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sh3lsh

  • one year ago

How many distinct ways can we arrange four A’s and two B’s around a circle? Explanation preferred.

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  1. perl
    • one year ago
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    You can do this by trial and error , or we can look up "circular permutations with repetitions allowed".

  2. sh3lsh
    • one year ago
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    Is there a concrete formula for this? I've found this site which referred to circular permutations by just \[(n-1)!\] http://www.ditutor.com/combinatorics/circular_permutations.html

  3. perl
    • one year ago
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    that is for n distinct objects , here we have reptition

  4. perl
    • one year ago
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    non distinct objects

  5. sh3lsh
    • one year ago
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    I understand. I'm trying to find other material online that will help with this question. I'm afraid I don't know what to do to account for the repeated elements.

  6. perl
    • one year ago
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    I don't see a simple formula online either :) We can do it out , list all the possibilities

  7. sh3lsh
    • one year ago
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    http://artofproblemsolving.com/community/c6h521144 Oh, should I just attempt these problems through trial and error? Doing so results in 3.

  8. sh3lsh
    • one year ago
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    Do you understand why they get the 2 in 5 choose 2?

  9. perl
    • one year ago
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    Because there are 2 arrangements that are different up to rotation.

  10. perl
    • one year ago
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    do you understand how they got this aabbb = abbba = bbbaa = bbaab ababb = abbab

  11. sh3lsh
    • one year ago
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    It's in a circle, so you can rotate the terms a bit to make it equivalent. Correct?

  12. perl
    • one year ago
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    right

  13. perl
    • one year ago
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    Lets find the number of ways to arrange AAAABB , we know this is $$ \Large \frac {6! } {4!~2! }$$ then we can delete those that are the same up to rotation

  14. sh3lsh
    • one year ago
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    How do we find the duplicates without enumeration?

  15. perl
    • one year ago
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    you would have to use that formula here scroll down to the last problem on this page http://www.artofproblemsolving.com/community/c6h539781

  16. sh3lsh
    • one year ago
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    Oh gosh, this is too much. I'll just have to enumerate.

  17. perl
    • one year ago
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    they did all the enumerations there . they got 5

  18. perl
    • one year ago
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    Let B = '1' and A = '2' $$112222 \sim 221122 \sim 222211 \\211222 \sim 222112 \sim 122221 \\121222 \sim 221212 \sim 122212 \\ 212122 \sim 222121 \sim 212221 \\122122 \sim 221221 \sim 212212 $$

  19. sh3lsh
    • one year ago
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    Doesn't 212122 ~ 122212?

  20. sh3lsh
    • one year ago
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    Alrighty. I'm sort of understanding. I have to go through this example a little more. Thanks for the steps and help!

  21. perl
    • one year ago
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    a1 = 4 a2 = 2 define gcd (2,4) = d , which is 2 Let the circular arrangements be similiar if they only differ in rotation by N/d, here 6/2= 3. So the circular arrangements are similar if they different in rotation by 3 112222 and 221122 differ by 3 rotations , so they are similar. The point of making this equivalence class is then we can use that formula below.

  22. perl
    • one year ago
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    |dw:1434298495220:dw|

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