# Combinatorics-next problem with numbers

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.

Tom Mattson
Staff Emeritus
Gold Member

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.

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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?