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How many ways are there to arrange the letters $A$, $A$, $B$, $B$ around a circle?

My first go with was is to arrange $4$ objects around a circle with gets us $(4-1)!$ but we have to take care of the inner arrangements of $A$, $A$ and $B$, $B$ which leaves us with $$\frac 6 {2! \cdot 2!} $$ but that's obviously wrong. I have tried other methods too but they fail to generalize.

By counting by hand for both the $2A$'s and $B$'s case and the $3A$'s and $3B$'s case we get the answers $2$ and $4$ respectively (hopefully I didn't make any mistakes counting) the arrangments i got

  • what's wrong with my argument?
  • what's the right method for solving this?
  • how can this be generalized?
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Ak2399
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    Have you heard of burnside's lemma? – JMoravitz Sep 27 '20 at 18:26
  • no, I haven't heard of it and I'm looking at how it applies right now, but I also want to know why is my argument wrong? – Ak2399 Sep 27 '20 at 18:30
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    Why is your argument wrong? The guess of $\dfrac{(2k-1)!}{k!k!}$ for the number of circular arrangements of $2k$ objects half of which of one type and half of the other? Well, for starters, it should be clear that the guess isn't even an integer for several $k$, in particular for $k$ prime. Now... you could I suppose use this to count the arrangements of $k$ A's, $k-1$ B's and $1$ C as $\dfrac{(2k-1)!}{k!(k-1)!}$, and the logic of which is that you can use that $C$ as a reference point. You can't really do that when the intended reference point can be confused with other points however. – JMoravitz Sep 27 '20 at 18:35
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    Please see - https://math.stackexchange.com/questions/3810598/possible-combinations-of-4-pieces-of-a-and-4-pieces-of-b/3811108#3811108 – Math Lover Sep 27 '20 at 18:39
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    One of my own questions I posted that is related: When is $\binom{2n}{n}\cdot\frac{1}{2n}$ an integer? and the question that inspired me to ask it: arrangement of $n$ oranges and $n$ apples around a circle where another answerer (now deleted) makes the same mistake as you. – JMoravitz Sep 27 '20 at 18:41

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