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

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

~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 $\frac{25}{26}\cdot \frac{9}{10}=\frac{45}{52}$ thus we have $1-\frac{45}{52}=\frac{7}{52}$ so our answer is $7+52 = \boxed{59}.$

~Dhillonr25

Video Solution by OmegaLearn

https://youtu.be/jRZQUv4hY_k?t=98

~ pi_is_3.14

@@ Line 15: / Line 15: @@
 == 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{59}.</math>
 ~Dhillonr25

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