Problem Statement
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 $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
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Solution
$$\text{For a three-letter arrangement to be a palindrome, the last letter must be the same as the first letter}$$ $$\implies \text{The probability of a three-letter arrangement to be a palindrome is } \dfrac{1}{26}$$ $$\text{Similar logic can be applied to the three-digit number}$$ $$\implies \text{The probability of a three-digit arrangement to be a palindrome is } \dfrac{1}{10}$$ $$\therefore \text{The probability of either of these events happening is } \dfrac{1}{26} + \dfrac{1}{10} - \dfrac{1}{260} = \dfrac{35}{260} = \dfrac{7}{52}$$ $$\text{Let } m/n = 7/52 \implies m+n = 59$$ 