Question 1096948
THE SUM OF ALL THE DIGITS IS 9000, but I was solving the wrong problem.


If we could use zero for the first digit, we would have a
{{{10*9*8=720}}} 3-digit sequences with no repeated digits.
Because each of the 10 digits would be equally likely in each position,
and the average digit is {{{(0+9)/2=4.5}}} ,
the sum of all the digits on those {{{9*8*7}}} {{{3 - digit}}} sequences is {{{720*3*4.5=9720}}} .
Of those 3-digit sequences, the {{{9*8=72}}} that begin with zero do not count as 3-digit numbers.
The sum of their digits is the sum of the last two (non-zero) digits.
Those last two digits are from 1 to 9,with an average of {{{(1+9)/2=5}}} ,
and each of those digits appears with the same frequency.
So, the sum of those digits is
{{{72*2*5=720}}} .
Subtracting that sum
(the sum of digits of the 3-digit no-repeat sequences that are not valid 3-digit numbers)
from the sum of all digits of all 3-digit no-repeat sequences,
we get the answer as {{{9720-720=highlight(9000)}}} .