# How Many License Numbers Consist of $4$ digits and $4$ letters, where one letter or digit is repeated?

If license numbers consist of four letters and four numbers how many different licenses can be created having at least one letter or digit repeated?

$$26 \cdot 26 \cdot 26 \cdot 26 \cdot 10 \cdot 10 \cdot 10 \cdot 10 = 4569760000$$

Is this right?

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You should consider permutations of letters (L) and numbers (N). For example, LLLLNNNN, LNLNLNLN, NNNNLLLL, ... - are different. –  Oleg567 Aug 6 '13 at 3:23
@Oleg567 so you are saying the answer is? –  suffix Aug 6 '13 at 3:25
@Oleg567 Thank you for your comment. I didn't realize that. –  Alraxite Aug 6 '13 at 3:32
@Alraxite so my answer is not right? –  suffix Aug 6 '13 at 3:39
@suffix No, what you've calculated is the total number of licences which are of the form $\text{LLLLNNNN}$ where $\text{L}$ and $\text{N}$ denotes a letter and a number respectively. –  Alraxite Aug 6 '13 at 3:47

Assuming the licences are of the form $\text{LLLLNNNN}$ where $\text{L}$ and $\text{N}$ denote letters and numbers respectively:

Number of licences with no letters or numbers repeated:

$26\cdot 25\cdot 24\cdot 23\cdot 10\cdot 9\cdot 8\cdot 7$.

So, number of licences with at least one letter or number repeated:

$\text{Total number of licences} - 26\cdot 25\cdot 24\cdot 23\cdot 10\cdot 9\cdot 8\cdot 7$.

Where the total number of licences is $26^4\cdot 10^4$ which is what you calculated.

Assuming the letters and numbers can appear in any order:

You can obtain this answer by multiplying the answer obtained above by $\dfrac{8!}{4!4!}$ which is the number of permutations of $8$ things in which $4$ objects are of one kind and $4$ objects are of the other kind. This is the same as the answer given by Omnitic.

Please note that as pointed out by Omnitic, the following does not restrict the number of letters and digits to $4$:

Assuming the numbers or letters can be filled in any order (and are not restricted to be $4$):

The licences have $8$ places which can be filled in with a number or a letter. Thus each place can have one of $26+10=36$ values.

Now, number of licences with no letters or numbers repeated: $36\cdot 35\cdot 34\cdot\cdot\cdot 30\cdot29$.

So, number of licences with at least one letter or number repeated:

$\text{Total number of licences} - 36\cdot 35\cdot 34\cdot\cdot\cdot 30\cdot29$

The total number is given by $36^8$.

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If the licence plate is of the form $LLLLNNNN$ where L stands for letter and N for integer between 0 and 9 inclusive then no.

Number of ones with no repeats: $26*25*24*23*10*9*8*7=\frac{26!}{22!}\frac{10!}{6!}$ number with repeats= $26^4*10^4-\frac{26!}{22!}\frac{10!}{6!}$

However if the order of the characters is not restricted then it is $\binom{8}{4}26^4*10^4-\binom{8}{4}\frac{26!}{22!}\frac{10!}{6!}$ And it also depends on the alphabet you use and the number system ;)

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Shouldn't it be $26^4\cdot 10^4$ instead of $27^4\cdot 10^4$? –  Alraxite Aug 6 '13 at 3:42
yes, corrected, thanks –  Modded Bear Aug 6 '13 at 3:44
Your final expression's value, $\binom{8}{4}26^4*10^4-\binom{8}{4}\frac{26!}{22!}\frac{10!}{6!}$ doesn't match with my answer. –  Alraxite Aug 6 '13 at 4:44
Oh, you are right. Thank you. –  Alraxite Aug 6 '13 at 4:53