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

 

L

/ \

G G

\ /

L

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?

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

L-2-_-_

then you need to place a luck

L-2-L-_

and lastly you have tlhe last G

L-2-L-1

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;

 

NOW LETS TRY IT YOUR WAY.

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

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

C

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.