# Number of license plates formed by four digits and one letter, qualified.

I need some help with this question:

If a license plate for a vehicle consist of five characters: $4$ digits (the first of which cannot be $0$), followed by one letter of the alphabet (which cannot be $I$ or $O$), how many different license plates are possible?

• Are you familiar with the concept of a counting tree? – Jonas Dahlbæk Aug 12 '14 at 12:24

We have a license plate format:

• first digit from $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}:\;$ 9 choices
• second, third, fourth digit from $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}:\;$ 10 choices, for each place.
• fifth position: one of 24 letters (26 letters of alphabet, minus the two distinct letters not allowed) gives 24 choices.

Using the rule of the product, that gives us:

$$9 \times 10 \times 10 \times 10 \times 24 = 9 \times 10^3 \times 24 = 216,000\;\text{license plates available}$$

The first condition means that you have to choose from the $9000$ numbers between $1000$ and $9999$. The second one, assuming that your alphabet contains $25$ characters, means that you the letter can be any of $23$ different ones. With no further restriction, the number of combinations is $$9000 \cdot 23.$$ If you prefer, the first part you could also view as 4 separate digits, so that there are 9 combinations for the first one, and 10 for the remaining three, which nets you $$9 \cdot 10 \cdot 10 \cdot 10 \cdot 23$$ total combinations.

Edit: Okay, after seeing the other answers and counting thrice, I now see how to arrive at $26$ letters instead of $25$ ...

Imagine there are $4$ slots: $A$, $B$, $C$ and $D$ for each of the four digits of the license plate.

For $A$: There are $9$ digits to choose from since $0$ is omitted. There are $26$ alphabets and omitting $I$ or $O$ leave you with $24$. So by the fundamental principle of counting you have: $(9)(24)$ ways of selecting a number or an alphabet for Slot $A$.

For Slot $B$: $10$ ways (because you can choose a number from the 10 digits)

For Slot $C$: $10$ ways

For Slot $D$: $10$ ways

So in total you have: $(9)(24)(10)(10)(10)=216,000$ possible license plates by the fundamental principle of counting.