You’re not taking into account that the digit can occur in any of the six positions. There are $10\cdot36^5$ passwords that begin with a digit. There are also $10\cdot36^5$ passwords that end with a digit; some of these also begin with a digit and have already been counted, but some do not, so your figure of $10\cdot36^5$ is necessarily too small.
The easiest way to count the acceptable passwords is to note that there are $36^6$ six-character strings made up of upper-case letters and digits, and $26^6$ of them are made up entirely of letters, so there are $36^6-26^6$ that include at least one digit.
It is possible to count them directly, but the counting is more complicated. For each of the $6$ positions in the password there are $10\cdot36^5$ passwords having a digit in that position, so to a first approximation there are $6\cdot10\cdot36^5$ acceptable passwords. However, as noted in the first paragraph, this counts some passwords more than once. For each pair of positions in the password there are $10^2\cdot36^4$ passwords having digits in both of those positions, and all of these passwords have been counted twice. Since there are $\binom62$ pairs of positions, we must subtract $\binom62\cdot10^2\cdot36^4$ to get rid of the double-counting. Unfortunately, this overcompensates, and there are further corrections to be made. The net result, given by the inclusion-exclusion principle, is
$$\sum_{k=1}^6(-1)^{k+1}\binom6k10^k36^{6-k}\;.$$
Either way, the result is $1,867,866,560$.
(b): How many passwords are there minus how many passwords are there with no uppercase or no lowercase letter?
(c): How many passwords are there minus how many passwords are there with no uppercase or no lowercase letter or no number?
Notation:
$$\Omega := \{ p\ |\ p\text{ is an } 8 \text{-digit password from the alphabet } [A-Z,a-z, 0-9]\}\\
U := \{p \in\Omega\ |\ p\text{ contains an uppercase letter}\} \\
L := \{p \in\Omega\ |\ p\text{ contains a lowercase letter}\} \\
\neg U := \Omega \setminus U = \{p\ |\ p\text{ is an } 8 \text{-digit password from the alphabet } [a-z,0-9]\} \\
\neg L := \Omega \setminus L = \{p\ |\ p\text{ is an } 8 \text{-digit password from the alphabet } [A-Z,0-9]\}$$
We trivially have
$$|\Omega| = 62^8, |\neg U| = |\neg L| = 36^8$$
Example for (a):
$$|A| = |L| = |\Omega| - |\neg L| = 62^8 - 36^8$$
Example for (b):
Now there are $36$ non-uppercase and $36$ non-lowercase characters as well as $10$ non-letters so
$$|B| = |U\cap L| = |\Omega| - |\neg U| - |\neg L| + |\neg U \cap \neg L| = 62^8 - 36^8 - 36^8 + 10^8$$
by inclusion-exclusion.
Can you do (c) now?
Best Answer
Let's look at cases. The password is $6$, $7$ or $8$ characters long.
Case 1. Say the password is $6$ characters long. We have $26$ lowercase letters to choose from and $10$ digits to choose from. Therefore we have $26+10=36$ selections for each slot in our password. By the rule of product this means we have $36^6$ total possibilities for our password. But we are concerned with passwords that have at least one digit and surely some of these in our total have no digits.
To remove the passwords with no digits let's total the number of no-digit $6$ character passwords we have, which is $26^6$. Then there are $36^6-26^6$ passwords with at least one digit (for more on this principle https://brilliant.org/wiki/principle-of-inclusion-and-exclusion-pie/).
Now, figure out case Case 2 ($7$ character password) and Case 3 ($8$ character password) similarly in the same fashion and add together the result of all three cases, this is your solution. We add these events because they are mutually exclusive (i.e. you may not have a $6$ character password and also a $7$ character password at the same time, if unfamiliar with this principle: https://brilliant.org/wiki/rule-of-sum/).