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Four digit number permutations

  1. Dec 21, 2006 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

    3. 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?
  2. jcsd
  3. Dec 21, 2006 #2
    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: Dec 21, 2006
  4. Dec 21, 2006 #3
    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.
  5. Dec 22, 2006 #4
    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?
  6. Dec 22, 2006 #5
    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.
  7. Dec 22, 2006 #6
    maybe it'll help if I pointed out that 34 + 43 has the same sum as 33 + 44.
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