Question: Each user on a computer system has a password, which is six to eight characters long, where each character is an upper-case letter or a digit. Each password must contain at least one digit. How many possible passwords are there?
I'm in the Basic of Counting section of my Discrete Mathematics book, and I have a problem with my reasoning with this. I will give you my reasoning and the books reasoning. Both give different answers, but I don't see a difference in the train of thought, so I need someone to point out the difference.
My Attempt: Immediately I noticed 3 kinds of character length which allows me to break it down to three cases, respectively $P_6, P_7, P_8$, then add all of them. For $P_6$, $5$ will be made up of alpha numeric characters, and $1$ is made up of just digits due to the requirement of "at least one digit", thus
$$P_6 = (36)^5*10$$
Should be enough to show my train of thought, now the books solution.
Books Solution: The book did the same thing in dividing in 3 cases and adding them later so I'll go ahead and show you their train of thought for $P_6$.
$$P_6 = 36^6 - 26^6$$
Basically its the number of possible 6 alphanumeric minus just alpha numeric.
I know that both give different answers, but I still can't tell why.