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Subject: "permutations without repetition"     Previous Topic | Next Topic
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Conferences The CTK Exchange College math Topic #716
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qassim issa
guest
Jun-04-09, 10:42 AM (EST)
 
"permutations without repetition"
 
   hi,
do you have a method how to count Permutations from 1..n that do not
have the same combination in it: like
if we have the numbers 1,2,3,4
the Permutation of this is just
1234
4213


because if we take the rest of Permutations its have the combination

1243 (43 its apper up)
1324 (32 apear up)
1342( (34 apears up)
1423 (23 appear up)
and all of the rest appears up

qassim issa


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  Subject     Author     Message Date     ID  
  RE: permutations without repetition alexbadmin Jun-04-09 1
     RE: permutations without repetition qassim issa Jun-05-09 2
     RE: permutations without repetition qassim issa Jun-05-09 3
         RE: permutations without repetition alexbadmin Jun-05-09 4
     permutations without repetition qassim issa Jun-05-09 5
         RE: permutations without repetition alexbadmin Jun-08-09 6
             RE: permutations without repetition qassim issa Jun-10-09 7
             RE: permutations without repetition qassim issa Jun-10-09 8
                 RE: permutations without repetition alexbadmin Jun-10-09 9

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alexbadmin
Charter Member
2391 posts
Jun-04-09, 07:23 PM (EST)
Click to EMail alexb Click to send private message to alexb Click to view user profileClick to add this user to your buddy list  
1. "RE: permutations without repetition"
In response to message #0
 
   One thing that occured to me is that the problem with cycle permutations is easier. (In a cyclic permutation 1234, 1 follows 4.) What's below is just "thinking aloud."

Imagine a complete graph Kn with n nodes. There are n(n-1)/2 edges. These edges can be split into [n(n-1)/2/n] not intersecting cycles, each through all the nodes of the graph. If we think of the graph as directed with each edge traversable in both directions, this number is to be multiplied by 2.

If the nodes of the graph Kn are placed at the vertices of a regular n-gon, then the required number of cycles is obtained by drawing the polygon itself, then by drawing a plygonal star by skipping 1 node. Then another one by skipping 2 nodes, etc. If we come with a loop without visiting all the vertices, we are getting into a trouble, for this would break regularity of this construction and requiring an edge of different length. My estimate then is an upper bound and may not be exact.

So, for prime n, the number of (directed) cycles is (n-1). For n=4, it's two. For n=5, it's 4:

12345
13524
14253
15432

For n=6, it is at most 4. Say,

123456
135246
142536
154326

Appears 4, not less.


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qassim issa
guest
Jun-05-09, 06:52 AM (EST)
 
2. "RE: permutations without repetition"
In response to message #1
 
   13524 its the same 14253 because it have the set {2,5} and {5,2} therfoure we dont count it. its the same with 15432 and 12345 ,the set {5,4} and {5,4} its have to be out of our counter


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qassim issa
guest
Jun-05-09, 06:52 AM (EST)
 
3. "RE: permutations without repetition"
In response to message #1
 
   hi , thank for your reaction

the Permutations ,we have to take, must be with out of sets in other Permutations:
in your answer:
12345
13524
14253
15432
the set {4,5} in 12345 and {5,4} in 15432 for my question its duplicated there for we count it as one Permutation.
as the same with the set {5,2} in13524 {2,5} in 14253
for my question its duplicated there for we count it as one Permutation


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alexbadmin
Charter Member
2391 posts
Jun-05-09, 07:01 AM (EST)
Click to EMail alexb Click to send private message to alexb Click to view user profileClick to add this user to your buddy list  
4. "RE: permutations without repetition"
In response to message #3
 
   >the set {4,5} in 12345 and {5,4} in 15432 for my question
>its duplicated there for we count it as one Permutation.
>as the same with the set {5,2} in13524 {2,5} in 14253
> for my question its duplicated there for we count it as
>one Permutation

Well, for one, you have not stated this explicitly. Secondly, in that case, in your example for n=4

1234
4213

12 appears in the first permutation whilst 21 in the second.

With the new rule, the answer would be [(n-1)/2].


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qassim issa
guest
Jun-05-09, 10:23 PM (EST)
 
5. "permutations without repetition"
In response to message #1
 
   i think the way we count the permutations is
if we want,for example, to count the permutations without reptition of the subsets. we count the subsets in the range
1..12
{1,2},{1,3}..{1,12}= 11
{2,1},{2,3},{2,4},{2,5},{2,6},{2,7},{2,8},{2,9},{2,10},{2,11},{2,12}=11
{3,1},{3,2},{3,4},{3,5},{3,6},{3,7},{3,8},{3,9},{3,10},{3,11},{3,12}
=11
...
..
{12,1}...{12,11}=11
the nuber of set we have 12x11/2=66
we have 66 option of sets that we will arange in 12 nember(11 places)
66/11=6
there four we must have 6 permutations.

ithink this is the count, but i didnt 'till now' find amethod to arrange the sets.


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alexbadmin
Charter Member
2391 posts
Jun-08-09, 02:56 PM (EST)
Click to EMail alexb Click to send private message to alexb Click to view user profileClick to add this user to your buddy list  
6. "RE: permutations without repetition"
In response to message #5
 
   From what I suggested before, the count for n = 12 is not more than 5.


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qassim issa
guest
Jun-10-09, 06:26 AM (EST)
 
7. "RE: permutations without repetition"
In response to message #6
 
   we cant bulid the permutations of n when its an odd number like 9,15,21...
p1: 9 8 7 6 5 4 3 2 1
p2: 9 7 5 3 1 8 6 4 2
p3: 2 8 5 1 7 4 9 6 3
the 4th permutations in this method :
p4: 9 5 1 6 2 7 3 8 4
but we can sea that the set {1,5} had appeard in in "p3" as "5,1" therfour if i change the set {9,5} to {5,9} its will be agood new permutation:

5 9 1 6 2 7 3 8 4


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qassim issa
guest
Jun-10-09, 06:26 AM (EST)
 
8. "RE: permutations without repetition"
In response to message #6
 
   this is my answer to the question:
n=12
12 11 10 9 8 7 6 5 4 3 2 1
11 9 7 5 3 1 12 10 8 6 4 2
10 7 4 1 11 8 5 2 12 9 6 3
9 5 1 10 6 2 11 7 3 12 8 4
12 7 2 9 4 11 6 1 8 3 10 5

n=13:
13 12 11 10 9 8 7 6 5 4 3 2 1
13 11 9 7 5 3 1 12 10 8 6 4 2
13 10 7 4 1 11 8 5 2 12 9 6 3
13 9 5 1 10 6 2 11 7 3 12 8 4
13 8 3 11 6 1 9 4 12 7 2 10 5
13 7 1 8 2 9 3 10 4 11 5 12 6


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alexbadmin
Charter Member
2391 posts
Jun-10-09, 06:27 AM (EST)
Click to EMail alexb Click to send private message to alexb Click to view user profileClick to add this user to your buddy list  
9. "RE: permutations without repetition"
In response to message #8
 
   > n=13:

Yeap. You got the spirit.


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