Password Problem

Suppose the length of the password is $n$ .

First, let us forget about the restriction.

Every character of the password would then either be a letter (26 items) or a digit (10 items). That is a total of 36 items to choose from.

Since there are n characters, there are a possible of $36^n$ such passwords.

Now, consider the passwords without any digits at all.

Using a similar Logic, there are a possible of $26^n$ such passwords.

Now, look at this:

Number of passwords allowing everything = $36^n$ Number of passwords allowing no digits = $26^n$

We are interested in the once that has atleast one digit,

so, the number of that kind of passwords = $36^n-26^n$

Now, Krishna says that the password length varies from 6 to 8.

So, we have three cases, length 6, length 7 and length 8

Thus we need to evaluate $(36^6 - 26^6)+(36^7 - 26^7)+(36^8 - 26^8)$

You could do that with a calculator or maybe do this:

• Go to sagecell.sagemath.org
• Insert this

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n = var('n')
sum((36^n - 26^n),n,6,8)

• Press Evaluate

Or Even Simpler:

• Go to maxima-online.org
• Insert this

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sum((36^n - 26^n),n,6,8);

• Press Calculate

I enjoyed this problem.