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

$\frac{1}{26}+\frac{1}{10}-\frac{1}{260}=\frac{35}{260}=\frac{7}{52}\quad\Longrightarrow\quad7+52=\boxed{59}$

Solution 2

Using complementary counting, we count all of the license plates that do not have the desired property. 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 $7+52=\boxed{59}$.

~minor edit by Yiyj1

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 $7+52 = \boxed{59}.$

~Dhillonr25

Solution 4

We can find the probability of getting a letter and number palindrome through Solution One, which gives us $\frac{1}{26},$ and $\frac{1}{10},$ respectively. Then, we can use casework to solve the question. We begin by creating the cases:

 Case 1: The license plate includes only a letter palindrome, and no number palindrome Case 2: The license plate includes only a number palindrome, and no letter palindrome Case 3: The license plate includes both a number palindrome, and a letter palindrome

We know that the complement of these probabilities gives us the probability that the numbers and letters are NOT palindromes, so we can use that in our cases to get:

126910=9260Case 12526110=25260Case 2126110=1260Case 3

Finally, we can add them all together to get: $\frac{9 + 25 + 1}{260} = \frac{35}{260} = \frac{7}{52} = \frac{m}{n}.$ Thus, we have $m + n = \boxed{059}.$

~ Cheetahboy93

Video Solution by OmegaLearn

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

~ pi_is_3.14

See also

2002 AIME I (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
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All AIME Problems and Solutions

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