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February 8th, 2012, 04:24 AM  #1 
Newbie Joined: Feb 2012 Posts: 1 Thanks: 0  Permutation help
hi. I have a fairly long permutation I need help with for a thesis (I am just not math oriented but I try) To start off with I have six metrics, metric1, metric2, metric3, metric4, metric5, metric6. Each metric can have three values, but they are different for each metric. I believe the right answer is 729 but what is the right formula to arrive at it. (Just 3^6 ? ) Going on from there I have had to extend the number of metrics. I now have nine metrics [metric1..metric9]. Each may have a High, Medium or Low value, as above, and each value is different for and particular to, each metric. For example here is one permutation: m1 m2 m3 m4 m5 m6 m7 m8 m9  1H 2M 3L 4H 5M 6L 7L 8M 9M 1L 2L 3L 4M 5H 6M 7L 8H 9M I believe the total number of permutations is 19,683 having laboriously copied them out in an excel spreadsheet, but I need to confirm that, and to be able to state what formula I used to derive that number. Any help is much appreciated. Thanks, Tony 
February 8th, 2012, 05:36 AM  #2  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Re: Permutation help Hello, Tony! Quote:
That is the correct formula and the correct answer. [color=beige]. . [/color]You forget to give us the question. For each metric, we make a decision with 3 choices. And we make 6 such decisions. [color=beige] .[/color] Quote:
I don't understand your table. You have two permutations listed. If each metric can be , what are those numbers for?  
February 8th, 2012, 05:50 AM  #3  
Senior Member Joined: Dec 2011 Posts: 277 Thanks: 1  Re: Permutation help Quote:
If that is the case, then the number of way to permutate the three different values(e.g. a, b, c) in the first metric would be 3!. Likewise, the number of way to permutate another three different values (e.g. d, e, f) in the second metric would be 3!, and so on. Now, to calculate the total number of permutations (n) to arrange those 6 sets of 3 different values in those 6 metrics, we have Reason behind 3! and: 1. You need to know that there are n! different ways of arranging n distinct objects in a sequence. (The arrangements are called permutations.) 2. You need to also aware that suppose an event X can occur in m ways, and an event Y can occur in n ways, then the combinations of events X and Y can occur in ways. This is called the Product Rule Principle, one of the most basic and fundamental principle in the techniques of counting. Most importantly, this principle can be extended to more events. And in our case, we have six events. Did I understand you correctly, Tony? (EDIT: I'm sorry, Soroban. I didn't realize that you've posted your answer to this question because I didn't preview my post before posting! I'm really sorry for that.)  
February 8th, 2012, 06:30 AM  #4 
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Re: Permutation help Hello, Isbell! No apology necessary . . . You have an entirely different interpretation of the problem . . . refreshing1 
February 8th, 2012, 07:15 AM  #5  
Senior Member Joined: Dec 2011 Posts: 277 Thanks: 1  Re: Permutation help Quote:
 

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