The formula for sum of all numbers formed with all the given digits is:
(Sum of digits) (n-1)!(1111....ntimes)
n stands for number of digits.
For ex:
Sum of all numbers formed with 1,2,3 and 4 is
10 × 3! × 1111 = 66660.
Can anyone figure out how this was derived?
Best Answer
I figured the answer out.
If you fix 1 in the thousands place, all other numbers can be arranged in 3! Ways.
Therefore 1 occurs in thousands place 3! Times. Same for all the numbers. Therefore for thousands place, total sum= 1000(sum) 3!.
But same thing occurs for all other positions. Therefore sum= 1111...×(sum) ×(n-1)!