Permutations in a circle

[Deified]

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!

[Deified]

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!

[Deified]

Can no one help?

[Deified]

Can no one help?

[psi20]

Yeah, this is the problem my teacher and I also tackled. Let me look up the formula we found independently.

[psi20]

Yeah, this is the problem my teacher and I also tackled. Let me look up the formula we found independently.

[psi20]

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.

[psi20]

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.

[psi20]

What is your formula? What does is it supposed to tell you?

[psi20]

What is your formula? What does is it supposed to tell you?

[Deified]

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.

[Deified]

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.

[premjan]

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)!

[premjan]

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)!

[Deified]

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.

[Deified]

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.

[premjan]

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).

 

+-+-+-+-

++++----

++--++--

+--+-++-

+--++-+-

-++--+-+

[premjan]

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).

 

+-+-+-+-

++++----

++--++--

+--+-++-

+--++-+-

-++--+-+

[psi20]

Sorry Deified that we're not helping that much.

 

But I think there'd be more than just the 6 above.

 

+---++-+

+---+-++

+---+++-

++-++---

[psi20]

Sorry Deified that we're not helping that much.

 

But I think there'd be more than just the 6 above.

 

+---++-+

+---+-++

+---+++-

++-++---

[psi20]

"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.

[psi20]

"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.

[matt grime]

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.

[matt grime]

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.

[premjan]

the question is how to systematically eliminate duplicates (one + or - is as good as another), given that we cannot flip the circle over.