NIMCET 2013 Question Paper with solutions

Non-negative integers means the whole numbers starting from $0.$
If we consider all the one-digit, two-digit and three-digit numbers at once (i.e. we allow 0 at the extreme left):
The total number of numbers from 1 to 1000 :
$=(1$ possibility for thousands place) AND (10 possibilities for hundreds place) AND (10 possibilities for tens place) AND (10 possibilities for units place)
$=1\times 10\times 10\times 10={10}^3$
The total number of numbers WITHOUT the digit 1 from 0 to 999 :
$=(9$ possibilities for hundreds place) $\mathit{A}\mathit{N}\mathit{D}(9$ possibilities for tens place) $\mathit{A}\mathit{N}\mathit{D}(9$ possibilities for units place)
$=9\times 9\times 9=9^3$
$\therefore$ The number of numbers with a digit 1 will be ${10}^3-9^3$ .