How many positive, three digit integers contain atleast one 7? This is the Question: How many positive, three digit integers contain atleast one 7?
For these kind of questions I have always followed a technique of first taking care of the restriction provided in the question. The Restriction is contain atleast one 7
This is similar to the question:
In how many ways can the five letters J,K,L,M,N can be arranged such that L is not in the middle. Well for this question,taking care of the restriction,  I know that the middle letter stage can be accomplished in 4 ways since it cannot contain L, so the combination is 
$4\cdot3\cdot4\cdot2\cdot1 = 96$ ways.
But the above technique is not working for the question I asked. How to use the same technique where I first tackle with the restriction and then move on with the question.
What I tried:
Three Cases are possible:
1) Three Digits with atleast one seven: $1\cdot8\cdot7 = 56$ ways
2) Three Digits with atleast two seven: $1\cdot1\cdot8 = 8$ ways
3) Three Digits with atleast three seven: $1\cdot1\cdot1 = 1$ way
So, it should be $56\cdot8 = 448$ ways but thats wrong, The answer is 252 ways
So, how can I solve this question with the same strategy that I have followed of taking care of the restrictions first?

A simple way. 


*

*The first digit is 7. There are 100 of such numbers.

*The first digit is  not 7 but the second is. There are $8\cdot10=80$ such numbers.

*The first and second digit are not 7 andthe last one is 7. There are $8\cdot9=72$ (the first must be nonzero) such numbers, which gives the desired 252.
