Suppose you want to generate a password using ASCII characters ($128$ characters.)How many ways can you create a password of 10 characters long that has at least one lowercase letter (a-z) and at least one number ($0-9$)?
** MY ANSWER: $26*10*(128)^{10}$**
I don't understand why this is wrong? Can anybody help? Thanks!
On
The easiest way is to count all the passwords, then deduct those that don't meet the requirement. There are $128^{10}$ total passwords, ignoring the requirement for a lower case letter and a number. Of those $(128-26)^{10}$ have no lower case letter and $(128-10)^{10}$ have no number. But we have deducted twice those that have no lower case letter and no digit-an example of the inclusion-exclusion principle. So we add them back in-there are $(128-36)^{10}$ of those. So the final answer is $128^{10}-102^{10}-118^{10}+92^{10}=578747468760442009600$ per Alpha$
It is easiest to concentrate first on the bad strings, the ones that do not qualify. These are the strings that are missing a lower-case letter, or a number, or both.
There are $102^{10}$ strings with no lower-case letters, and $118^{10}$ with no numbers. However, if we add these two numbers we are double-counting the words that miss both letter and number. There are $92^{10}$ of these. It follows that there are $102^{10}+118^{10}-92^{10}$ bad strings. Thus the number of good strings is $$128^{10}-102^{10}-118^{10}+92^{10}.$$