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!

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!

Can no one help?

Can no one help?

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

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

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.

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.

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

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

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.

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.

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)!

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)!

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.

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.

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).

 

+-+-+-+-

++++----

++--++--

+--+-++-

+--++-+-

-++--+-+

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).

 

+-+-+-+-

++++----

++--++--

+--+-++-

+--++-+-

-++--+-+

Sorry Deified that we're not helping that much.

 

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

 

+---++-+

+---+-++

+---+++-

++-++---

Sorry Deified that we're not helping that much.

 

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

 

+---++-+

+---+-++

+---+++-

++-++---

"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.

"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.

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.

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.

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