Difference between revisions of "2002 AIME I Problems/Problem 1"
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== Problem == | == 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 <math>\dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math> | + | 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 <math>\dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math> |
== Solution == | == Solution == |
Revision as of 13:58, 2 August 2019
Contents
[hide]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 , where
and
are relatively prime positive integers. Find
Solution
Solution 1
Consider the three-digit arrangement, . There are
choices for
and
choices for
(since it is possible for
), and so the probability of picking the palindrome is
. Similarly, there is a
probability of picking the three-letter palindrome.
By the Principle of Inclusion-Exclusion, the total probability is

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 three digit non-palindromes, and there are
three letter non palindromes. As there are
total three-letter three-digit arrangements, the probability that a license plate does not have the desired property is
. We subtract this from 1 to get
as our probability. Therefore, our answer is
.
See also
2002 AIME I (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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