Difference between revisions of "2002 AIME I Problems/Problem 1"
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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 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>. | 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>. | ||
− | + | == 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 <cmath>\frac{25}{26}\cdot \frac{9}{10}=\frac{45}{52}</cmath> thus we have <math>1-\frac{45}{52}=\frac{7}{52}</math> so our answer is <math>7+52 = \boxed{059}.</math> | 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 <cmath>\frac{25}{26}\cdot \frac{9}{10}=\frac{45}{52}</cmath> thus we have <math>1-\frac{45}{52}=\frac{7}{52}</math> so our answer is <math>7+52 = \boxed{059}.</math> | ||
~Dhillonr25 | ~Dhillonr25 | ||
+ | |||
+ | == Solution 4 (PIE) == | ||
+ | |||
+ | The total number of possible license plates is <math>26^3 \cdot 10^3</math>. The number of license plates that contains at least <math>1</math> palindrome is <math>#(\text{license plate with a three-letter palindrome})+#(\text{license plate with a three-digit palindrome})-#(\text{license plate with a three-letter palindrome and three-digit palindrome})</math> by [Principle of Inclusion-Exclusion] | ||
+ | |||
+ | \boxed{\textbf{57}}$ | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
== Video Solution by OmegaLearn == | == Video Solution by OmegaLearn == |
Revision as of 09:02, 6 May 2023
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 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 .
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 thus we have so our answer is
~Dhillonr25
Solution 4 (PIE)
The total number of possible license plates is . The number of license plates that contains at least palindrome is $#(\text{license plate with a three-letter palindrome})+#(\text{license plate with a three-digit palindrome})-#(\text{license plate with a three-letter palindrome and three-digit palindrome})$ (Error compiling LaTeX. Unknown error_msg) by [Principle of Inclusion-Exclusion]
\boxed{\textbf{57}}$
Video Solution by OmegaLearn
https://youtu.be/jRZQUv4hY_k?t=98
~ pi_is_3.14
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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