My Math Forum Permutation help

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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
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Re: Permutation help

Hello, Tony!

Quote:
 To start off with, I have six metrics: metric1, metric2, metric3, metric4, metric5, metric6. Each metric can have three values: High, Medium, or Low. I believe the right answer is 729, but what is the right formula to arrive at it. (Just 3^6 ? )

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]$\text{Hence, there are }3^6\,=\,729\text{ possible outcomes.}$

Quote:
 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.[color=beige] . [/color][color=blue]What does this mean?[/color] For example here is one permutation: [color=beige]. . [/color]$\begin{array}{ccccccccc}m1 & m2 & m3 & m4 & m5 & m6 & m7 & m8 & m9 \\ \\ \hline \\ \\ 1H & 2M & 3L & 4H & 5M & 6L & 7L & 8M & 9M \\ \\ \\ 1L & 2L & 3L & 4M & 5H & 6M & 7L & 8H & 9M \end{array}$ I believe the total number of permutations is 19,683 [color=beige]. . [/color]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.

You have two permutations listed.

If each metric can be $H,\,M\text{ or }L$, what are those numbers for?

February 8th, 2012, 05:50 AM   #3
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Joined: Dec 2011

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Re: Permutation help

Quote:
 Originally Posted by Lucent 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 ? )
Do you really ask us to find the number of ways to permutate the three different values in each metric in which we have six metrics (metric1, metric2, metric3, metric4, metric5, metric6) and each metric can have three different values, and they are different for each metric? By permutate I mean (take for example) in the first metric of three different values (a, b, c), the number of ways to permutate these three values = 3!=6 $\rightarrow$ (a, b, c) (a, c, b) (b, a, c) (b, c, a) (c, a, b) and (c, b, a)

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 $n=3!.3!.3!.3!.3!.3!=46,656$

Reason behind 3! and$n=3!.3!.3!.3!.3!.3!=46,656$:
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 $m.n$ 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
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Re: Permutation help

Quote:
 Originally Posted by soroban Hello, Isbell! No apology necessary . . .
I'm glad to hear that, Soroban.

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