2002 AIME I Problems/Problem 1

Problem

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 $m+n.$

Solution 1

Consider the three-digit arrangement, $\overline{aba}$ . There are $10$ choices for $a$ and $10$ choices for $b$ (since it is possible for $a=b$ ), and so the probability of picking the palindrome is $\frac{10 \times 10}{10^3} = \frac 1{10}$ . Similarly, there is a $\frac 1{26}$ probability of picking the three-letter palindrome.

By the Principle of Inclusion-Exclusion, the total probability is

$\frac{1}{26}+\frac{1}{10}-\frac{1}{260}=\frac{35}{260}=\frac{7}{52}\quad\Longrightarrow\quad7+52=\boxed{059}$

Solution 2

Using complementary counting, we count all of the license plates that do not have the desired property. In order to not be a palindrome, the first and third characters of each string must be different. Therefore, there are $10\cdot 10\cdot 9$ three digit non-palindromes, and there are $26\cdot 26\cdot 25$ three letter non palindromes. As there are $10^3\cdot 26^3$ total three-letter three-digit arrangements, the probability that a license plate does not have the desired property is $\frac{10\cdot 10\cdot 9\cdot 26\cdot 26\cdot 25}{10^3\cdot 26^3}=\frac{45}{52}$ . We subtract this from 1 to get $1-\frac{45}{52}=\frac{7}{52}$ as our probability. Therefore, our answer is $7+52=\boxed{059}$ .

Solution 3

Note that we can pick the first and second letters/numbers freely with one choice left for the last letter/number for there to be a palindrome. Thus, the probability of no palindrome is $\frac{25}{26}\cdot \frac{9}{10}=\frac{45}{52}$ thus we have $1-\frac{45}{52}=\frac{7}{52}$ so our answer is $7+52 = \boxed{059}.$

~Dhillonr25

Solution 4 (PIE)

The total number of possible license plates is $26^3 \cdot 10^3$ . The number of license plates that contains at least $1$ palindrome is $#(\text{license plate with a three-letter palindrome})+#(\text{license plate with a three-digit palindrome})-#(\text{license plate with a three-letter palindrome and three-digit palindrome})$ (Error compiling LaTeX. Unknown error_msg) by [Principle of Inclusion-Exclusion]

\boxed{\textbf{57}}$

~isabelchen

Video Solution by OmegaLearn

https://youtu.be/jRZQUv4hY_k?t=98

~ pi_is_3.14

2002 AIME I (Problems • Answer Key • Resources)
Preceded by First Question	Followed by Problem 2
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