A vehicle licence plate number consists of $2$ letters followed by $2$ digits. How many different licence plates are possible?

Question

Why is $26 * 26 * 10 * 10$ the incorrect way to find the number of possible registration plates?

Answer 1

Unless specified differently, you have 26 choices for both letters and 10 choices for both digits, so your solution is correct. What they might mean is either 1) both numbers are diffferent 2) both letters are different 2) both letters and numbers are different. For 1), it's $\binom{26}{2} \times 2! \times 10^2$ (because order matters), for 2) it's $26^2 \times \binom{10}{2}2!$ (for the same reason). For 3) it's obviously $\binom{26}{2}\binom{10}{2}2!2!$. Also often you can't have $0$ in front. In such case you have $9 \cdot 10$choices for numbers.

Answer 2

Number of ways to choose 2 letters out of 26 $= \binom{26}{2} = 325$

Number of ways to chose 2 numerical digits from 0–9 $=\binom{10}{2}= 45$

Now the 2 letters must be followed by the 2 digits, each set can be arranged in four ways, So we can have, e.g., $AB12$ or $AB21$ or $BA12$ or $BA21$.

Therefore, number of different plates would be $325 * 45 * 4 = 58500$

I hope this was helpful..