how many 3 digit numbers are there such that at least one of its digit is 6? I know the answer using
Total-none of it is $6 = 900−648=252$
My doubt is that if we you it like
when unit digit is $6 = 9\cdot10\cdot1 = 90$
when tens place is $6 = 9\cdot1\cdot10 = 90$
when hundredth place is $6 = 1\cdot10\cdot10 = 100$
So total $= 90 + 90 + 100 = 280$
why my answer is not the same 
 A: If exactly one digit is a $6,$ there are $1\cdot 9\cdot 9+8\cdot1 \cdot 9+8\cdot 9 \cdot 1=225$ possibilities.
If exactly two digits are a $6,$ there are $9+9+8=26$ possibilities.
If exactly three digits are $6,$ there is only one possibility.
Thus there are $252$ possibilities in total.
Just to clear things up,
If the units digit is $6,$ then you should have $8\cdot9\cdot1$ possibilities, not what you got. 
If the tens digit is $6,$ then you should have $8\cdot1\cdot9=81$ possibilities. 
If the hundreds digit is $6,$ then there are $1\cdot9\cdot9$ possibilities. This is because you cannot include $0$ as the hundreds digit and you must subtract $1$ from $10$ since the other digits cannot be $6.$ Also you need to add the possibilities where there is more than one $6.$ 
A: Suppose that 'three-digit' means $abc$, where $a>0$.  
Now, we first count that there are $900$ of these numbers.
Of numbers without a $6$, then it's $8\times 9 \times 9$, since the first digit can be any of 1-5 or 7-9, and the rest 0-5 or 7-9. This gives $648$ numbers without a 6.
One then finds that there are $252 = 900-648$ numbers that contain at least one six (or any other specific non-zero digit).
A: your answer is different in second method because in first case when you have counted many no. 2 or 3 times like 666 is counted all three times so that ways counted multiple times must be subtracted and your answer will match
