the state I live in has licence plates that are 6 digits. three letters - three numbers., and three numbers and three letters.. -
recently there was a MVA and the witness was only able to remember the first three letters i.e. 588 but failed to recall or see, the last three digits which are letters. ( now if the Licence plate in question was three letters and three digits, and the witness
could only remember ABC but not the last three, the answer would be the easiest in the world.,, 999 or 1000 possible plates.)
Since only the three numbers (5880 was recalled. What would the possible combinations be??
588-AAA 9 is one, 588-AAB is two, 588-AAC is three.,etc.
I am trying to figure out the number ratio to use.
Would it be 588 or 'X' times 26x26x26 ??? (to the power of 3 or 9??)
any help ??
26 x 26 x 26
minus number of excluded three-letter combos
For example, I imagine SEX, GOD, FBI, etc. are specifically skipped
But surely the DMV database can just pull up all the registered combos matching 588[A-Z][A-Z][A-Z] and narrow it down considerably.
You're working with base-26 (if your only letters are A-Z) then it would be taken to the 3rd power. 26^3 = 17576 possible combinations.
Also you have base-10 for the first three digits, 1000 possible combinations (10^3) ...
What are the total possible combinations for this state's license plate? I'd guess 1000 x 17,576= 17,576,000 (for 000-ZZZ only) and 17576000 x 2 = 35,152,000 (for 000-ZZZ and ZZZ-000 formats)
It's probably slightly less than 26*26*26, since there are usually a few letters that aren't allowed because they're too similar to other (I, O, etc)
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