Overview
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A circular Permutation is the total number of ways in which n distinct objects can be arranged around a fixed circle. A permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements. The word “permutation” also refers to the act or process of changing the linear order of an ordered set.
Circular permutation is the total number of ways in which n distinct objects can be arranged around a fixed circle. In the circular permutation, there is nothing like a start or an end.
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Cases of Circular Permutation
There are 2 cases of circular permutations
The properties of circular permutation are somewhat similar to the properties of permutation. There are 2 types of Circular Permutation:
In circular permutation, one element is always fixed and all other elements are arranged relative to the fixed element. There is nothing like a start or an end. There is nothing like a beginning or end in circular permutation.
Rotation Doesn’t Change the Arrangement
In a circle, rotating all the objects together doesn’t create a new arrangement. That’s why circular permutation counts fewer arrangements than linear permutation.
Fixing One Object
To avoid counting duplicate rotations, we fix one object’s position and arrange the rest. That’s why the formula is (n – 1)! instead of n!.
Clockwise vs Counter-clockwise
If clockwise and counter-clockwise arrangements are considered the same (like in necklaces or bracelets), then the formula becomes:
(n – 1)! / 2
If direction matters (like seating at a round table), use (n – 1)! only.
Identical Items
If some items are repeated, divide by the factorial of the repeated items:
(n – 1)! / (p! × q! × ...)
where p, q… are the counts of repeated items.
Useful in Real-Life Arrangements
Circular permutation is commonly used in situations like:
Seating around a round table
Designing circular flower beds
Arranging beads in circular jewelry
Formula and Derivation of Circular Permutation
Let’s study the formula and derivation of circulation permutation with proof.
Formula 1. If Clockwise and Anticlockwise orders are different
When the clockwise and anticlockwise orders are different then: The number of circular permutations is.
\(P_n={nP_r\over{r}}\)
where n = Total number of objects.
r = Number of selected objects.
Pn = Circular permutation.
If n = r (selecting all the objects) then,
\(\begin{matrix}
P_n={nP_n\over{n}}\\
= {n\times(n−1)!\over{n}}\\
P_n=(n−1)!
\end{matrix}\)
Here the arrangement (1-2-3 and 1-3-2) is different when we see the images from clockwise and anticlockwise.
Suppose n things \((x_1, x_2, x_3,…, x_n)\) are to be arranged around in a circular fashion. There are n! ways in which they can be arranged in a row. On the other hand, all the linear arrangements depicted by
\(\begin{matrix}
x_1 , x_2 , x_3,…, x_n\\
x_n , x_1 , x_2, …. , x_{n − 1}\\
x_{n − 1} , x_n , x_1 , x_2 …. x_{n − 2}\\
……
x_2 , x_3 , x_4 , ….., x_1\\
\end{matrix}\)
will lead to the same arrangement for a circular table. Hence each circular arrangement corresponds to n linear arrangements (i.e. in a row). Hence the total number of circular arrangements of n persons is n!/n = (n − 1)!
The number of circular permutation is \(P_n={nP_r\over{r}}\)
r = Number of selected objects.
Pn = Circular permutation.
If n = r (selecting all the objects) then,
\(\begin{matrix}
P_n={nP_n\over{2n}}\\
= {n\times(n−1)!\over{2n}}\\
P_n={(n−1)!\over{2}}\\
\end{matrix}\)
Let us determine the number of distinguishable permutations of the letters ELEMENT.
Let us determine the number of distinguishable permutations of the letters ELEMENT.
Suppose we make all the letters different by labeling the letters as follows: \(E_1LE_2ME_3NT\)
Now, all the letters are different from each other. In this case, there are (7-1)! = 6! different circular permutations are possible.
Let us consider one such case of circular permutation, where the elements are arranged in the order below:
\(LE_1ME_2NE_3T\)
We can form new arrangements of permutations from this arrangement by only moving the E’s. Clearly, there are 3! or 6 such arrangements. We list them below.
\(\begin{matrix}
LE_1ME_2NE_3\\
LE_1ME_3NE_2\\
LE_2ME_1NE_3T\\
LE_2ME_3NE_1T\\
LE_3ME_2NE_1T\\
LE_3ME_1NE_2T\\
\end{matrix}\)
Because the E’s are not different, there is only one arrangement LEMENET and not six. This is true for every permutation.
From the properties of permutations, we already know that four persons A, B, C, and D can arrange themselves in 4!
In the circular permutation, there is nothing like a start or an end.
In circular permutation, one element is always fixed and all other elements are arranged relative to the fixed element. The number of ways in which the other three persons arrange themselves when one of them has a fixed position is (4 − 1)! = 3! = 6.
The application of circular permutations is as follows:
Example 1: In how many ways can 6 men be seated around a circular table?
Solution: 6 men can be seated around a circular table in (6-1)! = 5! = 5 × 4 × 3 × 2 × 1 = 120 ways.
Example 2: Find the number of ways in which 10 different beads can be arranged to form a necklace.
Solution: 10 different beads can be arranged in a necklace (circular form) =12×(10−1)!=12×(9)! ways.
Example 3: In a playground, 3 sisters and 8 other girls are playing together. In a particular game, how many ways can all the girls be seated in a circular order so that the three sisters are not seated together?
Solution: There are 3 sisters and 8 other girls in a total of 11 girls. The number of ways to arrange these 11 girls in a circular manner = (11– 1)! = 10!. These three sisters can now rearrange themselves in 3! ways. By the multiplication theorem, the number of ways that 3 sisters always come together in the arrangement = 8! × 3! Hence, the required number of ways in which the arrangement can take place if none of the 3 sisters is seated together: 10! – (8! × 3!) = 3628800 – (40320 * 6) = 3628800 – 241920 = 3386880.
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