Math Question For You! *problem solveing*
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Thread: Math Question For You! *problem solveing*

  1. #1
    Senior Member
    Join Date
    Jan 2002

    Math Question For You! *problem solveing*

    Though i said id be gone forever my actions were flawed and based on my life and as we all know life is UNPREDICTIBLE! thus i will be on AO once in a while but i will never fully be as active of a member as i was due to life and all its weird ways

    for my first post back heres my question

    how many possible combinations can u make out of every lettle in the alphabet, using any # of them from 1 - 8 at a single time, and capital and lower case counting as a diff letter such as abcdefgh being diff then abcDefgh

    Then the same question but ONLY USING LOWER CASE!


    leason and reasoning? this will show you why it is good to pick a password with random letters numbers and capitals

    By the way, most passwords do allowed numbers and weird ascii characters!

    **Odds of winning jackpot are 1:135,145,920 and thats picking 6 numbers out of 50 , with 8 and 26 letters should be more but close to that
    [shadow]i have a herd of 1337 sheep[/shadow]
    Worth should be judged on quality... Not apperance... Anyone can sell you **** inside a pretty box.. The only real gift then is the box..

  2. #2
    Senior Member
    Join Date
    Jun 2002
    maybe my thinking on this totally off - but i think the logic straightforward.

    a charset of 26 characters * 2 (upper and lower) = 52
    range of length [1..8]

    so this is = (52^8) + (52^7) + (52^6) + (52^5) + (52^4) + (52^3) + (52^2) + (52^1)


    if so, then i guess answer for #2 and #3 would be equal; calculating with a charset of only 26 characters, maintaining the same range of length.

    =(26^8) + (26^7) + (26^6) + ...(26^1)


    somehow, i get the feeling i'm totally misreading the problem? i guess i'm not sure if the question was intended to be rhetorical or not - oh well no harm in me trying.

  3. #3
    Junior Member
    Join Date
    Jul 2002
    This is exactly the type of question that you study in Discrete Math for a compsci degree, and actually it's one of the easiest questions you'll encounter.

    Anyway, the number of possible chars = 26 + 26 = 52 (uppercase + lowercase). Then using the product rule, where P is the total number of possible passwords, and Pn is the number of possible passwords of length n, where 1 < n < 8,

    P = P1 + P2 + P3 + P4 + P5 + P6 + P7 + P8

    P1 = the number of strings of length one = 52^1
    P2 = the number of strings of length two = 52^2
    P8 = the number of strings of length eight = 52^8

    Then you just add all Pn together to get:

    P = 54,507,958,502,660

    The second and third questions resolve to the same number of permutations, but the base will be 26 vice 52 this time:

    P1 = 26^1
    P2 = 26^2
    P8 = 26^8

    P = 217,180,147,158

    Quite an immense difference between the two!

  4. #4
    Junior Member
    Join Date
    Apr 2002
    i was wondering how they worked the odds of stuff like that out, thanx.
    just one thing tho.
    How do u calculate the odds of, say in lotto you have to pick 6 numbers out of 50 to win the jackpot, how do you work out how much the odds increase when u can pick 8 numbers but still only have to get 6 right?

  5. #5
    Junior Member
    Join Date
    Jul 2002
    Seeing you can pick the winning numbers in any order it is:

    (50*49*48*47*46*45)/(1*2*3*4*5*6) (a big number)

  6. #6
    Senior Member
    Join Date
    Jun 2002
    um, maybe ill get it right this time

    With an 8 game thing like you said:
    So it's like playing 28 games at once.
    So instead of odds of 1 / 15,890,700
    you get odds of 1 / 567,525.

  7. #7
    AntiOnline Senior Member souleman's Avatar
    Join Date
    Oct 2001
    Flint, MI
    Ok, now for the real question that wasn't being asked. Can you make this into a dictionary file on your hard drive. Well, if 1 character = 1 byte, then (we will assume you want both the upper and lower case)...
    (53459728531456 * 8) = 4.27678e+14
    +(01028071702528 * 7) = 7.1965e+12
    +(00019770609664 * 6) = 1.18624e+11
    +(00000380204032 * 5) = 1.90102e+9
    +(00000007311616 * 4) = 2.92465e+7
    +(00000000140608 * 3) = 421824
    +(00000000002704 * 2) = 5408
    +(00000000000052 * 1) = 52
    A really large number of bytes (somewhere around 430,000,000,000,000) so 430, TeraByte (my math might be off by a few thousand gig or so, but you get the idea) Although you could take off 427284 bytes, because most programs require more then 3 characters now....And that is just plain text. Don't forget about numbers, and symbols like !@#$%^&*(), etc. And now a lot of passwords can be 16 characters or more.....
    \"Ignorance is bliss....
    but only for your enemy\"
    -- souleman

  8. #8
    Old-Fogey:Addicts founder Terr's Avatar
    Join Date
    Aug 2001
    Seattle, WA
    But you could simply dynamically generate the entries instead of making a dictionary out of them, since you theoretically know them all without having them stored since you know how they are created... in which case it becomes brute-force all over again.

    I would like to point out that, in the miracle of probabilities, just because there are other possibilities doesn't mean that it is safe. The password "ILoveYou" has eight characters, lower and upper case, but does that mean the attacker has a 1 in 54,507,958,502,660 chance of guessing it? Nah uh.
    [HvC]Terr: L33T Technical Proficiency

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