Counting the total number of possible passwords I'm working from Kenneth Rosen's book "Discrete Mathematics and its applications (7th edition)". One of the topics is counting, he gives an example of how to count the total number of possible passwords.
Question: Each user on a computer system has a password, which is six to eight characters long, where each character is an uppercase letter or digit. Each password must contain at least one digit. How many possible passwords are there?
His solution: Let $P$ be the total number of possible passwords, and let $p_6, p_7$ and $p_8$ denote the number of possible passwords of length 6, 7, and 8, respectively. By the sum rule, $p = p_6 + p_7 + p_8$. 
$P_6 = 36^6 - 26^6 = 2,176,782,336 - 1,308,915,776 = 1,867,866,560$
Similar process for $p_7$ and $p_8$. 
This is where I'm confused, what is the logic for finding $p_6$? If I was given the question, I would have done as follows:
$p_6 = 36^5 * 10$, because 5 of the 6 characters can be a letter or a number, so 36 possible values for each character. One character has to be numerical, so it has 10 possible values. All multiplied together, gives you $p_6$. Obviously I'm wrong, but why is he right?
I'd just like to understand the thinking behind Rosen's solution, as he does not make that clear in the book.

This is the principle of inclusion-exclusion. We see here that it is much easier to count the number of cases without restriction and remove the undesirable cases.
Thus we get
$$(36^6-26^6)+(36^7-26^7)+(36^8-26^8).$$
In each case of a password of length $n$ where $n$ ranges over the integers from $6$ to $8$ inclusive we have removed the case where the password consists only of uppercase letters, leaving us with all possibilities for those passwords which contain at least one digit.
Although this is not always true, the words at least in a problem may imply that the method of inclusion-exclusion is the best way forward. Keep an eye out for similar verbiage in other problems and consider applying the same principle.
