permutation and/ or combination of california license plate

Question

just wondering

california's license plate is

0XXX0000

number, 3 alphabets and 4 numbers.

how many permutation / combinations of total different license plate it could be ? i learned this in discrete structure but I have forgotten

Answer 1

10^5 * 26^3

Replace "26" with the number of characters in the alphabet (more for special chars)

Since each "place" is independent of the next it's akin to a simple number.

That is in a three digit number there are 1,000 (10^3) combinations 000 through 999.

If we had a three digit hex number there would be 16^3.

The license plate combines 5 digits and 3 alphas, so it's a mix of digits and alphas, hence the 10^(number of digits places) * 26^(number of alpha places)

Answer 2

Each number can be 0, 1, ..., 9 (10 possibilities). Each letter can be A, B, ..., Z (26 possibilities). Since each position in the license plate is independent of all others, the number of possible valid plates is:

10 * 26 * 26 * 26 * 10 * 10 * 10 * 10 = 1757600000

The fact that the value in each position is independent is important. For example, choosing any number for the first position does not change or restrict what you can choose for the second, third, etc. positions.

Another way to think about it: for the first position, you have 10 possibilities. For each of these possibilities, you have 26 possibilities for the second position (e.g. you can have '0A...', '0B...', '0C...'), so for the first two positions, you have 10 * 26 combinations. This independence applies to every position on the entire plate.

Answer 3

If by number you mean digit, and by alphabets you mean letter, then the total number of combinations is

10 * 26 * 26 * 26 * 10 * 10 * 10 * 10 = 1757600000

Permutations are another thing - all the possible linear orders of a set. I don't think that's what you meant.

Answer 4

First digit can't be 0, and there are only three digits at the end, not four. Thus:

9 * 26 * 26 * 26 * 10 * 10 * 10 = 158184000

Before moving to the seven character plates, California issued ones with three digits followed by three letters. I believe those leading digits could be 0, so we have:

10 * 10 * 10 * 26 * 26 * 26 = 17576000

Commercial trucks have one non-zero digit, one letter, and five more digits:

9 * 26 * 10 * 10 * 10 * 10 = 2340000

Some government vehicles have all digits. The longest I've seen is six digits, so there are as many as another 1000000.

Vanity plates can have up to 7 characters (digits, letters, blanks). Presumably, there's a minimum number of non-blank characters. I'll assume 1.

37 * 37 * 37 * 37 * 37 * 37 * 36 = 92366150724

Vanity plates, however, cannot use a digit/letter pattern that matches a pattern used on serial plates. So we have to subtract all of the possible serial plates, giving a total of 92187050724 possible vanity plates.

I believe the special symbols on vanity plates (hand, star, etc.) are treated exactly like blanks. You cannot have FOO BAR and FOOBAR. I'm not sure, but it's possible that blanks are ignored altogether, which would lower the number. (It would be odd to issue both "ABC" and "__ABC".)