Given a 3-digit number containing exactly one "1". (ASSUME base-10 numbers, so digits are from 0-9)
Number contains exactly one "1", therefore the two other digits are limited to 2,3,4,....9,0. If the "1" is in the leftmost position, there are 9*9=81 ways to arrange the second and third positions. If the "1" is in the second or third position, then there are 8 choices for the leftmost digit (2-9), and 9 choices for the other digit for a total of 2*8*9=144 arrangements. So the total number of 3-digit numbers containing exactly one "1" is 81+144=225.
