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Beads on a Bracelet


Date: 05/09/97 at 10:10:38
From: Anonymous
Subject: Question

Dear Dr. Math,

I know that the number of circular permutations is (n - 1)! and that 
the number of circular permutations on a key ring is (n-1)!/2.

How many different arrangements of 3 red and 3 blue beads on a 
bracelet are there? The answer is 3 and I can demonstrate the 3, but 
cannot find a suitable expression using the factorial symbol. 

Can you help?

Thanks, 
Doug Buhler


Date: 05/09/97 at 17:11:31
From: Doctor Ceeks
Subject: Re: Question

Hi,

There is no simple formula for the number of different arrangements of 
N red and N blue beads on a bracelet.

The reason can be seen in a way which subsumes all three of the 
different questions you ask.  The way is this: answer the question for 
the non-circular analogue first, then deduce the circular answer.

Case 1: Circular permutations

Analogue: Permutations
Answer: n!

To deduce the circular permutations, one notes that if you write down 
all n! permutations, you can group them into sets of size n where each 
set of size n consists of permutations which give the same circular 
permutation.

Therefore, as you said, the number of circular permuations is
n!/n = (n-1)!.

Case 2:  On a key ring

Analogue: Permutations
Answer n!

To deduce the key ring arrangments, one notes that the n! permutations
can be grouped into sets of size 2n ...

Case 3: N blue and N red beads on a bracelet

Analogue: N blue and N red beads in a row
Answer: (2N)!/(N!N!)

Now try to group the set of row arrangements into subsets where each 
subset corresponds to the same bracelet arrangement. If 2N has many 
factors, then these subsets will have many different sizes.

Therefore, the answer will not be a simple expression in N.

However, if N is a prime number (as in your example where N = 3), then 
one can check that only subsets of sizes which divide 4N will occur 
(in fact, it seems that only subsets of sizes 2, 2N, or 4N can occur).  
It seems likely that there would exist a uniform formula for the 
number of arrangements in this case and I think it would be a 
challenge for junior high level students to find this formula.

-Doctor Ceeks,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Permutations and Combinations

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