If the password needs to be 18 random characters, where
Exactly one character is an upper-case letter (A-Z) Exactly one character is a number (0-9) All other characters are lowercase letters (a-z)
How many possible passwords are there?
I was thinking its just 26+26+10 combinations for each letter and so 62^18 as the answer but there being exactly one of the uppercase letters and exactly one number being there are throwing me off
Let $n$ be the size of the your alphabet. Assuming the password all valid passwords have precisely 18 characters, there are exactly $3060 \cdot n^{17} = (18 \cdot n) \cdot (17 \cdot 10) \cdot n^{16}$ valid passwords.
Firstly, there are $18$ possible choices for the position of the uppercase letter and $n$ choices for this letter. This is where the $18n$ factor comes from. Secondly, there are $17$ possible choices for the position of the number and $10$ choices for such a number, so we multiply by a factor of $17 \cdot 10$. Finally, there are $16$ positions left, each of which with $n$ possible choices. Hence we multiply by $n^{16}$. If you assume $n = 26$ is the size of the standard English alphabet you get $ 3060 \cdot 26^{17}$ valid passwords.
Since exactly one character is Uppercase and one character is a number, you don't have 62 possibilities for each character.
First, which one is Uppercase. $${18 \choose 1}\times 26$$ Second the number $${17 \choose 1}\times 10$$ Finally the rest could be any lowercase letter $$26^{16}$$ Complete answer $${18 \choose 1}\times 26\times {17\choose 1}\times 10 \times 26^{16}=18\times17\times10\times26^{17}$$
