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.
$\endgroup$2 Answers
$\begingroup$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.
$\endgroup$ 1 $\begingroup$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.
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