password question, need some help..

- anonymous

password question, need some help..

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- anonymous

##### 1 Attachment

- lopus

http://www.khanacademy.org/math/probability/v/probability-using-combinations

- anonymous

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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## More answers

- anonymous

Cinar, can you do this?

- anonymous

62^10

- anonymous

that is total only..

- anonymous

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?

- anonymous

10^52+10^36+10^36

- anonymous

listen I'll be right back ok, give me couple minutes, I need to do something..

- anonymous

\( (26 +26)^{10}\) Have no digit
\( (26+10)^{10} \) Have no lower case
\((26+10)^{10} \) Have no upper case

- anonymous

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

- anonymous

you'are right, I made a typo..

- anonymous

So we have \((10+26+26)^{10} - ((10+26)^{10} + (10+26)^{10} + (26+26)^{10} - (10^{10} + 26^{10} + 26^{10})) \)

- anonymous

Do you understand how I did that?

- anonymous

hold on..

- anonymous

it is like union of three set..

- anonymous

Yeah

- anonymous

Are you doing probability with set theory?

- anonymous

Discreet math?

- anonymous

Discreet math

- anonymous

So do you or don't you understand how I got what I got?

- anonymous

so, AUBUC=A+B+C-AIB-AIC-BIC+AIBIC
I=intersection

- anonymous

Yeah

- anonymous

got it..

- anonymous

how about part b, any idea..

- anonymous

Do you have any ideas where to start?

- anonymous

let see

- anonymous

preceded by a letter means a1 or 1a

- anonymous

I am confused..

- anonymous

I'm pretty sure it means a1

- anonymous

hmm, then password cannot be begun by a number..

- anonymous

Well, one thing to ask is, must all digits be preceded by a number, or just at least one digit?

- anonymous

and also no number is successive

- anonymous

22 or 555 not allowed

- anonymous

a2ddd3y8jj8i898 is ok

- anonymous

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.

- anonymous

so 1a23456789 would be ok too.

- anonymous

that' why I am confused..

- anonymous

listen I need some break to eat something..

- anonymous

I am back..

- anonymous

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.

- anonymous

So one way is 1 digit and then 9 letters, with at least one capital and one lowercase

- anonymous

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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