Rather than count the number of ways to put one or more #3#'s in a four-digit number, we can note that we can partition the set of all four-digit numbers into those which contain at least one #3#, and those which contain no #3#'s.
Thus, the number of four-digit numbers which contain at least one #3# is the difference between the number of four-digit numbers and the number of four-digit numbers containing no #3#'s.
The total number of four-digit numbers can be calculated as the number of integers from #1# to #9999# less the total number of integers from #1# to #9999# with fewer than four digits:
#"Total four-digit numbers" = 9999-999 = 9000#
The total number of four-digit numbers with no #3#'s can be calculated by noting that there are #8# possible choices for the first digit: #{1, 2, 4, 5, 6, 7, 8, 9}# and #9# for the each of the remaining digits: #{0, 1, 2, 4, 5, 6, 7, 8, 9}#. Thus, multiplying, we get
#"Total four-digit numbers with no 3's" = 8*9*9*9 = 5832#
Finally, subtracting:
#"Total four-digit numbers with">="one 3" = 9000-5832=3168#
