TITLE: How many three digit numbers have all three digits that are different?
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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 $9\times 9\times 8$ .