Four digit number permutations

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Homework Help Overview

The problem involves finding the sum of all four-digit numbers that can be formed using the digits 0, 1, 2, and 3. Participants are exploring the constraints of digit selection and the implications for valid four-digit numbers.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the total number of valid four-digit combinations and question whether the problem involves permutations or combinations. There is an exploration of the summation approach for calculating the total sum of the numbers formed, with some participants noting specific examples to clarify valid combinations.

Discussion Status

The discussion is ongoing, with participants providing insights and questioning assumptions about the digits that can be used. Some guidance has been offered regarding the frequency of digits in specific places, but no consensus has been reached on a definitive method for calculating the sum.

Contextual Notes

There is a noted constraint regarding the valid digits for four-digit numbers, particularly that the first digit cannot be zero. This has led to confusion about the total count and the validity of certain examples provided in the discussion.

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Homework Statement



Find the sum of all the four digit numbers that can be formed with the digits 0,1,2,3

Homework Equations





The Attempt at a Solution



The total number of numbers possible is 3*4*4*4=192.

Since the lowest number we can form is 1000, and the highest is 3333, the sum of all the digits should be summation (1-3333) - summation(1-1000).

The summation is given by [tex]\frac{n(n+1)}{2}[/tex] or in this case,
[tex]\frac{3333*3334-1000*1001}{2}[/tex] which gives 5055611... which is nt the given answer. What am I doing wrong?
 
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Wrt(1) Is that Permutation or combination you want? Because 1111 and 0123 are both 4digit numbers formed with the digits. I assume its a permutation by the title of your post though.

You know that the permutations is simply 4! = 24 when n=r ? and that sum of each is constant?
 
Last edited:
chaoseverlasting said:

Homework Statement



Find the sum of all the four digit numbers that can be formed with the digits 0,1,2,3

Homework Equations





The Attempt at a Solution



The total number of numbers possible is 3*4*4*4=192.

Since the lowest number we can form is 1000, and the highest is 3333, the sum of all the digits should be summation (1-3333) - summation(1-1000).

The summation is given by [tex]\frac{n(n+1)}{2}[/tex] or in this case,
[tex]\frac{3333*3334-1000*1001}{2}[/tex] which gives 5055611... which is nt the given answer. What am I doing wrong?

The number 1999, as an example, is included in your sum, but it is not one of the numbers that can be formed with only the digits 0,1,2,3.
 
Oh yeah... youre right... I completely overlooked that...

The sum would be 1000-1333 + 2000-2333+ 3000-3333... is there some simpler way to calculate it besides the summation formula?
 
Simpler? 6 times, the one's digit is a 3. 6 times, the one's digit is a 2. 6 times, the ones digit is a 1. 6 times, the 1's digit is a zero.

6 times, the ten's digit is a 3. 6 times, the ten's digit is a 2...

Think about it from there.
 
maybe it'll help if I pointed out that 34 + 43 has the same sum as 33 + 44.
 

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