2002 AIME I 第 1 题

Question

许多州的标准车牌格式是三个字母后接三个数字。假设每一种三字母三数字的排列等可能出现，这种车牌至少含有一个回文（三个字母的排列或三个数字的排列从左到右读和从右到左读相同）的概率为

\frac{m}{n}

，其中

m

和

n

是互质正整数。求

m + n

。

Many states use a sequence of three letters followed by a sequence of three digits as their standard license-plate pattern. Given that each three-letter three-digit arrangement is equally likely, the probability that such a license plate will contain at least one palindrome (a three-letter arrangement or a three-digit arrangement that reads the same left-to-right as it does right-to-left) is

\frac{m}{n},

where

m

and

n

are relatively prime positive integers. Find

m + n.

答案

答案：59

知识点：基本概率容斥原理回文数

难度评级：1890

小提示：

三个字母的排列是回文，当且仅当第三个字母等于第一个字母，所以其概率为

\frac{1}{26}

。

A three-letter arrangement is a palindrome exactly when the third letter equals the first, so that probability is

\frac{1}{26}

大提示：

用容斥合并字母事件和数字事件：

\frac{1}{26} + \frac{1}{10} - \frac{1}{26} \cdot \frac{1}{10}

。

Combine the letter and digit events by inclusion-exclusion:

\frac{1}{26} + \frac{1}{10} - \frac{1}{26} \cdot \frac{1}{10}

解答：

三个字母的排列是回文，当且仅当第三个字母与第一个字母相同，所以字母部分为回文的概率是

\frac{1}{26}

。同理，数字部分为回文的概率是

\frac{1}{10}

，且这两个事件相互独立。

由容斥原理，至少含有一个回文的概率为

\begin{aligned} &\frac{1}{26} + \frac{1}{10} - \frac{1}{26} \cdot \frac{1}{10} \\ &= \frac{10 + 26 - 1}{260} \\ &= \frac{35}{260} \\ &= \frac{7}{52} \end{aligned}

因此

m + n = 7 + 52 = 59

。

A three-letter arrangement is a palindrome exactly when the third letter matches the first, so the probability of a letter palindrome is

\frac{1}{26}.

Similarly, the probability of a digit palindrome is

\frac{1}{10},

and the two events are independent.

By inclusion-exclusion, the probability of at least one palindrome is

\begin{aligned} &\frac{1}{26} + \frac{1}{10} - \frac{1}{26} \cdot \frac{1}{10} \\ &= \frac{10 + 26 - 1}{260} \\ &= \frac{35}{260} \\ &= \frac{7}{52}. \end{aligned}

Thus

m + n = 7 + 52 = 59.

2002 AIME I 第 1 题

答案：59

难度评级：1890

小提示：

大提示：

解答：

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