TITLE: If all of the integers from $1$ to $99999$ are written down in a list, how many zeros will have been used?
QUESTION [6 upvotes]: If all of the integers from $1$ to $99999$ are written down in a list,
  how many zeros will have been used?

I just counted how many 5 digit numbers have 1, 2, 3 or 4 zero's and subtracted all the cases where the first digit is zero.
$\binom{5}{1}(9^4) + \binom{5}{2}(9^3\times 2) + \binom{5}{3}(9^2\times3)+\binom{5}{4}(9\times4) - 9(4) -90(3)-900(2)-9000(1)=38889$
This is a rather awkward sum though, and no calculator is supposed to be used, so I was wondering if there is a better/more elegant way of doing it.

REPLY [5 votes]: You may find this slightly more elegant (no calculator was needed):
The number of digits from 1 to 99999 is 
$$
9+90*2+900*3+9000*4+90000*5=488889
$$
The number of times a non-zero digit appears (in a particular place) is 
$$
10^4=10000
$$
Multiplied by five possible places is 
$$
10000*5=50000
$$
The total times all non-zero digits appear is
$$
50000*9=450000
$$
Now find the amount of zeros written by
$$
488889-450000=38889
$$