Assume all telephone numbers are 10 digits long, consisting of a 3-digit area code, then a 3-digit "exchange" number, followed by a 4-digit number. How many telephone numbers have no 0 in the area code and no 9 in the final 4-digit number?
I want to say the answer is $8^{10}$, but will that account for 0 also being the second or third number in the area code and 9 being the second, third or fourth number?
So let's work with the area code first consisting of 3-digits. If we had no constraint, then the total possible number of area codes is clearly $10^3 = 1000$ (starting from 000 up till 999).
Since no $0$ is allowed in the Area code ($AAA$), then we have nine allowed digits, hence a total number of area codes equal to $9^3$. So, $$\text{Total number of area codes} = 9^3$$
Similarly, the last digits has no 9 and using the same reasoning above, the total is $$\text{Total number of last 4-digits codes} = 9^4$$
The exchange number is left intact (no constraints) so: $$\text{Total number of last exchange numbers} = 10^3$$
So, the total number of possible phone numbers will be $9^3 \times 9^4 \times 10^3$