counting question for computer science?

Question

Let's say you are trying to create a password with the following specifications: It can be either 5, 6, or 7 alphanumeric characters (letters or digits 0.,,,,9 , without making the distinction between uppercase and lowercase letters). How could the multiplication and addition rules of counting be used to determine how many possible passwords could be created using this scheme? How many are possible?

Answer 1

You can break down your set of valid passwords into:
- X = Passwords with 5 alphanumeric chars
- Y = Passwords with 6 alphanumeric chars
- Z = Passwords with 7 alphanumeric chars
Now how X, Y, and Z are disjoint sets, so if we can count the size of X, Y and Z separately, we just need to add the sizes to get the set of valid passwords.

How do we count X? Each of the five characters of X can be one of 36 characters (10 numbers, and 26 characters). So the total possible number is 36*36*36*36*36, since there are 36 "choices" for each location. This is 36^5.

Similarly, we can count Y and Z to be 36^6 and 36^7.

So the total number of passwords is: 36^5+36^6+36^7.

Answer 2

Does without making the distinction mean that the password is not case-sensitive? If so, the answer is 36^5 + 36^6 + 36^7 = 80,601,412,608 possibilities. (26 letters + 10 numbers = 36 possible characters.) If the password IS case sensitive, then the answer is 62^5 + 62^6 + 62^7 = 3,579,330,974,624 possibilities. (26 CAPITAL LETTERS + 26 lowercase letters + 10 numbers = 62 possible characters.)

Answer 3

Well, for each character, you have 36 choices for its value. The choices are apparently independent. Therefore you have 36*36*...*36 possible combinations. (36 multiplied by itself n times)

So your answer for each length,n, is 36^(n). (Here, n is either 5 or 6 or 7. )

So you have 36^5 five-digit passwords, 36^6 six-digit passwords and 36^7 seven digit passwords. Add them together to get your total overall. You get something like (36^5)(1 + 36 + 36^2), if you feel the need to combine terms.

Answer 4

There are 36 possible letters or numbers which in the rest of this conversation we will call C. The number of places will be called N. The number of possible combinations (including blanks) is C^N so we have 78364164096. However since this includes passwords that are less than 5 characters in length, we need to subtract them. So we calculate 36^4 to determine the number of unusable passwords in the set. We get 1679616. Now we simply subtract 1679616 from 78364164096 and we get 78362484480 possible combinations.

Answer 5

So you have 36 choices for a character, if you choose 5 you have (36)*(36)*(36)*(36)*(36)=36^5 passwords.

36^6 for six characters and 36^7 for seven characters.

Depends how you look at it, but I think choosing 5 or 6 characters is a subset of choosing 7 characters, so I would say 36^7 is your number of possiblities.

Answer 6

There are 36 possible options for each character, so, the answer is:
36^5+36^6+36^7

Answer 7

26 letters + 10 digits = 36 options per character. 36^5 + 36^6 + 36^7 would be your answer. 80,601,412,608 options.

Answer 8

so many. use permutation and combination rules.