2002 AIME I Problems/Problem 1
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. 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) | ||
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Followed by Problem 2 | |
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