by Omnis
 P: 3 How do I work out password combination probabilities? For example, I'm looking for a password between 10-15 characters in length, the password can contain any alpha-numeric character (lower/upper alphabet character and 0-9 digits). How would I find out how many possible combinations there are?
 P: 15,319 What would you make for an answer as your first try? What logic did you use? Hint: How many possibilities are there for the first number in the sequence? How many possibilities are there for the second number in the sequence? Hw many times do you have to do this? What is a really compact way representing that as a generalized formula?
P: 3
 Quote by DaveC426913 How many possibilities are there for the first number in the sequence?
62

 Quote by DaveC426913 How many possibilities are there for the second number in the sequence?
62

 Quote by DaveC426913 How many times do you have to do this?
10-15 times depending on the password length

 Quote by DaveC426913 What is a really compact way representing that as a generalized formula?
I have no idea, thats why I came here, I'm wondering if its possible to recover a zip password on one of my files but I need to know how many possiblilities there are and I also like to know how to work it out for myself :)

 P: 15,319 Password Combination OK, well there's 62x62x62x62x62x62x62x62x62x62 + 62x62x62x62x62x62x62x62x62x62x62 + 62x62x62x62x62x62x62x62x62x62x62x62 + 62x62x62x62x62x62x62x62x62x62x62x62x62 + 62x62x62x62x62x62x62x62x62x62x62x62x62x62 + 62x62x62x62x62x62x62x62x62x62x62x62x62x62x62 or 62^10 + 62^11 + 62^12 + 62^13 + 62^14 + 62^15 or a little more than 781 septillion combinations. At one try per second, it will take you a mere 24 quadrillion years, a mere 1.8 billion times longer than the universe has existed to-date. Don't try to do them all in one sitting.
 Sci Advisor HW Helper P: 3,682 If you try ten billion passwords in a second with a given computer, and you use a million computers, it would take 1238 years on average.
HW Helper
PF Gold
P: 2,327
 Quote by DaveC426913 OK, well there's 62x62x62x62x62x62x62x62x62x62 + 62x62x62x62x62x62x62x62x62x62x62 + 62x62x62x62x62x62x62x62x62x62x62x62 + 62x62x62x62x62x62x62x62x62x62x62x62x62 + 62x62x62x62x62x62x62x62x62x62x62x62x62x62 + 62x62x62x62x62x62x62x62x62x62x62x62x62x62x62 or 62^10 + 62^11 + 62^12 + 62^13 + 62^14 + 62^15 or a little more than 781 septillion combinations. At one try per second, it will take you a mere 24 quadrillion years, a mere 1.8 billion times longer than the universe has existed to-date. Don't try to do them all in one sitting.
After the 10th character the user has the option not choose at all, so that is 63 choices.
 P: 22 isn't there a formula for that from allg or something. what is the equation not worked out like in a, b c, etc. form.
Math
Emeritus
Thanks
PF Gold
P: 39,682
The "formula" DaveC426913 was using earlier is often referred to as simply "The Fundamental Counting Principle": If you can do A in "n" ways and you can do B in "m" ways, then you can do A and B together in "mn" ways.

If you have 62 symbols you can use and there are no restrictions on how they can be used, then you have 62 ways you can write each of the, say, 10 symbols and so have

Altogether, then, there are a total of 6210+ 6211+6212+6213+6214+_6215 "passwords". That's the figure DaveC426913 gave.

 JasonRox]After the 10th character the user has the option not choose at all, so that is 63 choices.
But if the user chose no character for the 11th choice, he must make the same choice for the 12th, 13th 14th, and 15th. I don't think that simplifies anything.
 P: 22 ha good luck if this is really a problem not just a question.
P: 15,319
 Quote by JasonRox After the 10th character the user has the option not choose at all, so that is 63 choices.
What? No.

If the pw is 10 characters long, there are 62^10 possible combinations.
If the pw is 11 characters long, there are 62^11 possible combinations.
etc.

Add them together and you get all possible combinations within the parameters set.

So, your scenario where the user decides to not choose at all after the tenth character (or more accurataly: hits return), is simply the scenario where the pw is only 10 characters long.
P: 3
Thanks to everyone for the help

 Quote by ben328i ha good luck if this is really a problem not just a question.
It is a problem, I'm trying to recover a password for a zip file, but the luck is much appreciated
HW Helper
PF Gold
P: 2,327
 Quote by DaveC426913 What? No. If the pw is 10 characters long, there are 62^10 possible combinations. If the pw is 11 characters long, there are 62^11 possible combinations. etc. Add them together and you get all possible combinations within the parameters set. So, your scenario where the user decides to not choose at all after the tenth character (or more accurataly: hits return), is simply the scenario where the pw is only 10 characters long.
You're right!

P: 15,319
 Quote by Omnis Thanks to everyone for the help It is a problem, I'm trying to recover a password for a zip file, but the luck is much appreciated
If it's your password, that should eliminate about 99.999999% of the possibilities, shrinking it to ones you would have come up with.

 HW Helper PF Gold P: 2,327 Yeah, you can eliminate the characters you know for sure you didn't use!
 P: 172 Assuming order is important, ie:that "BAT" is not the same as "TAB" then the general answer is n!/r! or: 62!/ (pwd length)! Where "!" means the factorial function. Just like picking a lottery number. PS. It's easier to break the password encryption scheme than guess the actual password.
 P: 1 Figured I'd use this one. I'm trying to crack my password for something I got years ago, but I don't know how worth it it is. I need to know the number of possibilities for a 4 character password which has 6 characters to use. Basically 1111 to 6666. I'm good at maths, but that's only at a low level still. If someone can just work out what it is, that would be great. Cheers!!!
 P: 904 Use the Multiplication Principle of Counting: You have 6 choices for the first character, 6 choices for the second character, 6 for the third, and 6 for the fourth, so altogether you have 6*6*6*6 choices for a string of 4 characters. The principle comes from the following thought experiment. If you have a string of two characters with m choices for the first character and n choices for the second character, then first counting all possible strings starting with characters in the first space. For each character in the first space, there will be n possibilities for the second space. Ie., if the first character can be any letter in the alphabet, you would have sets like {A_, A_, ...}, {B_, B_, ...}, ... ,{Z_, Z_, ...} where the underscores are replaced by characters from the second set (of size n). Thus, you should have m sets of n strings, or m*n total strings.
Math
Emeritus