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
Best Answer
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).