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cinar

  • 2 years ago

password question, need some help..

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  1. cinar
    • 2 years ago
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  2. lopus
    • 2 years ago
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    http://www.khanacademy.org/math/probability/v/probability-using-combinations

  3. wio
    • 2 years ago
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    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.

  4. wio
    • 2 years ago
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    Cinar, can you do this?

  5. cinar
    • 2 years ago
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    62^10

  6. cinar
    • 2 years ago
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    that is total only..

  7. wio
    • 2 years ago
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    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?

  8. cinar
    • 2 years ago
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    10^52+10^36+10^36

  9. cinar
    • 2 years ago
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    listen I'll be right back ok, give me couple minutes, I need to do something..

  10. wio
    • 2 years ago
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    \( (26 +26)^{10}\) Have no digit \( (26+10)^{10} \) Have no lower case \((26+10)^{10} \) Have no upper case

  11. wio
    • 2 years ago
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    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

  12. cinar
    • 2 years ago
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    you'are right, I made a typo..

  13. wio
    • 2 years ago
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    So we have \((10+26+26)^{10} - ((10+26)^{10} + (10+26)^{10} + (26+26)^{10} - (10^{10} + 26^{10} + 26^{10})) \)

  14. wio
    • 2 years ago
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    Do you understand how I did that?

  15. cinar
    • 2 years ago
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    hold on..

  16. cinar
    • 2 years ago
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    it is like union of three set..

  17. wio
    • 2 years ago
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    Yeah

  18. wio
    • 2 years ago
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    Are you doing probability with set theory?

  19. wio
    • 2 years ago
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    Discreet math?

  20. cinar
    • 2 years ago
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    Discreet math

  21. wio
    • 2 years ago
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    So do you or don't you understand how I got what I got?

  22. cinar
    • 2 years ago
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    so, AUBUC=A+B+C-AIB-AIC-BIC+AIBIC I=intersection

  23. wio
    • 2 years ago
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    Yeah

  24. cinar
    • 2 years ago
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    got it..

  25. cinar
    • 2 years ago
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    how about part b, any idea..

  26. wio
    • 2 years ago
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    Do you have any ideas where to start?

  27. cinar
    • 2 years ago
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    let see

  28. cinar
    • 2 years ago
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    preceded by a letter means a1 or 1a

  29. cinar
    • 2 years ago
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    I am confused..

  30. wio
    • 2 years ago
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    I'm pretty sure it means a1

  31. cinar
    • 2 years ago
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    hmm, then password cannot be begun by a number..

  32. wio
    • 2 years ago
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    Well, one thing to ask is, must all digits be preceded by a number, or just at least one digit?

  33. cinar
    • 2 years ago
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    and also no number is successive

  34. cinar
    • 2 years ago
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    22 or 555 not allowed

  35. cinar
    • 2 years ago
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    a2ddd3y8jj8i898 is ok

  36. wio
    • 2 years ago
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    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.

  37. wio
    • 2 years ago
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    so 1a23456789 would be ok too.

  38. cinar
    • 2 years ago
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    that' why I am confused..

  39. cinar
    • 2 years ago
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    listen I need some break to eat something..

  40. cinar
    • 2 years ago
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    I am back..

  41. wio
    • 2 years ago
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    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.

  42. wio
    • 2 years ago
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    So one way is 1 digit and then 9 letters, with at least one capital and one lowercase

  43. wio
    • 2 years ago
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    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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