sh3lsh
  • sh3lsh
How many distinct ways can we arrange four A’s and two B’s around a circle? Explanation preferred.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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perl
  • perl
You can do this by trial and error , or we can look up "circular permutations with repetitions allowed".
sh3lsh
  • sh3lsh
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
perl
  • perl
that is for n distinct objects , here we have reptition

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perl
  • perl
non distinct objects
sh3lsh
  • sh3lsh
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.
perl
  • perl
I don't see a simple formula online either :) We can do it out , list all the possibilities
sh3lsh
  • sh3lsh
http://artofproblemsolving.com/community/c6h521144 Oh, should I just attempt these problems through trial and error? Doing so results in 3.
sh3lsh
  • sh3lsh
Do you understand why they get the 2 in 5 choose 2?
perl
  • perl
Because there are 2 arrangements that are different up to rotation.
perl
  • perl
do you understand how they got this aabbb = abbba = bbbaa = bbaab ababb = abbab
sh3lsh
  • sh3lsh
It's in a circle, so you can rotate the terms a bit to make it equivalent. Correct?
perl
  • perl
right
perl
  • perl
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
sh3lsh
  • sh3lsh
How do we find the duplicates without enumeration?
perl
  • perl
you would have to use that formula here scroll down to the last problem on this page http://www.artofproblemsolving.com/community/c6h539781
sh3lsh
  • sh3lsh
Oh gosh, this is too much. I'll just have to enumerate.
perl
  • perl
they did all the enumerations there . they got 5
perl
  • perl
Let B = '1' and A = '2' $$112222 \sim 221122 \sim 222211 \\211222 \sim 222112 \sim 122221 \\121222 \sim 221212 \sim 122212 \\ 212122 \sim 222121 \sim 212221 \\122122 \sim 221221 \sim 212212 $$
sh3lsh
  • sh3lsh
Doesn't 212122 ~ 122212?
sh3lsh
  • sh3lsh
Alrighty. I'm sort of understanding. I have to go through this example a little more. Thanks for the steps and help!
perl
  • perl
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.
perl
  • perl
|dw:1434298495220:dw|

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