 > > Permutations and Circular Permutations

# Permutations and Circular Permutations

Suppose you have three pens which are blue, black, and red in colour and two pencils. The two pencils are such that one of them is of company A and the other is of company B. Now your friend asks you to choose any of one of the items. In how many ways you can do so? The answer to this problem is five as you can choose any one of the given items at a time. Now your friend asks you to arrange them in a box. In how many ways you can do so? These types of problems based on arrangement come under the category of permutations. In this section, we will learn about permutations and circular permutation.

### Suggested Videos        Cases of Permutations Circular Permutations Properties of Combinations Formula ## Permutations

A permutation is basically an arrangement of items in a certain order out of which a few or all of them are taken at a time. In a permutation, we count the number of ways in the arrangement can occur.

Consider the above example of the pens and pencils. If we need to find the number of ways in which these items can arrange themselves in a box, how can we find it? There are two cases either we can select a pencil first or a pen first and then we can arrange them in any way possible.

Suppose the denotation of pens is the letter A and that of pencils is B. We have A1, A2, and A3 for the three pens and B1, and B2 for the pencils. The numbers of ways the arrangement can be done are

### Case 1

The first choice is of the pencil ### Case 2

The first choice is of the pen In both the cases, the number of ways is 6 = 2 × 3 or 3 × 2. Here, 6 = 3!

In factorial notation, n! = n × (n – 1) × (n – 2) × … × 2 × 1.

### Notation

Suppose we have n items out of which we have to make an arrangement for r of them. The numbers of ways in which the arrangement can take place are given by permutation as nPr = $$\frac{n!}{(n-1)!}$$n! ⁄ (n – r)!, 0 ≤ r ≤ n.

Check out yourself the above example with this denotation.

Note 0! = 1.

## Circular Permutation

The above-discussed arrangements are linear in nature. There are some arrangements which are circular in nature. For example consider the roundtable conference, making of a necklace with different coloured beads. These are like arranging the items in a closed loop.  The number of ways of counting associated with the circular arrangement gives rise to a circular permutation.

Suppose we have four chairs around the roundtable and we need to make arrangement for four persons A, B, C, and D. Any of the persons can take any of the positions. But as soon as one of them takes a chair, the number of options reduces by one for the other three persons. And at last, there is only one chair for the last person.

## Arrangement in Circular Permutation

For the above situation, four persons A, B, C, and D can arrange themselves in 4! ways if they are to be arranged in a row. As 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 persons in the arrangement. But in the circular permutation, there is nothing like a start or an end.

In the circular permutation, we consider that one person or object is fixed and the remaining persons are to be arranged. Suppose the position of A is fixed. The number of ways in which the other three persons arrange themselves When one of them has a fix position is (4 − 1)! = 3! = 6. So there are six possible ways in which the four persons A, B, C, and D can arrange themselves. This is also true if we fix the position of B or that of C or of D.

In the above example, the position of the person depends on the clockwise or anticlockwise arrangement. The number of arrangement reduces to $$\frac{1}{2}$$ (n – 1)! = $$\frac{1}{2}$$ (4 – 1)! = $$\frac{3!}{2}$$ = 3 if there is no such dependency. This will be the case if the position of the person does not depend on the order of the arrangement. It is like the arrangement of beads of the same colour in a necklace.

## Solved Examples for You

Problem: Find the number of ways in which 10 beads can be arranged to form a necklace.

Solution: Let us fix the position of one bead. Now, we are left with the arrangement of the remaining, 10 − 1 = 9 beads. These nine beads can arrange themselves in 9P9 = 9! ways. As there is no dependency on the position of beads in a clockwise or anticlockwise manner. The required number of ways = $$\frac{1}{2}$$ (9!) = 181440.

Problem: Find the number of ways in which four girls and three boys can arrange themselves in a row so that none of the boys is together? How is this arrangement different from that in a circular way?

Solution:

#### Case 1: Linear arrangement

Let us first seat the four girls. The girls can seat in 4P4 = 4! = 24.

_ G1 _ G2 _ G3 _ G4 _

For this type of arrangement, the boys can only sit on the five blanked ( _ ) position. Three boys can arrange themselves in 5 P 3 = 5! ⁄ 2! = 60. The required number of ways = 24 × 60 = 1440. (By multiplication theorem)

#### Case 2: Circular arrangement

Since the condition is that none of the boys can sit together or adjacent to each other. We can get the required number of ways if we subtract the ways in which the three boys can seat up together from the total number of arrangement. The total number of ways in which the four girls and three boys can sit around the table = (7 – 1)! = 6!.

Let us assume that the three boys sit together. They are considered as one unit now. Here, we need to arrange only four girls and a unit of boy i.e., 4 + 1 = 5 persons. In the circular arrangement the required number of ways = (5 – 1)! = 4!. These three boys can now rearrange themselves in 3! ways. By the multiplication theorem, the number of the ways = 4! × 3!.

The number of ways in which the arrangement can take place if none of the boys is seated together is 6! – (4! × 3!) = 720 – 144 = 576. Here we see that in the circular arrangement the number of ways is less than those in the row or linear arrangement.

Share with friends

## Customize your course in 30 seconds

##### Which class are you in?
5th
6th
7th
8th
9th
10th
11th
12th
Get ready for all-new Live Classes!
Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes.  Ashhar Firdausi
IIT Roorkee
Biology  Dr. Nazma Shaik
VTU
Chemistry  Gaurav Tiwari
APJAKTU
Physics
Get Started

## Browse

##### Permutations and Combinations Subscribe
Notify of

## Question Mark?

Have a doubt at 3 am? Our experts are available 24x7. Connect with a tutor instantly and get your concepts cleared in less than 3 steps. 