Number of possible passwords - combinatorics This question has been asked and answered before:

An online portal requires a user login with a password which must be from 6 to 8 characters long, where each character is a lowercase letter or a digit. A password must contain at least 1 digit. How many possible passwords are there?

All answers so far either use the inclusion exclusion pie or sum the number of words with exactly 1 digit (and thus 26 choices for all other characters) plus the words with exactly 2 digits and so on.
When I saw this problem in class my first thought was: we are only going to deal with words of length $6$ and then add the cases where the word length is $7$ or $8$ (as these choices are mutually exclusive). We can choose from $26+10=36$ options for each character except for one in each word, since one character must be a digit so we have $10$ choices for this specific character and $6$ positions in which this character can be. Thus we must have $6\cdot10\cdot36^5$ possible passwords, which I understand is wrong since all possible passwords with length 6 without the at least one digit constraint are $36^6$ which is less than what we calculated before.
Our professor emphasized the fact that we have at least one digit and not exactly one. However I fail to understand where I am wrong intuitively. I can see the numbers do not make sense but I cannot exactly understand where I am miscounting (maybe counting some combinations twice?).
 A: The expression $6\cdot10\cdot36^5$ has a particular "special" character picked out as "the character which had to be a digit".
In a password like misha1 there is a unique character which is a digit, which must be that particular "special" character, so there is no problem.
But a password like 23fish has two digits, and either one of might be the special character. You count it once when you pick the first character as the digit character, put a 2 there, and fill in the rest with 3fish. You count it again when you pick the second character as the digit character, put a 3 there, and fill in the rest with 2_fish. A password like 456789 gets counted $6$ times.

You could correct your approach. If you think of $6 \cdot 10 \cdot 36^5$ as
$$
    10 \cdot 36^5 + 10 \cdot 36^5 + 10 \cdot 36^5  + 10 \cdot 36^5  + 10 \cdot 36^5  + 10 \cdot 36^5  
$$
where the $i^{\text{th}}$ term counts the passwords where the $i^{\text{th}}$ character is a digit, it's easy to see that there's overlap. Instead, to avoid overlap, you could make the $i^{\text{th}}$ term count the passwords where the $i^{\text{th}}$ character is the first digit that occurs. Then we get
$$
   10 \cdot 36^5 + 26 \cdot 10 \cdot 36^4 + 26^2 \cdot 10 \cdot 36^3 + 26^3 \cdot 10 \cdot 36^2 + 26^4 \cdot 10 \cdot 36 + 26^5 \cdot 10
$$
and this is another expression for the correct number of $6$-character passwords with at least one digit.

I'm also going to include the "standard" approach for the sake of anyone else who finds this question, though I understand that you're already aware of it.
That's to count all $6$-character passwords, then subtract the ones which don't have any digits, leaving only the ones that do. We get $36^6 - 26^6$ as the answer with this approach.
