How many three digit numbers have all three digits that are different? How do I solve this?
Please explain in detail. I am told the answer is $9\times9\times8$ but I do not understand how that would lead to the solution.
 A: Since a three digit number cannot have a zero at hundredth place , therefore the first digit from left can be chosen from any one of $1$-$9$ digits ,i.e., hundredth place can be filled in $9$ ways . The tenth place can have a zero, but not the digit at the hundredth place,i.e., there are $9$ ways the tenth place (i.e. the middle digit) can be filled with . Similarly , the units place can have a zero as a digit but not the digits at hundredth and tenth place and hence there are $8$ ways the units place can be filled with. 
Now, by the principle of counting, total no. of ways are $9\times 9\times 8$ .
A: Imagine a three digit number $\overline{abc}$. For it to have three digits, then $a$ can't be zero, so you have $9$ possibilities for $a$ ($1$ - $9$). For $b$ you have $9$ possibilities as well, since you can now use $0$ but can't use the number in $a$. Finally, for $c$ you can't use the number in $a$ and $b$, so you are left with $8$ possibilities. The number of numbers that match all this is $9\times 9\times 8$.  Why are we multiplying?  Because for every a you have $9$ possible values of $b$ and 8 possible values of $c$, etc.
