# Circular arrangements at a round table

Q: If six people, designated as $A,B,...,F,$ are seated about a round table how many different circular arrangements are possible, if arrangements are considered the same when one can be obtained from the other by rotation?

Attempt: There are 6! ways to arrange; 6 at one spot, 5 at the next, ... 1 possibility at the end.

Since arrangements are considered repetitive when it can be obtained from rotation (in this case, there are 6 cases, (1) ABCDEF (2) BCDEFA (3) CDEFAB ... (6) FABCDE

Then comes in the part I don't quite get an intuition for. We have $\frac{6!}{6} = 5!$ , which is hard for me to understand. My knee jerk response was to subtract 6 from 6!, because the total number of cases is 6! and the repetitions that we want to get rid of is 6, I thought simply: total number of cases - repetitions = permutation without repetition.

Why am I wrong? Why do we need to divide instead of subtract?

Any input would be appreciated.

You have to divide by $6$ since for each table position, there are $5$ other table positions that are merely rotations. As an example, the following positions are equivalent (imagine the line loops around):

$$ABCDEF$$

$$BCDEFA$$

$$CDEFAB$$

$$DEFABC$$

$$EFABCD$$

$$FABCDE$$

But consider the following equivalent rotations (which are different from the first set):

$$FEDCBA$$

$$EDCBAF$$

$$DCBAFE$$

$$CBAFED$$

$$BAFEDC$$

$$AFEDCB$$

So you see that any initial position of $A$, $B$, $C$, $D$, $E$, and $F$ in a line, there are five other equivalent rearrangements if you sit them at a table. Therefore, you divide by $6$.

Another explanation would be for you to consider arranging the people one at a time around a table. $A$ can sit anywhere at the table. $B$ has $5$ remaining seats to choose from, $C$ has $4$ remaining seats to choose from, etc. This gives for $n$ people, $(n-1)!$ ways of seating them around a table, since the first person can sit wherever they want.

• Thank you. That made perfect sense. – Corp. and Ltd. Oct 18 '16 at 6:08

For the sake of explanation, suppose there are four people.

If we list out all $24$ permutations and group together permutations that are the same up to rotation, then we get $6$ groups of $4$.

$$ABCD, BCDA, CDAB, DABC$$

$$ABDC, BDCA, DCAB, CABD$$

$$ACBD, CBDA, BDAC, DACB$$

$$ACDB, CDBA, DBAC, BACD$$

$$ADBC, DBCA, BCAD, CADB$$

$$ADCB, DCBA, CBAD, BADC$$

It should now be clear why we divide $4!$ by $4$ instead of subtracting $4$: each unique table arrangement appears in this list of permutations $4$ times, i.e., each row corresponds to a unique table arrangement.