# Possible 4 character passwords involving a letter and a digit.

A password consists of 4 characters, each of which is either a digit or a letter of the alphabet. Each password must contain at least ONE digit and AT LEAST ONE letter. How many different such passwords are there?

I am not sure how to go about this one. At first I did 4 blank digits: XXXX. I did lowercase and uppercase separately since it doesn't specify above. 36^4=1,569,616 for just digits and lowercase. 14,776,336 for lowercase, uppercase, and digits (62^4).

There are two different ways I can go about this and i'm not sure which one.

Originally I did 26*36*36*10 (reserved a spot for the digit, and one for the lowercase) and got 336,60 which seems a bit low.

Then I took the total and subtracted it from the things it could not be (10^4 for digits, and 26^4 for letters only) and got 1,212,640.

I essentially did the same with uppercase, but wasn't sure if I should use 26^4 twice (for lowercase and uppercase) or use 52^4.

Which method should I be using, and if possible please explain why?

Thank you!!!

We might as well assume case doesn't matter - if it does, you just need to adjust the numbers slightly. The total number of strings of characters is, as you pointed out, $36^4$. However, not all of these are passwords. If all characters are letters ($26^4$ possibilities) or all characters are numbers ($10^4$ possibilities, none of which doubles up with a case where all characters are letters) then our string is not a password. In all other cases the string IS a password. Hence there are $$36^4-(26^4+10^4)=1\ 212\ 640$$ possible passwords. If the password is case sensitive, the answer becomes $$62^4-(52^4+10^4)=7\ 454\ 720.$$

Passwords are almost always case sensitive, for a given letter there are $26 \times 2=52$ possibilities, and for the digits we have $10$ possibilities.

Given the restrictions, there are three scenarios;

1. 3 letters, 1 digit
2. 2 letters, 2 digits
3. 1 letter, 3 digits

Consider the first scenario. There are 4 ways to arrange where the digit goes, or equivalently where the letters go. Let L denote a letter, and D denote a digit. The 4 "choices" are LLLD, LLDL, LDLL, DLLL. 4 ways of choosing where one thing belongs (in this case, we are choosing where the letter goes, but it it is equal to choosing where 3 letters go) is $\binom{4}{1}$. Thus, if we want to count the total possibilities for creating a password with 3 letters and 1 digit we compute $$\binom{4}{1}52^310^1$$

So now we can extend this to the other scenarios and sum it all up. We find: $$\binom{4}{1}52^310^1+\binom{4}{2}52^2 10^2+\binom{4}{3}52^1 10^3 = 7,454,720$$