"An ID number is created as follows: The first part must consist of at least one letter and at most of four letters (the alphabet is assumed to have 26 letters). The second part must consist of at least five numbers (0, to 9) and at most 10 numbers. The last part must consist of two letters and one number (0 to 9). Examples might be: A-12345-BB7 or ABCD-12345678-5GJ".

How many ID numbers exist?

My approach is:

Let $\Omega:= \{\Omega_1 \times \Omega_2 \times \Omega_3\}$ where $\Omega_1:=\{(\omega_1), (\omega_1, \omega_2), (\omega_1, \omega_2, \omega_3), (\omega_1, \omega_2, \omega_3, \omega_4) \vert \omega_i \in \{A, ..., Z\}\}$ $\Omega_2:=\{(\omega_1, \omega_2, \omega_3, \omega_4, \omega_5), ..., (\omega_1, ..., \omega_{10}) \vert \omega_i \in \{0, ..., 9\}\}$
$\Omega_3:=\{(\omega_1, \omega_2, \omega_3)\vert $ not sure how to rigorously write down this set?$\}$

The cardinalities are:
$\Omega_1= \sum^{4}_{i=1}26^{i}$
$\Omega_2= \sum^{10}_{i=5}10^{i}$
$\Omega_3= 26^2 \cdot 9 \cdot 3!$
Hence, $\vert \Omega\vert= \left(\sum^{4}_{i=1}26^{i} \right)\cdot \left( \sum^{10}_{i=5}10^{i}\right) \cdot 26^2 \cdot 9 \cdot 3!$

Is this right?


You were close. $\Omega_1$ and $\Omega_2$ are correct. $\Omega_3$ however is incorrect for two reasons.

First, there are ten digits, not nine, so you should have used a $10$ in place of the $9$.

Next, when arranging the selected characters after selecting them in $3!$ ways, you have overcounted for when you swapped the "first" chosen letter's position with the "second" letter. It should only have been $3$ ways to rearrange.

Consider phrasing this in a different way: First, pick which of the three positions will be used by the digit. Then, pick what that digit is. Then, pick what letter appears in the left-most still available position. Finally, pick what the letter is that appears in the final remaining position.

$|\Omega_3|$ then is $3\cdot 10\cdot 26^2$


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