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Question: Find the sum of all distinct four digit numbers that can be formed using the digits 1; 2; 3; 4; and 5, each digit appearing at most once.

I have no clue as to where to begin this question. I know how to find the sum of the units place. However, how will I find this sum?

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HINT: Show that the possibilities for each of the four places are the same. Thus, if $s$ is the sum of the digits in the units’ place, $s$ is also the sum of the digits in the tens’ place, the hundreds’ place, and the thousands’ place. The units’s place contributes $s$ to the sum of these numbers. If the tens’ digits also add up to $s$, how much do they contribute to the sum of the numbers? Reason similarly for the other two positions.

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  • $\begingroup$ Aha! I get it. I was basically on the right track. Thanks! $\endgroup$ – Gummy bears May 7 '16 at 12:55
  • $\begingroup$ @Gummybears: You’re welcome! $\endgroup$ – Brian M. Scott May 7 '16 at 12:57
  • $\begingroup$ And also, thanks for providing a hint rather than the complete solution. All the other answers give me a final solution. $\endgroup$ – Gummy bears May 7 '16 at 13:01
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See get a number and you'll always see a number which will add with the first number to give $66666$ now total pairs are $60$ as total numbers are $120$ as exactly $2$ numbers give $66666$ so answer is $66666\cdot 60=3999960$

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  • $\begingroup$ This is actually a good method. I did it via Brian's method, which is longer. $\endgroup$ – N.S.JOHN May 7 '16 at 13:08
  • $\begingroup$ It's getting the answer not method ;) hoping it wasn't a competitive exam problem $\endgroup$ – Archis Welankar May 7 '16 at 13:11
  • $\begingroup$ Wow man, that's totally wrong. Almost ALL my math teachers told me it is about method and not about getting answer:) $\endgroup$ – N.S.JOHN May 7 '16 at 13:13
  • $\begingroup$ I am in 12 th grade and preparing for some exam so in a competitive one its do or die . So getting to correct answer seems more important. But yes method should be correct I agree $\endgroup$ – Archis Welankar May 7 '16 at 13:16
  • $\begingroup$ What exactly happened here?... I'm guessing the number 66666 is some sort of constant? $\endgroup$ – Gummy bears May 7 '16 at 13:18
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Firstly, there are 120 possible permutations of numbers.

$(5*4*3*2)$

For any number generated, we can match up a pair which corresponds 1 to 5, 2 to 4, ...

e.g. (2354, 4312), (1234, 5432)

If we add up the pairs, we always get 6666 and using symmetry, the answer is $6666*(120/2)=6666*60$

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  • $\begingroup$ No, using symmetry, you count each occurrence of 3 twice. $\endgroup$ – steven gregory May 7 '16 at 13:02
  • $\begingroup$ It doesn't really matter whether you count 3 twice since the symmetry would mean exchanging two numbers. $\endgroup$ – Sinpoint May 7 '16 at 13:04

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