A $\textit{palindrome}$ is a positive integer which reads the same forward and backward, like $12321$ or $4884$.

How many $4$-digit palindromes are there?
I know that a $4$-digit palindrome must have the same first and last digit, and the same second and third digit. So, the first digit can be any of the nine nonzero digits, from $1$ to $9$. The second digit can be any of the ten digits, from $0$ to $9$. The third digit must be the same as the second digit, and the last digit must be the same as the first digit. Therefore, the number of $4$-digit palindromes is the product of the number of choices for the first and second digit, which is $9 \times 10 = 90$. 