
August 6th, 2002, 09:21 AM
#1
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!
Then ONLY USING CAPITALS!
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..

August 6th, 2002, 11:32 AM
#2
Senior Member
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)
53459728531456
01028071702528
00019770609664
00000380204032
00000007311616
00000000140608
00000000002704
00000000000052
+============
54507958502660
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)
208827064576
008031810176
000308915776
000011881376
000000456976
000000017576
000000000676
000000000026
+==========
217180147158
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.

August 7th, 2002, 02:51 AM
#3
Junior Member
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!

August 7th, 2002, 04:24 AM
#4
Junior Member
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?

August 7th, 2002, 11:02 AM
#5
Junior Member
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)

August 7th, 2002, 11:19 AM
#6
um, maybe ill get it right this time
With an 8 game thing like you said:
(8*7*6*5*4*3)/(6*5*4*3*2*1)=28
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.

August 7th, 2002, 02:05 PM
#7
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

August 7th, 2002, 08:57 PM
#8
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 bruteforce 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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