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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.
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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perl
 one year ago
Best ResponseYou've already chosen the best response.3You can do this by trial and error , or we can look up "circular permutations with repetitions allowed".

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.1Is there a concrete formula for this? I've found this site which referred to circular permutations by just \[(n1)!\] http://www.ditutor.com/combinatorics/circular_permutations.html

perl
 one year ago
Best ResponseYou've already chosen the best response.3that is for n distinct objects , here we have reptition

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.3I don't see a simple formula online either :) We can do it out , list all the possibilities

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.1http://artofproblemsolving.com/community/c6h521144 Oh, should I just attempt these problems through trial and error? Doing so results in 3.

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.1Do you understand why they get the 2 in 5 choose 2?

perl
 one year ago
Best ResponseYou've already chosen the best response.3Because there are 2 arrangements that are different up to rotation.

perl
 one year ago
Best ResponseYou've already chosen the best response.3do you understand how they got this aabbb = abbba = bbbaa = bbaab ababb = abbab

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.1It's in a circle, so you can rotate the terms a bit to make it equivalent. Correct?

perl
 one year ago
Best ResponseYou've already chosen the best response.3Lets 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
 one year ago
Best ResponseYou've already chosen the best response.1How do we find the duplicates without enumeration?

perl
 one year ago
Best ResponseYou've already chosen the best response.3you would have to use that formula here scroll down to the last problem on this page http://www.artofproblemsolving.com/community/c6h539781

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.1Oh gosh, this is too much. I'll just have to enumerate.

perl
 one year ago
Best ResponseYou've already chosen the best response.3they did all the enumerations there . they got 5

perl
 one year ago
Best ResponseYou've already chosen the best response.3Let 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
 one year ago
Best ResponseYou've already chosen the best response.1Doesn't 212122 ~ 122212?

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.1Alrighty. I'm sort of understanding. I have to go through this example a little more. Thanks for the steps and help!

perl
 one year ago
Best ResponseYou've already chosen the best response.3a1 = 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.
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