。由于 $N$ 需要最大化，我们必须最小化移除的数，很快发现取出两个数即可。考虑取出 $5$ 和 $10$。剩余集合为 $\{1,2,3,4,6,7,8,9,11,12,13\}$，不包含 $5$ 个连续整数。只有 $3$ 种取出两个整数使得结果集合满足条件的方式（$5$ 和 $10$、$5$ 和 $9$ 或 $4$ 和 $9$），而总方式数 ${\binom{13}{2}}=78$。因此概率为 $\boxed{\dfrac{1}{26}}$。 Minor edits ~aashrithm29

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