How many eight character passwords are there if each character is either an uppercase letter A-Z, a lowercase letter a-z, or a digit 0-9, and where at least one character of each of the three types is used?
I believe this question is about using the technique of 'counting the complement'. The complement of "one character of each" is "either uppercase, lowercase, or digits is NOT used".
Total possible passwords : $62^8$
Total possible passwords w/ out uppercase letters: $36^8$
Total possible passwords w/ out lowercase letters: $36^8$
Total possible passwords w/ out digits : $52^8$
I think my trouble is from fully understanding what the 'complement' means. From what I understand it is (very loose explanation) the opposite of the restriction placed on our original set. So in our case the original set is $62^8$ and the complement of the restriction placed on that set, one of each character, is all the possibilities where types of characters are not there. Which are what I typed above this paragraph. However, I feel like I am missing something. Otherwise I would say that the answer is: $62^8 - [36^8 + 36^8 + 52^8]$.
Main problem: Not understanding how to compute the complement.