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 $m+n$.
Source: 2002 AIME problem 1.
For the three letters, the first and the last letter needs to be same in order to be a palindrome. So there are $26^2$ possibilities for the three letters to be a palindrome. Similarly, for the three digits there are $10^2$ possibilities. The probability of picking the palindrome in a license plate is $\dfrac{1}{10} + \dfrac{1}{26} = \dfrac{9}{65}$.
I found in a solution that they are subtracting $\dfrac{1}{260}$ from $\dfrac{9}{65}$. I am unable to find where I am overcounting that I have to subtract. Can anyone give me an explanation?
Best Answer
Hi there this is an interesting question.
I have come up with a solution to this.
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 $\frac{10\cdot 10\cdot 9\cdot 26\cdot 26\cdot 25}{10^3\cdot 26^3}=\frac{45}{52}$.
Now we subtract this from 1 to get $1-\frac{45}{52}=\frac{7}{52}$ as our probability. Therefore, our answer is $7+52=59$.
To answer your doubt, P(A∪B)=P(A)+P(B)−P(A∩B) we use this property. Here $\frac{1}{260}$ is the probability of having two palindromes(P(A∩B)). Also this result can be obtained by the Principle of Inclusion-Exclusion.
I hope this was helpful.