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Question: If six people, designated as A,B,C,D,E,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?

Note: Since I don't know how to post images here/ too lazy to post them, you will have to figure out this problem by imagining the diagram or drawing it out on your own.

My attempt: Even though this question deals with circular arrangements, we can still use the permutation formula of indistinguishable objects dealing with linear arrangements: n!/nr!

  • First, there are 6! ways to arrange the positions of the six letters A-F; rotating each letter one at a time to a new position of the table. When a letter, say A, returns to the position it started on, that is considered a repetition. Therefore, 6! is the total number of possible circular arrangements.

  • That makes the n! = 6!. So the bottom, nr!, is equal to 6, because say, 1)ABCDEF, 2)BCDEFA, 3)CDEFAB, 4) DEFABC, 5) EFABCD, 6) FABCDE, are all the same arrangements. Therefore, nr! = 6, because 6 is the number of indistinguishable objects.

  • Dividing 6!/6 we get 120 , which is the correct answer.

My problems with this question:

  • From my way of doing the question above, is it the right way to do it? Although nr! = 6, why isn't it 6! ? Shouldn't it be 6! for the bottom, since that's how the formula is stated?

  • Another way of doing this is (n-1)! Why is this also correct? Why is 1 subtracted from n?

  • Also another way is to fix the starting position of any letter, say A. That will give 5! which is 120. What does fixing one of the letters have to do finding the possible arrangements? Really confused...
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  • $\begingroup$ Really really confused. I am totally lost while looking at the other two methods of solving this question. $\endgroup$ – harold232 Sep 23 at 5:28
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The idea of a circular table is that all the seats are interchangeable until someone sits down. You seat A in one seat, which orients the table. All positions for $A$ are equivalent. Now you can seat the rest in $5!$ ways.

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  • $\begingroup$ So basically, when A is fixed and when the 5 others exchange seats, the person sitting beside A on the left skips over A and exchanges seats with the person sitting on A's right hand side? Thus, making every 5 arrangement distinct? $\endgroup$ – harold232 Sep 23 at 5:32
  • $\begingroup$ I think a better way to see it is to say you are distributing five people to five seats. You don't need to think about people exchanging seats, but it amounts to the same thing. $\endgroup$ – Ross Millikan Sep 23 at 5:34
  • $\begingroup$ In other words, what I was trying to say is that if A is fixed, skipping over A when exchanging seats in a clock-wise direction, will result in 5 distinct arrangements? For example, if I make A fixed, the arrangements of the 5 other people/letters skipping over A, will not result in something like 1)ABCDEF, 2)BCDEFA, 3)CDEFAB, 4) DEFABC, 5) EFABCD, 6) FABCDE? $\endgroup$ – harold232 Sep 23 at 5:37
  • $\begingroup$ Also for the approach of dividing the permutation of total objects by the permutation of indistinguishable objects 6!/6, why is 6 not 6! ? The formula shows that the denominator is supposed to also be permutated $\endgroup$ – harold232 Sep 23 at 5:40
  • $\begingroup$ Once you seat A, you just line up the other people and seat them clockwise from A. There are $5!$ ways to line them up. Alternately, you can pretend the seats are labeled, seat everybody in one of $6!$ ways, then note that there are $6$ rotations of the table that are equivalent, so you divide by $6$. $\endgroup$ – Ross Millikan Sep 23 at 14:38
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Let $B,C,D,E,F$ form a sequence (e.g. $CFDBE$) and let them hold the hand of their neighbours. Only the left hand of outmost left and the right hand of the outmost right stay unpossessed.

There are $5!=120$ possibilities for this.

Now let $A$ come in, taking the two hands that are not possessed yet.

A circle is made now (seats and table are somehow irrelevant).

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First note that $$ \frac{6!}{6}=\frac{6\cdot5\cdot4\cdot3\cdot2\cdot1}{6}=5\cdot4\cdot3\cdot2\cdot1=5!=(6-1)!, $$ so all expressions agree.

The expression $6!/6$ is justified by the observation that the $6!$ ordinary permutations come in groups of six that are equivalent as circular permutations. So on the list of ordinary permutations, each circular permutation is represented six times. Dividing $6!$ by $6$ corrects for this overcounting.

There’s a general principle here that we might call the division principle: when a counting method over counts each of the things we want to count by the same factor $f$, dividing the count by $f$ gives the correct count.

The formula you propose using, $n!/n_r!$, is not applicable here. That formula also derives from the division principle, but It is to be used when there are $n$ slots and $n$ objects, but $n_r$ of the objects are considered interchangeable and may be permuted amongst themselves. An example would be the number af arrangements of the letters of BUBBLE. If the letters are written on cards, there are $6!$ ways of permuting the physical cards, but each letter arrangement occurs $3!$ times among the permutations of cards because there are $3!$ ways of permuting the B cards. Hence dividing by $3!$ gives the correct count. Despite the similarities, this is simply not relevant to your problem.

The other method, using $(6-1)!$, comes about when one avoids the overcounting in the first place, by counting only permutations relative to one of the individuals, say A. In this method, we are placing individuals not in physical chairs, but in the chair to the right of A, the chair two seats to the right of A, and so on. There are now only $6-1=5$ individuals to permute, which gives the result.

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