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Permutations Question

Posted on 2012-03-13
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Last Modified: 2012-08-13
My friend and I are discussing a question on permutations.

A grid of 49 letters, numbers and symbols is shown. A random character is taken from the grid. This is done six times. The idea is for this to randomly generate a password. What are the total number of passwords that could be generated?

My thought is that there is a 1/49 chance of any character being chosen, therefore the total number of passwords is 49^6.

My friend says that because this is a permutations question, the formula for permutations makes it (49!)/(49-6)!

However I'm pretty sure this is when you have the "balls in a sack" type question, and each time you take a ball out of the sack you don't return it, so it doesn't apply in this case as a valid password could be "111111" (i.e. the same character repeated each time).

Our problem is that the question is assigned 10 points, and that so far we can't find any similar examples in the course material of how they want the question answered. My friend points out that you don't get a 10 point question whose answer is 49^6.

So I'd appreciate some feedback - who is right in solving the problem? Any suggestions what working to include in order to get all 10 of those points??
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Question by:purplesoup
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by:rowansmith
rowansmith earned 100 total points
ID: 37713999
Because this is an assignment it would be wrong of me to give you the answer.  This website has the information you are looking for: http://www.mathsisfun.com/combinatorics/combinations-permutations.html

Also I would strongly suggest that you look at how the question is worded.  It's not clear to me from your description if the letter is "returned" to the rack after it is selected or if it then becomes unavailable.  eg, the letter X can only occur ONCE in the password or can it occur multiple times.

Good luck.
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by:petr_hlucin
petr_hlucin earned 100 total points
ID: 37714004
Hello,

the right answer is 49^6, because each of the times you can select exactly one character out of 49 and this is done 6 times. Permutations needn't be employed, because the number of permutations is already included in 49^6.

Let's consider a simple example where you can select a 2 characters long password out of 3 characters (a, b, c). All possible passwords are listed below:"
aa
ab
ac
ba
bb
bc
ca
cb
cc

The number of all possibilities is 9 = 3^2 = size_of_grid^lenght_of_password != (3!)/(3-2)! = 6/1 = 1.
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by:rowansmith
rowansmith earned 100 total points
ID: 37714028
Hmmm... (3!)/(3-2)! = 6/1 = 6 not 1 - which would be correct in the event that a letter can not be reused.

So depending on how the question is worded depends on which answer will be correct.

For extra points you could possibly give both answers with workings explaining that the question is Ambiguous and can be interpretted in two ways.  Teachers love a smart arse :-)
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by:Otto_N
Otto_N earned 300 total points
ID: 37714046
Both options (49^6 and 49!/43!) is 6 multiplications, and your friend's argument of a more complex answer for the points value doesn't hold water. (And this is a math question, not a psychology one ;-) ).

One additional condition that you can look out for, is uniqueness within the grid.  If the same character is repeated in the grid, the total number of options are reduced by the repeated characters:  So, if 3 characters are repeated, the options are reduced to 46, and the correct answer can be 46^6 or 46!/40!.
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Author Comment

by:purplesoup
ID: 37714941
thanks, good point, I've been trying to list as many points to make in the answer as possible, let me know if you think any of these are wrong or I've missed something:

- there are 49 unique characters in the table
- we want to identify the number of possible passwords choosing characters randomly from the table, where the passwords are exactly six characters long
- a character can appear more than once
- the chances of choosing any character are the same: 1/49.
- we have 49 different characters to choose from for the first character, the same 49 for the second character and so on
- therefore the number of combinations is 49 x 49 x 49 x 49 x 49 x 49 or 49^6
- this equates to 13 841 287 201
- or 1.38 x 10^10 to 3 sf
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by:
Otto_N earned 300 total points
ID: 37715505
I see no problem with your answer.

Note that the probability of a particular character being chosen (1/49) is not relevant to the total number of passwords that can be generated. If the question was worded a bit differently (asking for the number of passwords that can be generated before the probability of a unique password being generated fall below a specified threshold) this would start playing a role (or rather, 1/13 841 287 201 will).  This answer could be worth much more than the 10 points on offer, so I wouldn't worry too much about it, if I were you.
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Author Closing Comment

by:purplesoup
ID: 37737222
thanks everyone for your help
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