Question

password question, need some help..

Answer 1

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

Answer 2

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.

Answer 3

Cinar, can you do this?

Answer 4

62^10

Answer 5

that is total only..

Answer 6

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?

Answer 7

10^52+10^36+10^36

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

you'are right, I made a typo..

Answer 12

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

Answer 13

Do you understand how I did that?

Answer 14

hold on..

Answer 15

it is like union of three set..

Answer 16

Yeah

Answer 17

Are you doing probability with set theory?

Answer 18

Discreet math?

Answer 19

Discreet math

Answer 20

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

Answer 21

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

Answer 22

Yeah

Answer 23

got it..

Answer 24

how about part b, any idea..

Answer 25

Do you have any ideas where to start?

Answer 26

let see

Answer 27

preceded by a letter means a1 or 1a

Answer 28

I am confused..

Answer 29

I'm pretty sure it means a1

Answer 30

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

Answer 31

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

Answer 32

and also no number is successive

Answer 33

22 or 555 not allowed

Answer 34

a2ddd3y8jj8i898 is ok

Answer 35

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.

Answer 36

so 1a23456789 would be ok too.

Answer 37

that' why I am confused..

Answer 38

listen I need some break to eat something..

Answer 39

I am back..

Answer 40

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.

Answer 41

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

Answer 42

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)