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Circular Permutations and stuff.


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According to what I know, if you have a circular permutation, you divide by the number of objects in the arrangement if there is no clasp. You also divide by 2 if there are reflections.


Suppose you had a bracelet of 2 types of charms. One kind of charm is for luck, the other for gift. There are 2 lucks and 2 gifts arranged on this bracelet. This bracelet can not have a luck next to another luck and a gift next to another gift. It's superstitious and will give you the opposite. So you have one way to arrange it



/ \


\ /


But according to what I know, and what my book did, you should have (2! * 2!)/(2*4)


The 2!'s come from the ways to arrange the Lucks and the ways to arrange the Gifts. The 2 in the denominator comes out because it's a bracelet. The 4 in the denominator is from the total number of charms. If this is so, then there is 1/2 of a combination. The heck?

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not a stats question, its a combinatorics question..in your case maybe finite.


but i think your entire calculation is wrong.


In a circular permutation you always fix one point as always...

so lets say you fix 1 luck.


L- _ - _ - _


then you no 2 lucks are sidebyside so *2 for the G


then you need to place a luck


and lastly you have tlhe last G


but you have to remember you can permutate the G's by themselves so you must divide by nG! ...inthis case 2!

which gives you 1*2*1*1/2! = 1;



2!*2! for the permutations of the L and G.../2 cuz its a bracelet...not really sure about this one

so you get 2!2!/2 = 2


you shouldn't be dividing by 4. you should be diiving by n! where n is the number of one of the 2 types. in etiher case 2!

so 2!*2!/2!/2 = 1

but double check why you divide by to for it being a bracelet

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  • 4 weeks later...

You divide out the 2 because it's a bracelet and can be flipped. For example, you have 3 keys on a chain. A B


If you set A as the fixed point, then A _


A 2

1 which means there are 2! ways, or 2 ways. But really, there is only one way. You can flip the key chain around. The A will always be adjacent to B and C. The B will always be next to C and A.

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