Finding Number Of Cases,Simple Counting Question How many positive integers less than 1000 have at least one decimal digit equal to 9?
Please correct me:
Answer = (1) +
{(1*10) + (9*1)}+ 
{(1*10*10) + (9*10*1) + (9*1*10)}
==> 9 + 9x + x9 + 9xx + xx9 + x9x
I think I counted 99 twice and 999 three times, are there more cases I over counted?
 A: It looks as if you counted numbers like $399$, or $919$, or $991$ more than once.
It is easier to count the numbers that don't have a $9$. There are $8$ from $1$ to $9$. For $10$ to $99$, the first digit can be any of $8$, and for each choice the second digit can be any of $9$, for a total of $8\cdot 9$. Similarly, for $100$ to $999$, the first digit can be any of $8$, and for any first digit, there are $9\cdot 9$ ways to choose the rest, for a total of $8\cdot 9\cdot 9$. 
Add up, subtract from $999$. 
Another way: It is slightly easier to insert if necessary lead $0$'s to make every number from $0$ to $999$ a three-digit number. Then there are $9^3 -1$ "nine-free" numbers (we do not allow $000$). Subtract from $999$. 
A: The numbers you are interested in are of the form $$\times \times \times$$ where $\times \in \{0,1,2,\ldots,9\}$. $$\text{Number of numbers that have at-least one digit as $9$ =}$$
$$\text{(Number of numbers of the form $\times \times \times$)}$$
$$ - \text{(Number of numbers of the form $\times \times \times$ with none of the digits as $9$)}$$
$$\text{Number of numbers of the form $\times \times \times$} = 10 \times 10 \times 10 = 1000$$
$$\text{Number of numbers of the form $\times \times \times$ with none of the digits as $9$} = 9 \times 9 \times 9$$
Hence,
$$\text{Number of numbers that have at-least one digit as $9$} = 10^3 - 9^3 = 271$$
