Combinatorics-next problem with numbers

  • Thread starter Jurij
  • Start date
  • #1
Jurij
14
1
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.
 

Answers and Replies

  • #2
Jurij
14
1
Could anybody help me? Please, give me at least hint.
 
  • #3
quantumdude
Staff Emeritus
Science Advisor
Gold Member
5,575
23
Let's start with this: Why do you think the answer is 56?
 
  • #4
Jurij
14
1
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:
  • #5
neurocomp2003
1,366
3
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?" ??!??!
 
  • #6
Jurij
14
1
Yes. you're right. My english isn't very good. But it should be clear now. Could anyone help me?
 
  • #7
Jurij
14
1
Is there something that you can't understand in the problem or you don't know how to do it?
 

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