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February 11th, 2012, 08:21 AM   #1
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PERMUTATION

Hi to everyone....

I would like to know which is the permutation 4661 of the elements G = {1,2,3,4,5,6,7}
I have search for the formula around the internet but I could not find it...
So I hope any one can help about this solution.
Thank you, and I hope I will get any answer ASAP
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February 11th, 2012, 01:25 PM   #2
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Re: PERMUTATION

Hello, toli!

Quote:
I would like to know which is the permutation 4661 of the elements G = {1,2,3,4,5,6,7}.

Your question is not clear, but I suspect I know what you mean.

Suppose we write all 5,040 permutations of {1,2,3,4,5,6,7} in increasing order.
What is the 4661st number of the list?

Is that it?

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February 11th, 2012, 03:01 PM   #3
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Re: PERMUTATION

when we calculate the 7! "FACTORIAL" it is true that we have 5040 permutations and that permutation number is 7654321...
so I need to find the permutation 4661 " LET'S SUPPOSE IT IS 6453217 " i guess it is a little bit better written.

I have tried to find the way out of the solution but I couldn't find any Formula on my book or in the internet...
so I hope you can help me with this exam... thank you once again soroban
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February 11th, 2012, 04:56 PM   #4
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Re: PERMUTATION

Hello, toli!

You're still not very clear.
Can you tell me whether my interpretation is correct?


Quote:
When we calculate the 7! (factorial), it is true that we have 5040 permutations
and that permutation number is 7654321... [color=beige] . [/color][color=blue]Does this mean the FIRST one is 7654321 ?[/color]

So I need to find the permutation 4661.[color=beige]. . [/color][color=blue]Does this mean the 4661th permutation on the list?[/color]
Let's suppose it is 6453217. i guess it is a little bit better written.

I have tried to find the way out of the solution,
[color=beige]. . [/color]but I couldn't find any Formula on my book or in the internet...
So I hope you can help me with this exam.[color=beige] . [/color][color=blue]This is an exam equation?[/color]
Thank you once again, soroban

IF my interpretation is correct, and we are looking for the 4661th number on the list of permutations,
[color=beige]. . [/color]there is a procedure for determining this number.
It was discovered by a high school student in the 1960's and is very complicated.

But I won't explain it if I'm misreading the problem.

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February 11th, 2012, 05:03 PM   #5
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Re: PERMUTATION

You can solve it very easily with PARI/GP if soroban's interpretation is right. Look at numtoperm.
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February 12th, 2012, 04:22 AM   #6
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Re: PERMUTATION

CRGreathouse:

thank you for giving me your program, I am downloading it now and I will let you know
if it works or no... which I think it will...
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February 12th, 2012, 04:38 AM   #7
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Re: PERMUTATION

Soroban I will give you one of the exam that's on my book I hope I will explain it as best as I can so maybe you do have any solution
because I really have no Idea about this one...

Determine permutation 604 of the elements 1,2,3,4,5,6.

It is known that form 6 element's we have 6! = 720 Permutation's
Permutation from 1 to 120 number 1 is the first digit, similar act is with the other permutation;
With other words, Permutation 601 is:

612345 >>>>>>> permutation 601

then we have

612354 >>>>>>> permutation 602

612435 >>>>>>> permutation 603

612453 >>>>>>> this is the permutation that we need for these exam...

BUT WHEN IT COMES TO THE QUESTION FOR PERMUTATIONS 4661 FOR THE ELEMENTS G = {1,2,3,4,5,6,7} I AM STUCK AND I DON'T KNOW HOW TO FIGURE IT OUT...

I HOPE THIS IS HELPFULLY FOR YOU Soroban.
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February 12th, 2012, 05:05 AM   #8
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Re: PERMUTATION

Soroban I have found the Permutation of 4661 that corresponds to the elements of 1,2,3,4,5,6,7
and that permutation is 7,3,6,1,5,2,4 but what I need is the steps on how to find the solution of these given question...
I hope you know what to do...
thank you.
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February 12th, 2012, 05:10 AM   #9
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Re: PERMUTATION

CRGreathouse:

I have tried to find the short digits for the program you advice me to use it but I really could
not find any solution maybe just because I didn't know how to use
otherwise thank you a lot for help I appreciate what you have done for me.
Thank you again.
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February 12th, 2012, 01:13 PM   #10
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Re: PERMUTATION

Code:
> ?numtoperm
numtoperm(n,k): permutation number k (mod n!) of n letters (n C-integer).

> numtoperm(7, 4661)
%1 = [4, 1, 6, 2, 3, 5, 7]
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