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password question, need some help..

Mathematics
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1 Attachment
http://www.khanacademy.org/math/probability/v/probability-using-combinations
a) Let's try first by calculating the total number of passwords without requirements, and then subtracting the total number of passwords which DON'T meet the requirements.

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Other answers:

Cinar, can you do this?
62^10
that is total only..
Now how many don't meet the 'at least one digit' rule, how many don't meet the 'at least one lower case letter', and how many don't meet the 'at least one upper case letter' requirement?
10^52+10^36+10^36
listen I'll be right back ok, give me couple minutes, I need to do something..
\( (26 +26)^{10}\) Have no digit \( (26+10)^{10} \) Have no lower case \((26+10)^{10} \) Have no upper case
We still double count the ones who break two or more rules, so we have to get rid of them. \(10^{10} \) break both upper and lower case rule \(26^{10} \) break both upper case and digit rule \(26^{10} \) break both lower case and digit rule \(0\) break all three rules, because it must contain at least a digit or letter
you'are right, I made a typo..
So we have \((10+26+26)^{10} - ((10+26)^{10} + (10+26)^{10} + (26+26)^{10} - (10^{10} + 26^{10} + 26^{10})) \)
Do you understand how I did that?
hold on..
it is like union of three set..
Yeah
Are you doing probability with set theory?
Discreet math?
Discreet math
So do you or don't you understand how I got what I got?
so, AUBUC=A+B+C-AIB-AIC-BIC+AIBIC I=intersection
Yeah
got it..
how about part b, any idea..
Do you have any ideas where to start?
let see
preceded by a letter means a1 or 1a
I am confused..
I'm pretty sure it means a1
hmm, then password cannot be begun by a number..
Well, one thing to ask is, must all digits be preceded by a number, or just at least one digit?
and also no number is successive
22 or 555 not allowed
a2ddd3y8jj8i898 is ok
I think a123456789 is also okay, because it has one digit immediately preceded by a letter. They didn't say ALL must follow the rule, just one.
so 1a23456789 would be ok too.
that' why I am confused..
listen I need some break to eat something..
I am back..
Ok. So one thing we can take into consideration is that for a password to have a digit and be invalid, it would have to have to be 1... 11.... 111.... basically a bunch of digits in order with no letters in between.
So one way is 1 digit and then 9 letters, with at least one capital and one lowercase
So find how many ways we can have a password that is valid in (a) and not valid in (b) and subtract that from (a)

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