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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!

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Your answer is the amount of passwords you could generate that starts with a lower case letter, followed by a number, followed by 10 other ASCII characters. So, it's a string of length 12, and you don't take into consideration the different permutations. –  kba Dec 17 '12 at 3:50
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ASCII characters 0 through 31 are control characters, so for practicality you would never use these in a password. Same goes for 127, the delete character. And character 32 is the space character, which is questionable in a password. So if this is not just an exercise, you might want to limit your available characters to 94 0r 95. –  alex.jordan Dec 17 '12 at 7:08

2 Answers 2

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$

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Is the power of 12 correct? –  Amzoti Dec 17 '12 at 5:05
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@Amzoti: No, I was fixing it while you looked. –  Ross Millikan Dec 17 '12 at 5:06

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 $$128^{10}-102^{10}-118^{10}+92^{10}.$$

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