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 

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 $6.$ 
