Combinatorics Question: 8-Letter Passwords

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Homework Help Overview

The problem involves determining the number of eight-letter passwords that can be formed using the letters A-Z, with the condition that up to one letter may be repeated. Participants are exploring combinatorial methods to approach this problem.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss breaking the problem into cases based on the number of repetitions allowed, including scenarios with no repetitions and varying degrees of letter repetition. Some question the interpretation of "up to one letter" and suggest using inclusion/exclusion principles, while others express uncertainty about their approach.

Discussion Status

The discussion is ongoing, with participants sharing their reasoning and calculations. Some have provided detailed breakdowns of their approaches, while others seek clarification and validation of their methods. There is no explicit consensus on the best approach, but several lines of reasoning are being explored.

Contextual Notes

One participant notes that they have not learned inclusion/exclusion yet, which may influence their approach to the problem. There is also a concern about whether the interpretations of the problem statement are aligned among participants.

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Homework Statement



How many eight-letter passwords using the letters A-Z are there in which up to one letter is allowed to be used more than once?


Homework Equations





The Attempt at a Solution



I broke the problem up based on repetition of one letter: (26)8 ways with no repetition, (8 choose 2)*26*(25)6 ways with letter repeated once, (8 choose 3)*26*(25)5 ways with letter repeated twice, and so on, until 26 ways with letter repeated 7 times filling all 8 spaces. Adding these 8, gives the total ways.
 
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Do you mean _at most_ one repretition?

Maybe inclusion/exclusion would help: Count the number of ways in which you can get exactly 1 repetition, then exactly two repetitions,..., exactly 8 repetitions, and then use inclusion/exclusion starting with 26^8 possible combinations.
 
That's the exact question from the book. It means either no letters are repeated or one letter can be repeated as many times as you want.

I didn't learn inclusion/exclusion yet so I don't think we are supposed to use it.

So, I counted the ways in which there is no repetition, there is exactly one repetition, exactly 2 repetitions, and so on. Then, I added these up to give me the total ways in which one letter is allowed to be used more than once.
 
Well, if no letters are repeated, then you are just choosing 8 letters out of 8.

And then you could consider separately cases in which exactly one letter is repeated. You seem to be on the right track on your first post; I can't tell where you're stuck.
 
Sorry, I'm not really stuck. I just can't ever be sure that I'm even on the right track with these kind of problems. So, I just wanted to get some input to make sure I'm not way off. Thanks for your help.
 
No problem; if you post your answer in detail, maybe we can see better what (may be) wrong.
 
I did (26)8 + (8 choose 2)*(26)7 + (8 choose 3)*(26)6 + (8 choose 4)*(26)5 + (8 choose 5)*(26)4 + (8 choose 6)*(26)3 + (8 choose 7)*(26)2 + 26

This is based on no repetition + exactly 1 rep. + exactly 2 rep. + exactly 3 rep. + exactly 4 rep. + exactly 5 rep. + exactly 6 rep. + exactly 7 rep.

For no rep., there are obviously (26)8 ways.
For 1 rep., choose 2 places for repeating digits. Repeating digits can be chosen in 26 ways and the rest of the digits can be chosen in (25)6 ways. Similarly, for the rest up to exactly 7 rep. in which there are of course 26 ways to have the password contain all of the same character.
 

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