# License to Count

When I’m stuck in traffic, I have random thoughts racing through my mind since my stereo/radio is out of commission. After all, what else is there to do when your car is moving at such a slow pace?

This morning, while I was crawling along with the throng of cars, I began wondering just how many possible license plate number combinations exist as I glanced from one vehicle to another.

In California, there are seven spots available for the license plate number. The usual combination is a number, three letters and three more numbers. So for example, a typical license plate would read 1AAA111.

To satisfy my curiosity, I decided to calculate the permutation.

Since there are ten available numbers–zero through nine–and twenty-six letters in the alphabet, that gives me a total of thirty-six characters. I do not account for personalized plates that have fewer than seven numbers/letters or state vehicles that have plates fewer than seven numbers/letters. I calculated the permutation as follows: 36!/29!.

So what is the final number? It turns out there are 42,072,307,200 different possible license plate combinations for California automobiles. I sure hope we never see forty billion cars in California because that would make traffic even worse!

**Update:** My friend, Sam, pointed out to me that this problem should not be solved with a permutation. Instead, it can be solved by simple multiplication. He left notes below regarding how it can be solved correctly based on the DMV criteria. However, I have seen trucks where the license plate’s fourth spot is occupied by a number. Given this criteria and making no exceptions when it comes to letters (i.e., I will allow the letters I, O and Q to be used in any of the spots where numbers usually occupy. Also, I will allow numbers to be used in the fourth spot. That brings the number of combinations to 10*26*26*36*10*10*10, which equals 243,360,000. However, if we include the DMV’s restriction and allow for the numbers to occupy the 4th spot, we have 10*23*26*33*10*10*10, which equals 197,340,000 combinations.

I disagree. this should be a simple multiplication (as each digit is independent of others, so long as it is in the correct place on the plate), rather than a permutation. Also i don’t think they use all the letters. Yep, DMV site says:

“Exclusions: The letters I, O, and Q are not used in the first or third alpha positions of the alpha-numeric series. ” – http://dmv.ca.gov/pubs/plates/standardplates.htm (for passenger vehicles, other categories differ).

so the first numerical position would be 0-9 (10 possibilities)

second position would be 23 (alpha, but not i,o,q)

third position would be 26 (all alpha)

fourth position would be 23 (same as second)

fifth position, sixth, seventh (same as first, numbers only)

so, 10 * 23 * 26 * 23 * 10 * 10 * 10 = 137,540,000

population of Cali ~ 37 million

.. so we’re good till we all have 4 personal vehicles or the population goes up drastically.

Realistically though, there are other restrictions i believe (no leading ‘0’ i think, etc) and also other characters for vanity plates. Without all of that info, i couldn’t say for sure how many.

ah, thanks for bringing that to my attention. i think where i may have misled you is that i was trying to calculate the possible number of combinations based on the available characters, i.e. the letters and numbers, with no restrictions at all on the positioning or limitations on which letters are used. so, the first position is allowed to be a letter (A-Z) or number and so forth. i should have prefaced that somewhere in my entry. the reason for my not being rigid with the limited placement and letters is due to seeing state-issued plates (non-vanity plates) in which there were numbers in places that would usually be reserved for letters.

Combinatorics is fun!

License plates are wrong! and cars are worse!

everyone should walk to work!