Beads on a BraceletDate: 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/ |
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