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 $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution

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 $7+52=\boxed{059}$.

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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