how many 3 digit numbers are there such that at least one of its digit is 6?

Question

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

Answer 1

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 $6.$

Answer 2

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).

Answer 3

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