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 November 7th, 2016, 04:15 PM #1 Newbie   Joined: Nov 2016 From: America Posts: 1 Thanks: 0 Complex equation I'm not sure if factorials should be in use or not. if n = 3 and p = 3 then x = 39 if n = 2 and p = 2 then x = 6 if n = 2 and p = 3 then x = 13 if n = 3 and p = 2 then x = 12 ------------- So this is for passwords. n is the amount of allowed characters in a password. So let's say a password is only allowed the digits 0-9 to be used, this would mean n = 10. p is the maximum length of the password. x is the amount of possible combinations. so if p = 3, and n = 10, then some passwords could be: 012 648 729 824 820 091 etc... So does anyone know of an equation that can be used to find x (given n and p)?
 November 7th, 2016, 05:46 PM #2 Senior Member     Joined: Sep 2015 From: CA Posts: 1,105 Thanks: 576 $n$ is the alphabet size $p$ is the maximum password length This implies that lengths $1$ through $p-1$ are valid as well The total number of combinations is the sum of the number of combinations of each length up through $p$ $x = n + n^2 + \dots + n^p=\dfrac{n^{p+1}-1}{p-1}-1=\dfrac{n \left(n^p-1\right)}{n-1}$

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