Question 567696
There are 901 numbers between 100 and 1000 (inclusive). Instead of counting how many numbers have at least one 2, we can count how many numbers do not have any 2's at all, and subtract from 901 (since this is the complement of what we want).


For each of the three digits, there are 8 choices for the first digit (anything except 0, 2), 9 choices for the second digit (anything except 2) and 9 choices for the third digit. Hence the number of 3-digit numbers with no 2's is 8*9*9 = 648. Including the number "1000" we have 649. Subtract this from 901 to get 901 - 649 = 252.


Note that it does not matter if we include "100" or "1000" in our set since we'll be subtracting them anyway (instead of 901 - 649 we'll get 900 - 648 or 899 - 647 = 252).