Permutation question with a dial

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SUMMARY

The discussion focuses on calculating the number of ways to connect numbers on a dial with 12 positions arranged in a circle. The user initially determined that there are 66 possible lines by forming 6 pairs of numbers, each offset by one position, but sought a more mathematical approach using factorials. The user attempted the formula 12!/(6!^2 * 2^6) but did not arrive at the expected result of 66. The conversation emphasizes the need for a clearer understanding of permutations and combinations in circular arrangements.

PREREQUISITES
  • Understanding of permutations and combinations
  • Familiarity with factorial notation and calculations
  • Basic knowledge of circular arrangements in combinatorial mathematics
  • Experience with mathematical problem-solving techniques
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  • Study the concept of circular permutations in combinatorics
  • Learn about the application of the multinomial coefficient in partitioning sets
  • Explore the use of generating functions for combinatorial problems
  • Investigate the relationship between pairing and factorial calculations in combinatorial contexts
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Students studying combinatorial mathematics, educators teaching permutations and combinations, and anyone interested in solving mathematical problems involving circular arrangements.

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Homework Statement



So, you have a dial with 12 numbers (1 through 12), and you're wondering how many ways can you connect an number to another. So its therefor asking how many lines can you make with 1 - 12 in a circle.

Homework Equations





The Attempt at a Solution



I got the solution, I made 6 pairs one number to the one 3 numbers away from it (in a circular manner) and I knew if I were to offset everything by one id next a combination, and I could do this 11 times.

6 pairs * 11 offsets = 66 possible lines.


But I was wondering how to do this in a more mathematical manner (with factorials.) And I watched this interresting video on the enigma code "http://www.youtube.com/watch?v=G2_Q9FoD-oQ".

So I tried the same logic on the permutation the man used for the 26 letters. I attempted
12!/(6!^2 * 2^6) and didn't get 66.

How would you solve this using the logic that video used on the Engima code. I would simply like to know other ways to solve this equation
 
Physics news on Phys.org
You want the number of ways of partitioning 12 things into pairs? Pick one item at random. How many ways of choosing its pair? Now pick one of the remaining 10. How many ways to choose its pair? ...
 

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