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I Did not Understand Arrangement around a Circular Table

(i) If Clockwise and Anticlock order are Different

(ii) If Clockwise and Anticlock order are not Different

for (1)

I Have Understand like that arrangement of letters $A,B,C,D,E$

Arrangement around a circular table of letters $A,B,C,D,E = 5!$

Arrangement around a circular table of letters $B,C,D,E,A = 5!$

Arrangement around a circular table of letters $C,D,E,A,B = 5!$

Arrangement around a circular table of letters $D,E,A,B,C = 5!$

Arrangement around a circular table of letters $E,A,B,C,D = 5!$

Now If we taken as one Direction (Like for clockwise) We have seen that all $5$ arrangements are the same.

So My Question is How can I Count Circular permutation of $5$ Letters

Similarly How can I calculate $(II)$ part, Thanks

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For 2, if you can't differentiate start and end points, means, A B C D E is same as B C D E A. – Ram Jan 29 '13 at 5:48
up vote 5 down vote accepted

Here is a way of visualizing the situation, in your case of $n=5$.

There are $5$ chairs, evenly spaced around a circular table. These chairs are labelled $1,2,3,4,5$, say counterclockwise.

We want to seat $5$ people $A,B,C,D,E$, so that seating arrangements that differ by a rotation are considered the same. The string $BAEDC$ means $B$ is at $1$, $A$ at $2$, $E$ at $3$, $D$ at $4$, and $C$ at $5$. This is considered the same as $AEDCB$, also $EDCBA$, $DCBAE$, and $CBAEDE$.

Call these $5$ arrangements a family. Any arrangement belongs to a unique family of $5$.

There are $5!$ arrangements of the letters $A,B,C, D, E$ in a line. We ask how many families there are. Well, since each family contains $5$ strings, there are $\dbinom{5!}{5}=4!$ families. Thus there are $4!$ circular permutations of our $5$ people.

With $n$ people, exactly the same idea works, and there are $\dfrac{n!}{n}=(n-1)!$ circular permutations.

Here we considered two arrangements, one of which is clockwise, and the other anti-clockwise, as different.

But if we consider two such arrangements as being the same, then for example $ABCDE$ and $EDCBA$ are considered the same. This means (in our case) that now each family has $10$ objects, twice as many as before. So the number of "really different" arrangements in this case is $\dfrac{5!}{(2)(5)}$.

This generalizes nicely to $n$ people, with a couple of minor exceptions. If $n\ge 3$, the number of circular arrangements with clockwise and anti-clockwise considered the same is $\dfrac{n!}{2n}=\dfrac{(n-1)!}{2}$.

If $n=1$, obviously there is only $1$ arrangement, whatever the conditions. And if $n=2$, the one and only arrangement is the same, whether clockwise or anticlockwise are considered the same or not.

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Thanks André Nicolas – juantheron Jan 29 '13 at 17:56

Your choice of arrangements ABCDE, BCDEA, etc is not where I would start.

(i) Pick any person, put them anywhere. Now you have $4!$ ways of choosing next four clockwise positions starting with that person. So the answer is $4!$

(ii) If clockwise and anticlockwise are different, there are is still $4!$ total arrangements, as each anti-clockwise permutation is the same as a different clockwise permutation, so these add no extra. Answer is still $4!$

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Thanks gnometorule – juantheron Jan 29 '13 at 17:54

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