For this problem we are given that a password is a string of $6$ characters. The password must contain exactly $6$ characters and can contain uppercase and lowercase letters of the alphabet, numbers 0 through 9, and an underscore.
How many passwords cannot have a number character?
Solution: We know that there are $26$ upper and lower case words, so $52$ letters to choose from, and we can include our underscore. Therefore, we must have $53^6$ passwords.
How many passwords have exactly one underscore and that is not at the beginning or the end of the password.
Solution: We have one underscore in between the start and the end of the password. $62\cdot 63\cdot 63\cdot 63\cdot 63\cdot 62\cdot = 62^2 \cdot 63^4$
Password must have at least one number.
Solution: We take all the passwords that have a number and subtract out the number of passwords that don't have a number, $63^6-53^6$
Is my thinking on this correct? I am new to counting and was wondering if my intuition on these problems are on the right track or completely wrong.