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A senate committee has 5 democrats and 5 republicans. Each of the democrats and republicans are indistinguishable from the other democrats and republicans respectively. In how many ways can they sit around a circular table?

I know that they can be arranged in $\frac{10!}{5! 5!}$ ways if they are not placed in a circle. Each circular arrangement is then overcounted 10 times, so the answer should be $\frac{10!}{10 (5! 5!)}$. Unfortunately, this answer comes out to be a fraction. How should this problem be approached?

1110101001
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  • Yes, I know so that is why I divided by 10 to account for the 10 possible rotations. – 1110101001 Apr 05 '14 at 19:56
  • Yeah, sorry, was a bit to quick. – M.B. Apr 05 '14 at 19:57
  • What would you get, using the same method, if you counted the ways of arrange 10 undistinguishable democrats in a circle? – Mariano Suárez-Álvarez Apr 05 '14 at 19:59
  • @user2612743: try to look at the case with 2 democrats and 2 republicans. Do you see anything special? How do the arrangements split into equivalences classes? Are they of the same size? – M.B. Apr 05 '14 at 20:03
  • @M.B. In the 2 democrat and 2 republican case there are only two possible arrangements, right? I don't see how that helps with the 5 democrat and 5 republican case... – 1110101001 Apr 05 '14 at 20:05
  • Correct, but the equivalence classes are not of the same size. One has four elements, the other 2. That's why you simply cannot divide by 10. – M.B. Apr 05 '14 at 20:06
  • You can't divide by 10. This is what's known as a "necklace" question, it is quite a bit more complex than regular combinatorical ones. I will give it a thought :) – Oria Gruber Apr 05 '14 at 20:09
  • @M.B. I don't quite understand what you mean by equivalence classes. For 2 republicans and 2 democrats there are 4!/(2!2!) = 6 arrangements not in a circle. In a circle there are only 2. I see that you cannot simply divide by 4 to get from one to the other but I don't understand why. – 1110101001 Apr 05 '14 at 20:09
  • @OriaGruber Could you please explain why dividing by 10 doesn't work? – 1110101001 Apr 05 '14 at 20:09
  • @user2612743: in the case for 2 and 2 you have the following: {XXYY, YXXY, YYXX, XYYX} and {XYXY, YXYX}. – M.B. Apr 05 '14 at 20:10
  • @M.B. I see that it gets split into those but I still don't understand "why" this happens. How come in the regular, distinguishable circular case of XXYY each permutation is over-counted four times whereas in the indistinguishable version it is split into four and two times? – 1110101001 Apr 05 '14 at 20:17
  • I am writing a highly inelegant answer now :) – M.B. Apr 05 '14 at 20:17
  • A similar question came up here: arrangements of $n$ oranges and $n$ apples around a circle where I derived the general solution. The general solution is also given by https://oeis.org/A003239 – JMoravitz Mar 15 '16 at 15:58
  • @JMoravitz Ah thanks for linking me to your excellent response :D – 1110101001 Mar 16 '16 at 00:27

2 Answers2

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Added below: A solution using Burnside’s Lemma, which is the way to generalize and regularize what I did, as asked about in a comment. I didn't use Burnside's Lemma in my answer, because I wanted to give an answer that was elementary.

I know that they can be arranged in $\tfrac{10!}{5!5!}$ ways if they are not placed in a circle. Each circular arrangement is then overcounted 10 times, so the answer should be $\tfrac{10!}{10\cdot5!5!}$.

Each arrangement is not overcounted by a factor of 10.

For example, while the ten linear arrangements $RRRRRDDDDD, DRRRRRDDDD,\dots RRRRDDDDDR$ all lead to the same seating arrangement (all party members together), $RDRDRDRDRD$ and $DRDRDRDRDR$ are the only two linear arrangements that lead to strictly alternating seating around the circle.

Each circular arrangement is counted a full 10 times only if its rotations by tenths of a circle yield different linear sequences when reading from the top clockwise. If an arrangement is unchanged by a rotation smaller than $2\pi$, there are fewer than 10 linear arrangements. The only possible smallest nonzero rotations by tenths that fix a circular pattern are $\tfrac{2\pi}{10}$ (pattern counted 1 times), $\tfrac{2\pi}{5}$ (pattern counted 2 times), and $\tfrac{2\pi}{2}$ (pattern counted 5 times).

For the question at hand, $\tfrac{2\pi}{5}$ is the only possibility. (If the pattern were fixed by a one-tenth rotation, all symbols would have to be the same. If it were fixed by a one-half rotation, the number of democrats and republicans would have to both be even.)

The only arrangement fixed by a $\tfrac{\pi}{5}$ rotation is the one I mentioned above, which is counted twice. Every other arrangement is counted ten times.

The number of arrangements is then $\frac{\tfrac{10!}{5!5!}-2}{10}+1$, or 26.

Added: Briefly, Burnside’s lemma (which is easy to google) says that given a collection of patterns, you can find the number of them that are distinguishable up to a group of motions (such as rotations here) by finding the average number of patterns fixed by one of the motions, which happens to be the same number. (This is not immediately obvious!)

So in this case (5 each of $R$s and $D$s at a round table), Look at the set of $\tfrac{10!}{5!5!}$ seating arrangements without considering two the same if one is a rotation of the other. Then consider the group of all rotations. There are 10 different rotations, by $0$, $\tfrac{\pi}{10}$, $2\tfrac{\pi}{10}$, ... $9\tfrac{\pi}{10}$. For each of these 10 rotations, count the number of seating arrangements “fixed” by the rotation. All $\tfrac{10!}{5!5!}$ are fixed by the $0$ rotation. None are fixed by the $k\tfrac{\pi}{10}$ for $k=1,3,5,7,9$, and $2$ patterns each are fixed by the $2\tfrac{\pi}{10},4\tfrac{\pi}{10},6\tfrac{\pi}{10}$ and $8\tfrac{\pi}{10}$ rotations. So the average number of arrangements fixed by a rotation is $\tfrac{1}{10}(\tfrac{10!}{5!5!}+0+2+0+2+0+2+0+2+0)$

JMoravitz
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Steve Kass
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  • This is an excellent answer. Maybe the answer would benefit from emphasizing why only rotations by the divisors of 10 have to be considered. Also, I think it should be $\pi$ not $2\pi/5$ for pattern counted 2 times. – M.B. Apr 05 '14 at 21:03
  • I meant 5 times. – M.B. Apr 05 '14 at 21:09
  • @M.B. Why are only the divisors of 10 counted for the rotations? – 1110101001 Apr 05 '14 at 22:38
  • And what is meant by "fixed"? – 1110101001 Apr 05 '14 at 22:48
  • And is this possible to extend this to three groups of people (democrats, republicans, tea party) or do you have to use burnsides lemma for that? – 1110101001 Apr 05 '14 at 22:54
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    @user2612743: rotation by 3 would mean that all elements were equal. Why? Obviously 0 = 3 = 6 = 9 = 2 = 5 = 8 = 1 = 4 = 7 = 0. What you need is that $nk \equiv 0\pmod{10}$ for $n<10$ implying that $k|10$. – M.B. Apr 05 '14 at 23:14
  • user2612743: “Fixed” means indistinguishable from the starting arrangement. If $RDRDRDRDRD$ were the seating arrangement and everyone moved two chairs clockwise (one-fifth of the way around the table), the result would be $RDRDRDRDRD$, so the $RDRDRDRDRD$ arrangement is fixed by a one-fifth rotation. @M.B.: $\tfrac{2\pi}{5}$ is correct as the minimum rotation for seatings counted twice. If the minimum rotation is $\pi$, the seating, its $\tfrac{\pi}{5}$, $\tfrac{2\pi}{5}$, $\tfrac{3\pi}{5}$, and , $\tfrac{4\pi}{5}$ rotations are five different linear patterns and are all counted. – Steve Kass Apr 05 '14 at 23:59
  • Yes, I updated my comment above, but it was after five minutes so I was unable to make a proper edit. – M.B. Apr 06 '14 at 00:05
  • @M.B. How would you extend this to more than two groups, such as having 2 blue, 2 red, and 2 yellow objects? – 1110101001 Apr 06 '14 at 01:55
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    This is too long to answer in a comment. I'll append a response to my answer. – Steve Kass Apr 06 '14 at 02:32
  • For the burnside's lemma, how do you find the number of fixed elements given the rotation number? – 1110101001 Apr 06 '14 at 02:47
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    You have to just "work it out." For example, for a seating of ten around a table to be fixed by a half-circle rotation, the seating arrangement would have to look like $abcdeabcde$ (first five match second five). How many arrangements have this form depends on the particular situation, and it may be easy or hard to find. If you had two each of blue, red, and yellow objects to seat at a table for six, and you want to know how many arrangements are fixed by a half-circle rotation, how can you have $xyzxyz$? Well, the $x$, $y$, and $z$ are blue, red, and yellow in any order, so there are six. – Steve Kass Apr 06 '14 at 03:00
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    For the record, the solution using burnside's is slightly incorrect. Rotation by 4, and 6 steps does have fixed arrangements noting that the element in position $0$ and rotating by an amount of $4$ must match that in positions $0,4,8,2,6$ and there are $2$ such arrangements. I'm taking the liberty of correcting the answer above. – JMoravitz Mar 15 '16 at 16:07
  • why do you consider the clock wise arrangement same as anti clock wise arrangement? – parvin Aug 09 '17 at 17:03
  • @parvin I don’t understand your question. Can you give a specific pair of arrangement that would be considered the same or different depending on whether clockwise and anti-clockwise were considered the same? – Steve Kass Aug 10 '17 at 00:42
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This is not a very elegant solution, but it works. Perhaps more clever ways exist, e.g. using Burnside's lemma.

I divide the problem into five pieces and I label democrats by X and republicans by Y.

Case 1) They split into two: XXXXXYYYYY

Case 2) There are four democrats in a row. Hence, XXXXY----Y, where we will put exactly one X amongst the four dashes. This gives four possibilites.

Case 3) Three democrats in a row: XXXY-----Y. There are five dashes and a total number of ${5\choose 2} = 10$ arrangements.

Case 4) Two, and no more than two, in a row: XXY------Y. Six dashes, but slight more tricky. You compute this guy.

Case 5) XYXYXYXYXY.

Now add up.

M.B.
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  • Where do you account for the possible rotations? – 1110101001 Apr 05 '14 at 20:24
  • @user2612743: simply by putting the longest chain of republicans in the beginning of the enumeration. (Except for case 4 which is a bit more tricky.) – M.B. Apr 05 '14 at 20:27