TITLE: Find the sum of all 4-digit numbers formed by using the digits 1, 2, 3, 4, 5
QUESTION [5 upvotes]: ...and where no digit is to be repeated in any 4-digit number.
Yes, I am aware of similar questions on this site, but none of the answers gave me insight as to how this works.
I would like an answer in ridiculously simple terms, like explaining it to a 5 year old.
One answer:

Total numbers formed using those 5 digits is 5P4. i.e. 120 numbers.
Since all digits will equally occupy the unit's place, therefore 1
  will occur 24 times in unit's place. Similarly 2 will occur 24 times
  in unit's place. So on and so forth. Therefore sum of all digits in
  unit's place will be equal to 24 x (1+2+3+4+5)=24 x 15.
Similarly sum of all digits in ten's place will be equal to 24 x 150.
Therefore total sum of all numbers =24 x (15+150+1500+15000)=399960 .

Why is there the sum (1 + 2 + 3 + 4 + 5)? Am I misreading the question? Does "no digit is to be repeated in any 4-digit number" mean that there shouldn't be a number like 4432, or does it mean that a number should not be repeated in the same "unit slot"?

REPLY [2 votes]: It's just a shorthand for adding up the $120$ possible values of the ones place.
The digit $1$ occurs $24$ times, as does $2, 3, 4,$ and $5$,
If your digits were $(2,4,5,7,9)$ then the sum would be $24 \times (2+4+5+7+9)$.