2002 AIME I Problem 1

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

Answer: 59

Concepts:basic probability inclusion-exclusion palindrome

Difficulty rating: 1890

Small hint:

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

\frac{1}{26}

Big hint:

Combine the letter and digit events by inclusion-exclusion:

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

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

\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.