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November 7th, 2016, 04:15 PM   #1
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Question 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

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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)?
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November 7th, 2016, 05:46 PM   #2
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$n$ 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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