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Password Combination 
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#1
Nov807, 08:28 AM

P: 3

How do I work out password combination probabilities?
For example, I'm looking for a password between 1015 characters in length, the password can contain any alphanumeric character (lower/upper alphabet character and 09 digits). How would I find out how many possible combinations there are? 


#2
Nov807, 08:44 AM

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? 


#3
Nov907, 05:24 PM

P: 3




#4
Nov907, 05:49 PM

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 todate. Don't try to do them all in one sitting. 


#5
Nov907, 10:14 PM

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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.



#6
Nov1007, 12:12 AM

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#7
Nov1007, 12:51 AM

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. 


#8
Nov1007, 07:00 AM

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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 62^{10} possible 10 letter "passwords". Similarly, there are 62^{11} possible 11 letter "passwords", 62^{12} 12 letter "passwords", 62^{13} l13 letter "passwords", 62^{14} 14 letter passwords,, and, finally, 62^{15} 15 letter "passwords". Altogether, then, there are a total of 62^{10}+ 62^{11}+62^{12}+62^{13}+62^{14}+_62^{15} "passwords". That's the figure DaveC426913 gave. 


#9
Nov1007, 01:57 PM

P: 22

ha good luck if this is really a problem not just a question.



#10
Nov1007, 02:14 PM

P: 15,319

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. 


#11
Nov1207, 09:35 PM

P: 3

Thanks to everyone for the help



#13
Nov1207, 10:08 PM

P: 15,319

If it's not your password... well, that's raises an intriguing question... 


#14
Nov1207, 10:10 PM

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Yeah, you can eliminate the characters you know for sure you didn't use!



#15
Nov1307, 12:49 AM

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. 


#16
Nov1508, 09:29 AM

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!!! 


#17
Nov1608, 08:21 AM

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. 


#18
Nov1608, 01:02 PM

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