Question 394292
The 9 occurs in the units digits for the numbers 9, 19, 29, ..., 249 and in the tens place 90, 91, ..., 99 and 190, 191, ..., 199.


We can count each case separately since they are mutually disjoint and do not affect each other (we can count 99 and 199 in both sets since the 9 occurs twice in each number). The number of elements in the set {9, 19, ..., 249} is 25 and the number of elements in the sets {90, 91, ..., 99}, {190, 191, ..., 199} is 10 times 2, or 20. The total number of 9's is 25+20, or 45.