Question

If a state issued license plates using the scheme of four letters followed by two digits, how many plates could it issue?

Answer 1

For each letter place, how many possible choices are there?
How about digits?

Answer 2

that is the whole question. we are learning combinations and permutations

Answer 3

Correct, but I mean to say this:

Lets imagine our license plate only consists of a single letter, how many choices do I have?

Answer 4

26

Answer 5

Okay, now lets extend it to 2 letters:

We have 26 for the first choice. For example, lets say we choose A for the first choice: We then have:
\[A ? \]
How many choices do we have for simply the list of A's...well another 26:
\[ AA \\ AB \\ AC\\ AD \\ ...\]

So for every choice in slot one, we have 26 combinations. Therefore for every first choice we have 26. So the combinations possible of a 2 letter plate is \( 26*26=26^2 \).

Can we extend this reasoning to 3 letters, or four?

Answer 6

yes it has to be 4

Answer 7

so 26 to the 4th power?

Answer 8

Correct, but for four letters, we would have 26 for the first choice, 26 for the second, 26 for the third, etc.

That is correct.

Now, for each digit how many choices do we have?

Answer 9

2?

Answer 10

Well we must choose two digits, but imagine we have to choose just one first. How many options do we have to choose from?

Answer 11

26??

Answer 12

Assuming we are in base 10, our digits are \(0-9\). So, we have 10 choices for the first digit, and 10 choices for the second. How many choices total do we have for a 2 digit number?

Answer 13

20

Answer 14

thanks for the help

Answer 15

Not quite, if we only had 20, the highest 2 digit number we could construct would be 19 (00-19).

We actually have 100 2 digit numbers (00-99).

So our total license plate is \(26^4 * 10^2\)

Answer 16

so 456976 x 100?

Answer 17

Notice to calculate this, given any amount choices, the total amount of combinations is the the multiplication of the number of choices for EACH choice. IE:
\[\underbrace{26}_{\text{first numeral}} \times \underbrace{26}_{2^{nd}\text{numeral}} \times \underbrace{26}_{3^{rd}\text{numeral}} \times \underbrace{26}_{4^{th}\text{f numeral}} \times \underbrace{10}_{1^{st}\text{fdigit}} \times \underbrace{10}_{12^{nd}\text{fdigit}} \]

That is correct:.

Answer 18

I dont feel like retyping that but the first four should be 1-4th letters, not numerals

Answer 19

can i ask you another question