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Difference between revisions of "2002 AIME I Problems/Problem 1"

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== Solution ==
 
== Solution ==
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The [[Principle of Inclusion-Exclusion]] can be used to solve this problem:
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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. 
  
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<math>\frac{1}{26}+\frac{1}{10}-\frac{1}{260}=\frac{35}{260}=\frac{7}{52}\quad\Longrightarrow\quad7+52=\boxed{059}</math>
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By the [[Principle of Inclusion-Exclusion]], the total probability is
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<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>
  
 
== 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}}
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[[Category:Intermediate Combinatorics Problems]]
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[[Category:Intermediate Probability Problems]]

Revision as of 22:32, 24 April 2008

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

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

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

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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