Difference between revisions of "2002 AIME I Problems/Problem 1"
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== Solution == | == Solution == | ||
+ | === Solution 1 === | ||
Consider the three-digit arrangement, <math>\overline{aba}</math>. There are <math>10</math> choices for <math>a</math> and <math>10</math> choices for <math>b</math> (since it is possible for <math>a=b</math>), and so the probability of picking the palindrome is <math>\frac{10 \times 10}{10^3} = \frac 1{10}</math>. Similarly, there is a <math>\frac 1{26}</math> probability of picking the three-letter palindrome. | Consider the three-digit arrangement, <math>\overline{aba}</math>. There are <math>10</math> choices for <math>a</math> and <math>10</math> choices for <math>b</math> (since it is possible for <math>a=b</math>), and so the probability of picking the palindrome is <math>\frac{10 \times 10}{10^3} = \frac 1{10}</math>. Similarly, there is a <math>\frac 1{26}</math> probability of picking the three-letter palindrome. | ||
By the [[Principle of Inclusion-Exclusion]], the total probability is | By the [[Principle of Inclusion-Exclusion]], the total probability is | ||
<center><math>\frac{1}{26}+\frac{1}{10}-\frac{1}{260}=\frac{35}{260}=\frac{7}{52}\quad\Longrightarrow\quad7+52=\boxed{059}</math></center> | <center><math>\frac{1}{26}+\frac{1}{10}-\frac{1}{260}=\frac{35}{260}=\frac{7}{52}\quad\Longrightarrow\quad7+52=\boxed{059}</math></center> | ||
− | + | === 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 <math>10\cdot 10\cdot 9</math> three digit non-palindromes, and there are <math>26\cdot 26\cdot 25</math> three letter non palindromes. As there are <math>10^3\cdot 26^3</math> total three-letter three-digit arrangements, the probability that a license plate does not have the desired property is <math>\frac{10\cdot 10\cdot 9\cdot 26\cdot 26\cdot 25}{10^3\cdot 26^3}=\frac{45}{52}</math>. We subtract this from 1 to get <math>1-\frac{45}{52}=\frac{7}{52}</math> as our probability. Therefore, our answer is <math>7+52=\boxed{059}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2002|n=I|before=First Question|num-a=2}} | {{AIME box|year=2002|n=I|before=First Question|num-a=2}} |
Revision as of 17:10, 10 August 2010
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 |