# Counting Functions and Sets

Counting Lists With Repetition

## Homework Statement

How many ways can you create an 8 letter password using A - Z where at most 1 letter repeats?

## The Attempt at a Solution

I'm not sure how to attack this problem but first I thought that A-Z considers 26 letters so with no restrictions on passwords we can create 268 passwords. I'm thinking it's 268 - X, where X is a term or a series of terms, but I'm not sure how to determine them, or if this is even the correct setup.

Well there are two cases given by the restrictions as follows:
A) No letter repeats in which we have a k list without repetition which is given by (n)k = n!/(n-k)!

B) One letter repeats in which case I think it's 26*[(n-1)!/(n-k-1)!].

And of course in this case n = 26 k = 8. Is this correct? If not could someone give me a hint?

Last edited:

## Answers and Replies

In the case of taking an 8-list of 26 letters where at most 1 letter repeats, would it just be the sum of case A and case B as I wrote?

i think you almost nailed it. i would separate it to 2 sections.
one - no repeating. in this particular case, we choose 8 letters from 26 and count all the ways we can order them, which is what you've mentioned $\frac{26!}{(26-8)!}$

two - one letter repeats. in this one, we choose only 7 letters from 26 and count all the way we can order 8 letters and subtract the number of ways we order the letters when the two similar letters are together.

why is that? i'll let you ponder about it. :)

then - as you've already typed, you will add what you've gotten from the 1st section and add it to the result of the 2nd section and that's the