Circular Permutations: Definition, Factorial, Arrangements, Examples

Suppose you have three pencils that are blue, black and yellow and two pens. The pens are such that one of them is brand A and the other is brand B. Now, if your friend asks you to choose any one of them, in how many ways can you do so? The answer to this problem is five ways as you can choose any one of the given items at a time. Now your same friend asks you to arrange all the items in a box. In how many ways can you do so? These types of problems based on arrangement come under the category of permutations.

In this article, we will learn about permutations and circular permutations in detail, along with some examples.

Factorial Notation Formula

The continued product of first \(n\) natural numbers is called \(n\). It is denoted by \(n!\) And is given by

\(n! = 1 \times 2 \times 3 \times 4 \times  \ldots  \times (n – 1) \times n\)

Permutations

A permutation is basically an arrangement of objects in a particular order and out of which one or more are taken out at a time.

Consider the same example of pens and pencils. If we need to find the number of ways in which these items can be arranged in a box, how can we find it?

Initially, there are two cases, either we can select a pencil first or a pen first. Then, we can arrange them in any way possible.

Let’s denote the three pencils as \({P_1},\,{P_2}\) and \({P_3}\) denote pens as \({A_1}\) and \({A_2}\). The various ways of arrangement are as explained below.

Case (i): If the first choice is a pencil, then,

In this case, the number of ways \(6 = 2 \times 3\) or \(3 \times 2\).

Here, \({\rm{6 = 3!}}\)

Case (ii): If the first choice is a pen,

In this case, the number of ways \(6 = 2 \times 3\) or \(3 \times 2\).

Here, \({\rm{6 = 3!}}\)

Thus, in both cases, the arrangement can be made in \(6\) ways.

Note: Suppose we have \(n\) objects out of which we have to make an arrangement for \(r\) of them, the number of ways of the arrangement is given by,

\(^n{P_r} = \frac{{n!}}{{(n – r)!}},\,0 \le r \le n\)

Learn more about Circles here

Circular Permutations

The above-discussed arrangements are linear in nature. There are some arrangements that are circular in nature.

Example: Consider the round-table conference, making of a necklace with different coloured beads. These are like arranging the objects in a loop. The number of ways of counting associated with the circular arrangement gives a circular permutation.

Suppose we have three chairs around the round table, and we need to make the arrangements for three people \(A,\,B\) and \(C\). Anybody can take any of the positions. But, as soon as one of them takes the chair (it is fixed position), then the number of people is reduced by two. And at last, there is only one chair for the last person.

Arrangements in a Circle

For the above situation, \(A,\,B\) and \(C\) can be arranged in \(3!\) ways if they are arranged in a row. In a linear permutation, i.e., in a row arrangement, there is a start, and there is an end. We need to take into consideration the position of all the people in the arrangement. But in a circular permutation, there is nothing like a start or an end.

In a circular permutation, we consider that one person or object is fixed, and the remaining objects are to be arranged. Suppose object \(A\) is fixed, then, the number of ways in which the other two people arrange themselves when one of them has a fixed position is

\({\rm{(3 – 1)! = 2! = 2}}\)

There are two possible ways in which three people \(A,\,B\) and \(C\) can arrange themselves. This is also true if we fix the position of \(B\) or \(C\).

Here, the position of the person depends on the direction of the arrangement: clockwise or anticlockwise. So, if there is no such dependency, the number of arrangements reduces to,

\(\frac{1}{2}(n – 1)! = \frac{1}{2}(3 – 1)! = \frac{{2!}}{2} = 1\)

This holds if the position of the person does not depend on the order of the arrangement. It is similar to the arrangement of beads of the same colour in a necklace.

Suppose four people \(A,\,B,\,C\) and \(D\), are to be arranged along a circle, it is as shown below.

Shifting \(A,\,B,\,C\) and \(D\) one position in the anticlockwise direction, we get the arrangements as follows.

On further shifting, we get,

Note that the arrangements are not different as relative positions of \(A,\,B,\,C\) and \(D\) remains unchanged. But, in the case of linear permutations, the arrangements are given as shown below.

Thus, corresponding to a single circular arrangement of four different objects, there will be \(4\) distinct linear arrangements. Let the number of different objects be \(n\), and the number of their circular permutations be \(x\).

For a circular permutation, the number of linear arrangements is \(n\).

Then, for \(x\) circular permutations, the number of linear arrangements \( = n{x_ \ldots }(i)\)

The number of linear arrangements of \(n\) different objects \( = n! \ldots (ii)\)

Now, from \((i)\) and \((ii)\), we get \(nx = n{\rm{!}}\)

\( \Rightarrow x = \frac{{n!}}{n}\)

\( \Rightarrow x = \frac{{n(n – 1)!}}{n}\)

\( \Rightarrow x = (n – 1)!\)

Suppose \(n\) people \(\left( {{a_1},\,{a_2},\,{a_3},\, \ldots \ldots ,\,{a_n}} \right)\) are to be seated around a circular table. There are \(n!\) different ways they can be seated in a row (linear permutations) as given below.

\({a_1},\,{a_2},\,{a_3},\, \ldots \ldots ..,\,{a_n}\)

\({a_n},\,{a_1},\,{a_2},\, \ldots \ldots \ldots ,\,{a_{n – 1}}\)

\({a_n},\,{a_1},\,{a_2},\, \ldots \ldots ..,\,{a_{n – 1}}\)

\( \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .\)

\( \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .\)

\({a_2},\,{a_3},\,{a_4},\, \ldots \ldots \ldots ,\,{a_1}\)

We know that a circular permutation corresponds to \(n\) unique linear permutations. Hence, the total number of circular arrangements of \(n\) people or objects is \(\left( {n – 1} \right){\rm{!}}\) In other words, the linear arrangement has a beginning and an end, but there is nothing like a beginning or an end in circular arrangements. So, in circular permutations, we fix one object, and the remaining objects are arranged in \(\left( {n – 1} \right){\rm{!}}\) ways.

Clockwise vs Anticlockwise Arrangements

Let us see the following circular arrangements.

In figure \((i)\), the order is clockwise, whereas, in figure \((ii)\), the order is anticlockwise. Note that these two are different arrangements. When a distinction is made between clockwise and anticlockwise arrangements of \(n\) different objects around a circle, then the number circular arrangements is \(\left( {n – 1} \right){\rm{!}}\)

But, if no distinction is made between the clockwise and anticlockwise arrangements of \(n\) different objects around a circle, then the number of circular arrangements is \(\frac{{(n – 1)!}}{{2!}}\).

The arrangements of distinct beads on a necklace as shown in figures \((iii)\) and \((iv)\).

Now, look at \((iii)\) having three beads \({x_1},\,{x_2},\,{x_3}\) as shown. Flip \((iii)\) over to its right. We get the arrangement \((iv)\). Although \((iii)\) and \((iv)\) are the outcomes of an arrangement, they are considered as two different arrangements during calculation. To nullify this redundancy, the actual number of distinct arrangements is taken as

\(\frac{{(n – 1)!}}{{2!}}\)

Note 1:

When the positions are numbered, circular arrangements are treated as a linear arrangement. Whereas, In linear arrangements, it does not make any difference whether the positions are numbered or not.

Note 2:

Number of circular permutations of \(n\) different objects taken \(r\) at a time.

Solved Examples – Circular Permutations

Q.1. Find the number of ways in which \(6\) beads can be arranged to form a necklace.
Ans: Given: \(6\) beads
Arrangement: to form a necklace (circular permutation)
Let us fix the position of the first bead.
Now, we have left with \(5\) beads, which can be arranged in \({\rm{5}}!\) ways.
As there is no dependency on the position of beads in a clockwise or anticlockwise direction.
Required number of ways \( = \frac{1}{2}(5!) = 60\)

Q.2. Find the number of ways in which \(4\) girls and \(3\) boys can arrange themselves in a row such that no boys are together? How is this arrangement different from that in a circular way?
Ans:
Linear Arrangement:
Let us first seat \(4\) girls.
Girls can seat in \(4! = 4 \times 3 \times 2 \times 1 = 24\) ways

For this type of arrangement, the boys can sit in five blank positions.
So \(3\) boys can arrange in five blank places as \(^5{P_3} = \frac{{5!}}{{2!}} = \frac{{120}}{2} = 60\) 
By using the fundamental principle of counting,
Required number of ways \( = 24 \times 60 = 1440\)
Circular Arrangement:
Since the condition is that none of the boys can sit together or adjacent to each other, we can find the number of ways if we subtract the ways in which the three boys can sit up together from the total number of arrangements.
The total number of ways in which the four girls and three boys can sit around the table \({\rm{ = (7 – 1)! = 6!}}\)
Now, let us assume that three boys sit together. They are considered as a single entity. We need to arrange four girls and a single entity, i.e. \({\rm{4 + 1 = 5}}\).
In a circular arrangement, this can be done in \({\rm{(5 – 1)! = 4!}}\) ways.
These three boys can be arranged in \(3!\) ways.
Number of ways \(4! \times 3!\)
Therefore, the number of ways in which the arrangement can take place if none of the boys are seated together is \(6! – (4! \times 3!) = 720 – 144 = 576\).
Thus, we can observe that the number of ways in a circular arrangement is less than the number of ways in linear or row arrangement.

Q.3. In how many ways six people be seated at a circular table?
Ans: Given: \(6\) people are to be arranged in a circular table
Let us fix the position of one person.
Now, we are left with \(5\) people.
Number of arrangements is \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\) ways

Q.4. \(30\) people are invited to a party. In how many ways can they be seated in a circle such that two particular people sit on the sides of the host?
Ans: Total number of people at the party \( = 31\) 
As two people say \(A\) and \(B\) are to be seated beside the host, say \(H\), so the number arrangements are \(AHB\) and \(BHA\). Now taking \([A,\,H,\,B]\) as a single element
Remaining number of guests \( = 29\)
Hence, the number of arrangements \( = 28!\)
Thus, the number of ways of arrangement besides the host \( = 28! \times 2\)

Q.5. In how many ways can \(30\) people be seated at a round table if there are \(8\) chairs?
Ans: As we know, the number of circular permutations of \(n\) different objects taken \(r\) at a time is \(\frac{{^n{P_r}}}{r}\), if clockwise and anticlockwise orders are taken as different.
For circular permutations, the clockwise and anticlockwise arrangements are different.
Hence, the total number of ways \( = \frac{{^{30}{P_8}}}{8}\)

Summary

The arrangement of objects can be made in two styles: linear and circular. In a circular permutation, we consider that one object is fixed, and the remaining is to be arranged. A circular permutation is the total number of ways in which \(n\) different objects can be arranged around a circle. The number of circular permutations of \(n\) distinct objects along with a fixed (cannot be picked up again if it is fixed) circle is \(\left( {n – 1} \right){\rm{!}}\) Circular permutation is of two types \((i)\). clockwise and anticlockwise orders are different \((ii)\). clockwise and anticlockwise orders are the same. Also, we observed that \(n\) linear permutations correspond to \(1\) circular permutation.

Frequently Asked Questions (FAQs)

Q.2. How do you find the number of circular permutations?
Ans: The number of circular permutations of \(n\) distinct objects along with a fixed (cannot be picked up again if it is fixed) circle is \(\left( {n – 1} \right){\rm{!}}\)

Q.3. What is the difference between circular permutation to linear permutation?
Ans: In a linear permutation, i.e. in a row arrangement, there is a start and an end. We need to take into consideration the position of all the objects in the arrangement. But, in a circular permutation, there is nothing like a start or an end, we have to fix one object, and the remaining objects can be arranged in a linear way.

Q.4. Why do we subtract \(1\) in circular permutation?
Ans: We need to take into consideration the position of all the objects in the arrangement. But, in a circular permutation, there is nothing like a start or an end. In the circular permutation, we consider one object is fixed, and the remaining people are to be arranged.

Q.5. How many types of circular permutations are there?
Ans: There are two types of circular permutations:

• Clockwise and anticlockwise orders are different.
• Clockwise and anticlockwise orders are the same.

Learn about Permutations and Combinations

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