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 09 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: 1,206 Thanks: 613 
$n$ is the alphabet size $p$ is the maximum password length This implies that lengths $1$ through $p1$ are valid as well The total number of combinations is the sum of the number of combinations of each length up through $p$ $x = n + n^2 + \dots + n^p=\dfrac{n^{p+1}1}{p1}1=\dfrac{n \left(n^p1\right)}{n1}$ 

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