# Finding Number Of Cases,Simple Counting Question

How many positive integers less than 1000 have at least one decimal digit equal to 9?

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?

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The question would be easier to answer if you explained how you counted. –  joriki Dec 3 '12 at 3:53
Yes, you're also double-counting three-digit numbers like $919$, $299$, etc. –  mjqxxxx Dec 3 '12 at 4:01

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 $999$.

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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, $$\text{Number of numbers that have at-least one digit as 9} = 10^3 - 9^3 = 271$$

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