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Give that a password can be 8-12 characters long and each character in the password can be ether composed of upper case letters, lower case letters, numeric digits, or one of the six special characters * < > ! + =

How many different password are available for this computer system? there are 5 cases for length of 8,9,10,11,and 12.

What I think: there are 68 different possibilities for each character slot there for for each length we aggregate up like below: Answer: 68^8 + 68^9 + 68^10 + 68^11 + 68^12

Is this correct?

Question 2: How many of these passwords contain at least one occurrence of at least one of the six special characters?

Dealing in cases again, I approached it by finding all the ways to construct the password (68^n. where n is for each case length 8,9,10,11,12) and subtracting all the ways to build a password without any special characters (62^n)

Add up all for each case to get answer.

so (68^8-62^8)+(68^9-62^9)+(68^10-62^10)+(68^11-62^11)+(68^12-62^12)

Is this correct?

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    $\begingroup$ Yes. Yes. ${}{}{}$ $\endgroup$ – dtldarek Dec 9 '12 at 23:05
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To add a formal answer: I agree with @dtldarek, your argument is good for both questions.

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