2003 AMC 10B Problem 10

Each letter position can be filled in $26$ ways and each digit position in $10$ ways, independently, so the number of plates is the product over positions.

Old scheme (one letter, four digits): $26 \cdot 10^4$.

New scheme (three letters, three digits): $26^3 \cdot 10^3$.

"By how many times" asks for the ratio new to old:
$$\frac{26^3 \cdot 10^3}{26 \cdot 10^4} = \frac{26^2}{10}.$$

The answer is $\boxed{\textbf{(C)}\ \frac{26^2}{10}}$.

Independent positions multiply, so the count of plates is a product of $26$'s and $10$'s, and swapping one digit slot for a letter slot multiplies the count by $\frac{26}{10}$. Two such swaps give a factor of $\left(\frac{26}{10}\right)^2 = \frac{26^2}{10^2}$, and the new scheme also drops one position entirely, contributing another factor of $10$: $\frac{26^2}{10^2}\cdot 10 = \frac{26^2}{10}$.

• Comparing the counts as a difference, or cancelling exponents incorrectly to get $\frac{26^2}{10^2}$ or $\frac{26^3}{10^3}$.
• Miscounting the old scheme as having three digits, which makes the ratio $\frac{26^2}{1}$ and matches nothing.