A hacker is attempting to access a high security computer but needs a password.

Question

A hacker is attempting to access a high security computer but needs a password. He uses a program to generate a password and evaluate if it's the right one. The hacker's programming skills are ironically quit limited and hence his program does not attempt to intelligently predict the password. Rather, it uses a brute force technique, and just attempts all possible passwords. The hacker gusses that the password has 10 unique characters and each character can be an alphabet, a numerical or one of 6 special characters. If the passwords are NOT CASE SENSITIVE; what is the number of possible passwords? Suppose the passwords ARE CASE SENSITIVE; what is the number of possible passwords? The hacker realizes that the security code is more complex than he initially thought and decides to consider the possibility that the character are not unique in addition to them being case sensitive; what is number of possible passwords? You can express the answers in a factorial form.

Explanation / Answer

Alright, the first step is to define our character-space for each password. In (a.) we have the 26-character lowercase alphabet, abcdefghijklmnopqrstuvwxyz, 10 numbers, 0123456789, and 6 special characters, let's use !@#$%^. For this I am only using lowercase, because the letter 'A' is treated as 'a' for case-insensitive passwords. This gives us a total of 26 + 10 + 6 = 42. In (b.) we have the same as (a.) with the addition of 26 uppercase letters, ABCDEFGHIJKLMNOPQRSTUVQXYZ. This gives us a total of 26 + 10 + 6 + 26 = 68. In (c.) we have the same character-space as (b.), but with a different restriction on uniqueness. (We still have the total 68 from (b.)). Because the characters in passwords for part (a.) and (b.) are unique, we can only use each character once. For (a.), we have 42 to choose from for the first character, 41 for the next, 40 for the third, ..., and 33 for the last. That gives us an equation of: 42*41*40*39*38*37*36*35*34*33. This can also be written as (42!)/(32!). Similarly, for (b.) we have 68 to start, 67 for the next position, 66 for the third, ..., and 59 for the last. That gives us an equation of: 68*67*66*65*64*63*62*61*60*59, or (68!)/(58!). For (c.), we don't have to have unique characters in the password, so each character can be picked independently of the previous ones. That gives us 68 to choose from for the first, second, third, ..., and last, for an equation of 68*68*68*68*68*68*68*68*68*68, or (68)^10. This problem is related to the subject of permutations (see http://en.wikipedia.org/wiki/Permutation). That is, when you are given a set of n = 68 things to choose k = 10 from randomly, the formula is (n!)/(n-k)! = (68!)/(68-10!) = (68!)/(58!), which is the same answer we got above. Good luck!

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