I'm new to this so hope this is the right place to ask!

I'm currently in the process of compiling a custom wordlist/password list and would like to know how many possible words can be produced.

All of the words will abide by the following rules:

  • Exactly 10 characters long
  • Will contain exactly 6 letters (a-f, all lowercase)
  • Will contain exactly 4 numbers (2-9)
  • The letters and numbers can appear anywhere in the word

I read in a forum somewhere that this problem could be solved using the following formula:


which is

(6 possible letters ^ 6 letters) * (8 possible numbers * 4 numbers) * password length

This results in: 1,911,029,760 combinations. Which seems low?

Any help with this would be greatly appreciated as it is for my Computer Science Dissertation!


The beginnig is right

(6 possible letters ^ 6 letters)* (8 possible numbers ^ 4 numbers) * (where you put your numbers in the password)

to choose where you put your numbers in the password, use the binomial coefficient

$$(6^6) \times (8^4) \times \binom{10}{4} = 40,131,624,960$$

  • $\begingroup$ Thats very helpful. Is there any way you could potentially include another rule which disallows characters from appearing more than twice in succession? So as to say 'c43a11abd3' would be ok, but 'ea4448bca2' would not? $\endgroup$ – Tom Oct 15 '15 at 20:38
  • $\begingroup$ @Tom I would suggest asking a new question. $\endgroup$ – N. F. Taussig Oct 16 '15 at 11:50

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