Question

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 $m + n.$

Answer

Solution:

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 $\frac{1}{26} + \frac{1}{10} - \frac{1}{26} \cdot \frac{1}{10} = \frac{10 + 26 - 1}{260} = \frac{35}{260} = \frac{7}{52}.$ Thus $m + n = 7 + 52 = 59.$

2002 AIME I Problem 1

Solution:

Problem 1 in Other Years