Question

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

Answer 1

You can do this by trial and error , or we can look up "circular permutations with repetitions allowed".

Answer 2

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

Answer 3

that is for n distinct objects , here we have reptition

Answer 4

non distinct objects

Answer 5

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.

Answer 6

I don't see a simple formula online either :)
We can do it out , list all the possibilities

Answer 7

http://artofproblemsolving.com/community/c6h521144

Oh, should I just attempt these problems through trial and error?
Doing so results in 3.

Answer 8

Do you understand why they get the 2 in 5 choose 2?

Answer 9

Because there are 2 arrangements that are different up to rotation.

Answer 10

do you understand how they got this
aabbb = abbba = bbbaa = bbaab
ababb = abbab

Answer 11

It's in a circle, so you can rotate the terms a bit to make it equivalent.
Correct?

Answer 12

right

Answer 13

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

Answer 14

How do we find the duplicates without enumeration?

Answer 15

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

Answer 16

Oh gosh, this is too much.
I'll just have to enumerate.

Answer 17

they did all the enumerations there . they got 5

Answer 18

Let B = '1' and A = '2'

$$112222 \sim 221122 \sim 222211
\\211222 \sim 222112 \sim 122221
\\121222 \sim 221212 \sim 122212
\\ 212122 \sim 222121 \sim 212221
\\122122 \sim 221221 \sim 212212
$$

Answer 19

Doesn't 212122 ~ 122212?

Answer 20

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

Answer 21

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.