How many 4-digit numbers can be formed with a limited number of digits?

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Homework Help Overview

The problem involves determining how many 4-digit numbers can be formed using no more than two different digits. The context suggests considerations around digit selection and placement, particularly regarding the inclusion of leading zeros.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the possibility of using the same digit in all four places, as well as configurations where three or two digits are the same. There is also a focus on the implications of digit selection from 0 to 9 and the potential restrictions on leading zeros.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem, particularly concerning the allowance of leading zeros. Some guidance has been provided regarding the cases to consider, but no consensus has been reached on the treatment of leading zeros.

Contextual Notes

There is ambiguity regarding whether leading zeros are permissible in the formation of 4-digit numbers, which affects the total count of valid combinations. Participants suggest considering both scenarios to fully address the problem.

xiphoid
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Homework Statement


how many 4 digit numbers are there which do not contain more than 2 different numbers?


Homework Equations






The Attempt at a Solution


all can contain the same digit
any 3 out of the 4 places can be occupied by the same digit
and
any 2 out of the 4 places can be occupied by the same digit
 
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xiphoid said:

Homework Statement


how many 4 digit numbers are there which do not contain more than 2 different numbers?


Homework Equations






The Attempt at a Solution


all can contain the same digit
any 3 out of the 4 places can be occupied by the same digit
and
any 2 out of the 4 places can be occupied by the same digit


You are on the right track, think about it this way. You have a choice of choosing the numbers from 1 to 9 and then you have three cases:

1. the case where all 4 numbers are the same number as you chose

2. the case where 3 numbers are the same number as you chose as well as you have to choose another number that's different from the other three

3. the case where 2 numbers are the same number as you chose and you have to choose 2 other numbers where they are different from each other and different from the other 2 you previously chose

Does that make sense?
 
leej72 said:
You have a choice of choosing the numbers from 1 to 9 and then you have three cases.
Shouldn't that be digits from 0 to 9 (a total of 10 possible digits)?
 
rcgldr said:
Shouldn't that be digits from 0 to 9 (a total of 10 possible digits)?

Probably yes, but (presumably) the first (left-most) digit cannot be 0. (Since this is not made clear, if I were doing the problem I would solve both versions.)

RGV
 
As you mention, it's not clear if leading zeroes are allowed, like a combination lock with 4 digits. I would assume they are allowed, since otherwise it eliminates all the combinations such as 0xxx, 00xx, 000x, 0000.
 
rcgldr said:
Shouldn't that be digits from 0 to 9 (a total of 10 possible digits)?

Yes that was a typo, I did mean from 0 to 9 but I am not too sure if the first number would be allowed to be a 0. I would personally think otherwise, not letting the first number to 0
 
as per the answer given, you will need to consider both the cases!
 

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