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 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{ | + | <center><math>\frac{1}{26}+\frac{1}{10}-\frac{1}{260}=\frac{35}{260}=\frac{7}{52}\quad\Longrightarrow\quad7+52=\boxed{59}</math></center> |
− | + | ||
− | Using complementary counting, we count all of the license plates that do not have the desired property. | + | == Solution 2 == |
+ | Using [[complementary]] counting, we count all of the license plates that do not have the desired property. 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{59}</math>. | ||
+ | |||
+ | ~minor edit by Yiyj1 | ||
+ | |||
+ | == 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{59}.</math> | ||
+ | |||
+ | ~Dhillonr25 | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/jRZQUv4hY_k?t=98 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
== 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]] | [[Category:Intermediate Combinatorics Problems]] | ||
[[Category:Intermediate Probability Problems]] | [[Category:Intermediate Probability Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 13:25, 2 August 2024
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
![$\frac{1}{26}+\frac{1}{10}-\frac{1}{260}=\frac{35}{260}=\frac{7}{52}\quad\Longrightarrow\quad7+52=\boxed{59}$](http://latex.artofproblemsolving.com/d/b/b/dbb32c8efe6a484d80478f3ff037a19c68f1a84a.png)
Solution 2
Using complementary counting, we count all of the license plates that do not have the desired property. 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
.
~minor edit by Yiyj1
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.