Count of 3-digit numbers with at least one digit as 9 
Find the number of $3$ digit numbers (repetitions allowed) such that at least one of the digit is $9.​$

I've posted my answer below. If there is a better way to solve this question, I would be glad to learn about that.
 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 $9$, then it's $8\times 9 \times 9$, since the first digit can be any of 1-8, and the rest 0-8.   This gives $648$ numbers without a 9.
One then finds that there are $252 = 900-648$ numbers that contain at least one nine (or any other specific non-zero digit).
A: If unit place is $9,$ hundredth place would have $9$ options $(1\,\text{to} \,9),$ tens place would have $10$ options $(0\, \text{to}\, 9).$ Total ways $90.$
When $9$ is fixed for tens place, unit place choices$=9 (0\, \text{to}\, 8),$ hundredth place choices$= 9 (1\, \text{to}\, 9).$ Total$=81.$
When $9$ is fixed for hundredth place, unit place choices$=9,$ tens place choices$=9.$ Total$=81$
Answer$=90+81+81=252$
A: The first digit $d_1$ can be $=9$ or one of $\{1,2,3,4,5,6,7,8\}$. In the first case we have $10^2=100$ choices for the  digits $d_2$ and $d_3$, since a $9$ is already here. In the second case the $9^2$ choices where both $d_2$ and $d_3$ are from $\{0,1,2,3,4,5,6,7,8\}$ are forbidden. Therefore
$$N=100+8\cdot(100-81)=252\ .$$
A: There are $\binom{3}{1}$ ways of having one $9$, and for each of these there are $9^2$ assignments of the remaining numbers.
There are $\binom{3}{2}$ ways of having two $9$s, and for each of these there are $9$ assignments of the remaining numbers.
There is exactly one way of having three nines.
Hence the answer is $\binom{3}{1} 9^2 + \binom{3}{2} 9 +1 = 271$.
Of course, a far simpler way is to note that there are $10^3$ possible 3 digit numbers and $9^3$ possible numbers with no 9 whatsoever, hence the answer is $10^3-9^3 = 271$.
If numbers starting with $0$ are disallowed, then we must remove the 19 numbers $009,019,..089, 090,...,099$. This would leave $271-19= 252$.
A: We can find this using Set theory.
We need to find the number of elements in union of 3 sets where the 3 sets are
Set 1) first digit contains 9
Set 2) second digit contains 9
Set 3) third digit contains 9
= n(S(first digit 9) ∪ S(second digit 9) ∪ S(third digit 9))
= n(first digit 9) + n(second digit 9) + n(third digit 9) - n(first two digits 9) - n(last two digits 9) - n(first and last digits 9) + n(all 3 digits 9)
= 100 + 90 + 90 - 10 - 9 - 10 + 1
= 252
Explanation:
3 digit numbers start from 100 and end at 999
Total no. of 3 digit numbers = 900
numbers in which 9 is the starting digit = 100 (Numbers from 900 - 999)
numbers in which 9 is the middle digit = 900/100*10 = 90 (Because every 100 has 10 numbers i.e. 90–99)
numbers ending with 9 = 900/10 = 90 (Because every 10 has a 9)
Numbers starting with 99 = 10 (990, 991, … 999)
Numbers ending with 99 = 9 (199, 299… 999)
Numbers starting at 9 and ending at 9 = 10 (909, 919,… 999)
Numbers with all digits as 9 = 1 (i.e. 999)
A: Total number of $3$-digit numbers $= 900$,
Total number of $3$-digit numbers that excludes $9 = 648$
Therefore required numbers containing $9$ at least once $= 900 - 648
                                                                                                   = 252$
