# Figuring out number of possible passwords with 8 characters

let's suppose that we have a total of 8 characters in which each character can be either an uppercase character of a numerical digit. I guess my question would be based on the product rule, wouldn't that mean there are a total of 36^8 possibilities for such a password.

For example: from A - Z there are 26 capital characters and 0 - 9 symoblized 10 digits. In an 8 character password, wouldn't that just equate to 36^8 possible passwords. I am looking at my text book currently and it is saying something completely different. For some reason, it states that to find the number of passwords with 8 characters with the rules where each character can only be a capital Letter or a numerical digit, we must first find the number of strings with 8 characters and then subtract the number of Strings with no digits (which would be 26^8). Why does this occur? They never bring up the fact in the textbook that you can then that if we take only numerical digits and no characters then there are 10^8 different possibilities. Thanks!

• Make absolutely sure that you have read the question correctly. An answer of $36^8-26^8$ is the number of 8 character strings using where each character is a letter or digit that must have at least one digit. If we ignore the "must have at least one digit" requirement, the answer is indeed $36^8$. – JMoravitz Oct 15 '16 at 18:55
• Thanks @JMoravitz, could you kindly add your answer to the answer section. Thanks! – Linuxn00b Oct 15 '16 at 18:57
• @Linuxn00b his comment is an answer – reuns Oct 15 '16 at 19:25
• I want to mark his answer as correct. You can not mark a response as correct. – Linuxn00b Oct 15 '16 at 19:32