Deified Posted November 17, 2004 Share Posted November 17, 2004 I came across a bit of a snag when I was working with permutations of things in a circle. The problem is that I have 8 particles, 4 negative, 4 positive(the particles are indistinguishable except by their charge) arranged in a circle. When I asked the question how many ways can I reorder these particles? I came up some interesting results. According to the formula I was taught: (8-1)!/4!*4! But this equals 8.75 permutations! Is this a special case? Is the formula wrong? Please help me out! Link to comment Share on other sites More sharing options...
Deified Posted November 17, 2004 Author Share Posted November 17, 2004 I came across a bit of a snag when I was working with permutations of things in a circle. The problem is that I have 8 particles, 4 negative, 4 positive(the particles are indistinguishable except by their charge) arranged in a circle. When I asked the question how many ways can I reorder these particles? I came up some interesting results. According to the formula I was taught: (8-1)!/4!*4! But this equals 8.75 permutations! Is this a special case? Is the formula wrong? Please help me out! Link to comment Share on other sites More sharing options...
Deified Posted November 19, 2004 Author Share Posted November 19, 2004 Can no one help? Link to comment Share on other sites More sharing options...
Deified Posted November 19, 2004 Author Share Posted November 19, 2004 Can no one help? Link to comment Share on other sites More sharing options...
psi20 Posted November 19, 2004 Share Posted November 19, 2004 Yeah, this is the problem my teacher and I also tackled. Let me look up the formula we found independently. Link to comment Share on other sites More sharing options...
psi20 Posted November 19, 2004 Share Posted November 19, 2004 Yeah, this is the problem my teacher and I also tackled. Let me look up the formula we found independently. Link to comment Share on other sites More sharing options...
psi20 Posted November 19, 2004 Share Posted November 19, 2004 Well, nevermind, we didn't tackle a problem like that. We did a one where there was an alternating pattern, meaning postive then negative then positive... in a circle. Does your problem have alternating positive/negative particles? I lost the formula for alternating patterns on circular permutations. However, I do remember one thing. Don't trust formulas , especially since mathematicians debate over this area of math. Link to comment Share on other sites More sharing options...
psi20 Posted November 19, 2004 Share Posted November 19, 2004 Well, nevermind, we didn't tackle a problem like that. We did a one where there was an alternating pattern, meaning postive then negative then positive... in a circle. Does your problem have alternating positive/negative particles? I lost the formula for alternating patterns on circular permutations. However, I do remember one thing. Don't trust formulas , especially since mathematicians debate over this area of math. Link to comment Share on other sites More sharing options...
psi20 Posted November 19, 2004 Share Posted November 19, 2004 What is your formula? What does is it supposed to tell you? Link to comment Share on other sites More sharing options...
psi20 Posted November 19, 2004 Share Posted November 19, 2004 What is your formula? What does is it supposed to tell you? Link to comment Share on other sites More sharing options...
Deified Posted November 19, 2004 Author Share Posted November 19, 2004 Well the formula I used was (n-1)!/x!*y! Where x and y the numbers of alike objects(indistinguishable). It is supposed to tell me how many possible different orders of particles I can have in a circle. One of the permutations would be an alternating pattern. Link to comment Share on other sites More sharing options...
Deified Posted November 19, 2004 Author Share Posted November 19, 2004 Well the formula I used was (n-1)!/x!*y! Where x and y the numbers of alike objects(indistinguishable). It is supposed to tell me how many possible different orders of particles I can have in a circle. One of the permutations would be an alternating pattern. Link to comment Share on other sites More sharing options...
premjan Posted November 19, 2004 Share Posted November 19, 2004 you seem to have confused two formulas: 1) the one for non-circular permutations n! 2) the one for combinations, n!/(n-r)!/r! The one for circular permutations is just (n-1)! Link to comment Share on other sites More sharing options...
premjan Posted November 19, 2004 Share Posted November 19, 2004 you seem to have confused two formulas: 1) the one for non-circular permutations n! 2) the one for combinations, n!/(n-r)!/r! The one for circular permutations is just (n-1)! Link to comment Share on other sites More sharing options...
Deified Posted November 19, 2004 Author Share Posted November 19, 2004 Because there are indistinguishable objects it works somewhat like a combination. The formula (I think) for indistinguishable permutations is n!/x! where x is the number of indistinguishable objects. You are taking away the permutations that look the same. eg: xxy xyx yxx Its 3!/2!=3 not 3!=6 because xxy and xxy are the same even if the x's switch. Link to comment Share on other sites More sharing options...
Deified Posted November 19, 2004 Author Share Posted November 19, 2004 Because there are indistinguishable objects it works somewhat like a combination. The formula (I think) for indistinguishable permutations is n!/x! where x is the number of indistinguishable objects. You are taking away the permutations that look the same. eg: xxy xyx yxx Its 3!/2!=3 not 3!=6 because xxy and xxy are the same even if the x's switch. Link to comment Share on other sites More sharing options...
premjan Posted November 19, 2004 Share Posted November 19, 2004 The right answer seems to be 6 though I am not sure what is the right formula to apply (if there is even a known one). +-+-+-+- ++++---- ++--++-- +--+-++- +--++-+- -++--+-+ Link to comment Share on other sites More sharing options...
premjan Posted November 19, 2004 Share Posted November 19, 2004 The right answer seems to be 6 though I am not sure what is the right formula to apply (if there is even a known one). +-+-+-+- ++++---- ++--++-- +--+-++- +--++-+- -++--+-+ Link to comment Share on other sites More sharing options...
psi20 Posted November 19, 2004 Share Posted November 19, 2004 Sorry Deified that we're not helping that much. But I think there'd be more than just the 6 above. +---++-+ +---+-++ +---+++- ++-++--- Link to comment Share on other sites More sharing options...
psi20 Posted November 19, 2004 Share Posted November 19, 2004 Sorry Deified that we're not helping that much. But I think there'd be more than just the 6 above. +---++-+ +---+-++ +---+++- ++-++--- Link to comment Share on other sites More sharing options...
psi20 Posted November 19, 2004 Share Posted November 19, 2004 "Because there are indistinguishable objects it works somewhat like a combination. The formula (I think) for indistinguishable permutations is n!/x! where x is the number of indistinguishable objects. You are taking away the permutations that look the same. eg: xxy xyx yxx Its 3!/2!=3 not 3!=6 because xxy and xxy are the same even if the x's switch." That's true for linear permutations that repeat, I believe. I'm not sure it applies to circular permutations that repeat. Link to comment Share on other sites More sharing options...
psi20 Posted November 19, 2004 Share Posted November 19, 2004 "Because there are indistinguishable objects it works somewhat like a combination. The formula (I think) for indistinguishable permutations is n!/x! where x is the number of indistinguishable objects. You are taking away the permutations that look the same. eg: xxy xyx yxx Its 3!/2!=3 not 3!=6 because xxy and xxy are the same even if the x's switch." That's true for linear permutations that repeat, I believe. I'm not sure it applies to circular permutations that repeat. Link to comment Share on other sites More sharing options...
matt grime Posted November 21, 2004 Share Posted November 21, 2004 Try counting: fix one element, say one of the pluses. Then you wish to know how many ways of writing the other elements, which is I believe 7choose3, as we only need to count the orderings relative to the fixed element. Link to comment Share on other sites More sharing options...
matt grime Posted November 21, 2004 Share Posted November 21, 2004 Try counting: fix one element, say one of the pluses. Then you wish to know how many ways of writing the other elements, which is I believe 7choose3, as we only need to count the orderings relative to the fixed element. Link to comment Share on other sites More sharing options...
premjan Posted November 21, 2004 Share Posted November 21, 2004 the question is how to systematically eliminate duplicates (one + or - is as good as another), given that we cannot flip the circle over. Link to comment Share on other sites More sharing options...
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