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

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SUMMARY

The discussion focuses on calculating the number of 4-digit numbers that can be formed using no more than two different digits. Three distinct cases are identified: (1) all four digits are the same, (2) three digits are the same and one is different, and (3) two digits are the same with two other different digits. The digits can range from 0 to 9, but the leading digit cannot be 0, which affects the total combinations. Participants emphasize the importance of clarifying whether leading zeros are permissible in the problem statement.

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  • Understanding of combinatorial counting principles
  • Familiarity with the concept of permutations and combinations
  • Knowledge of digit placement rules in number formation
  • Basic grasp of the implications of leading zeros in numerical representations
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  • Explore combinatorial counting techniques for digit arrangements
  • Learn about permutations with restrictions, particularly in number formation
  • Investigate the impact of leading zeros on numerical combinations
  • Study the principles of generating functions in combinatorial problems
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Mathematics students, educators, and anyone interested in combinatorial problems or number theory will benefit from this discussion.

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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