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qassim issa
guest

Jun0409, 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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alexb
Charter Member
2391 posts 
Jun0409, 07:23 PM (EST) 

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 K_{n} with n nodes. There are n(n1)/2 edges. These edges can be split into [n(n1)/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 K_{n} are placed at the vertices of a regular ngon, 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 (n1). 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
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Jun0509, 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

Jun0509, 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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alexb
Charter Member
2391 posts 
Jun0509, 07:01 AM (EST) 

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 [(n1)/2].


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qassim issa
guest

Jun0509, 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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qassim issa
guest

Jun1009, 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

Jun1009, 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 5n=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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