Combinatorics-next problem with numbers

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Homework Help Overview

The discussion revolves around a combinatorial problem involving the creation of 7-digit numbers from a set of 5 digits, with the condition that any two numbers must have at least 2 different digits. The original poster presents their interpretation of the problem and a proposed solution, while seeking validation and assistance in proving their reasoning.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to justify their answer of 5^6 by comparing it to smaller cases with 2-digit and 3-digit numbers, while others question the clarity of the problem statement and the reasoning behind the proposed solution.

Discussion Status

The discussion is ongoing, with participants seeking clarification on the problem's requirements and the original poster's reasoning. Some participants express confusion about the phrasing of the problem, indicating a need for further exploration of the assumptions involved.

Contextual Notes

There are indications of language barriers affecting communication, and some participants are questioning the clarity of the problem statement and the original poster's understanding of the task.

Jurij
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Hi. It's one more hard task from cominatorics
We have 5 digit. How many 7-digit numbers can we create that each two of them have at least 2 different digit?
Could you help me?
I think that the answer is 5^6 but don't know how to prove it.
 
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Could anybody help me? Please, give me at least hint.
 
Let's start with this: Why do you think the answer is 56?
 
OK.
I don't know if you understand the task. The numbers have to have at least 2 different digit on some position. For example when we have 5 digit: 1, 2, 3, 4, 5, numbers 1234512 and 1234545 or numbers 5555555 and 1551555 are good.

When we have 2-digit numbers we have 5^1=16 numbers and 5 that each two of them have at least 2 different digit.
When we have 3-digit numbers we have 5^3=125 numbers and 5^2=25 that each two of them have at least 2 different digit because each two of 2-digit numbers have at least 1 different digit and when we add third number we get 25.
So for 7-digit numbers we have 5^6.
 
Last edited:
HOw is this different from yiour last post? and your english is a bit off.

"can we create that each two of them have at least 2 different digit?" ??!??!
 
Yes. you're right. My english isn't very good. But it should be clear now. Could anyone help me?
 
Is there something that you can't understand in the problem or you don't know how to do it?
 

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