While doing homework today, the following question popped into my head: Can you easily calculate the amount of unique license plates consisting of 4 letters and 4 numbers in any order?

It doesn't seem to be easily possibly; $$ 26^3 * 10^3 * 8! $$ would include repeats (such as AAA123 and AAA123; As are in different positions)


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    $\begingroup$ Could you clarify? AAA123 and AAA123 are identical, and neither has $4$ letters or $4$ numbers. $\endgroup$ – Brian M. Scott Mar 28 '13 at 6:52

If you want license plates consisting of four numbers and four letters in any possible arrangement, then you must first choose $4$ of the $8$ spots for your numbers (or symmetrically, your letters). Then there are $10^4$ choices for your numbers and $26^4$ choices for your letters. This means that in total you have $\binom{8}{4}26^4\cdot 10^4$ distinct license plates.

If you make the restriction that the letters come before the numbers (or the numbers before the letters) then you must choose the first four spots for your letters and the last four spots for your numbers. This simply removes the binomial coefficient in the above, leaving $26^4\cdot 10^4$ distinct license plates.

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  • $\begingroup$ For your edification this is 7,677,196,800,000 or enough for about 1,000 vehicles for each person on earth. $\endgroup$ – Dale M Apr 5 '13 at 9:11

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