4-Digit Different Digit Patterns - 4536

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Homework Help Overview

The problem involves determining the number of 4-digit patterns where all digits are different. Participants are exploring the constraints of digit selection, particularly regarding leading zeros.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of factorials and permutations to calculate the total number of arrangements. There are questions about the validity of including zero as a leading digit and how that affects the total count.

Discussion Status

Some participants have pointed out the importance of not starting the number with zero, which seems to be a critical factor in reconciling the discrepancy between the calculated and expected answers. Multiple interpretations of the problem are being explored.

Contextual Notes

There is an assumption that the digits must be different and that the first digit cannot be zero, which is central to the discussion. The original poster's calculations do not account for this constraint, leading to confusion regarding the correct answer.

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Homework Statement



How many 4-digit patterns are there in which all the digits are different?

Homework Equations



10!/6! ??

The Attempt at a Solution



As shown above, I used 10!/6! and I get 5040. I get the same answer of course when I use 10x9x8x7. But my book shows the answer as 4536. What am I missing?
 
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:wink: the numbers that you want to arrange do not start with a 0 :wink:
 
rock.freak667 said:
:wink: the numbers that you want to arrange do not start with a 0 :wink:

Gadzooks! It never crossed my mind..
 
the total number of combinations of 4 digit numbers you can get is 9^4. Now just count how many contains same digits.
 
Choose any of the 9 digits (1 through 9) for the first digit. That leaves 9 digits for the second (because you are now allowed to use 0), 8 for the third digit, and 7 for the fourth.
 

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