# Circular Permutations: Demonstration, Examples, Solved Exercises

The **circular permutations ** are different types of groupings of all elements of a set, when they are to be arranged in circles. In this type of permutation the order matters and the elements are not repeated.

For example, suppose you want to know the number of distinct arrays of digits one through four, placing each number at one of the vertices of a rhombus. These would be 6 arrangements in total:

It should not be confused that the number one is in the upper position of the rhombus in all cases as a fixed position. Circular permutations are not changed by the rotation of the array. The following are a single or the same permutation:

__Demo and formulas__

__Demo and formulas__

In the example of the different 4-digit circular arrays located at the vertices of a rhombus, the number of arrays (6) can be found like this:

1- Any of the four digits is taken as a starting point at any of the vertices and advances to the next vertex. (it does not matter if it is turned clockwise or counterclockwise)

2- There are 3 options left to select the second vertex, then there are 2 options to select the third vertex and, of course, there is only one selection option for the fourth vertex.

3- Thus, the number of circular permutations, denoted by (4 – 1) P (4 – 1), is obtained by the product of the selection options in each position:

(4 – 1) P (4 – 1) = 3 * 2 * 1 = 6 different 4-digit circular arrays.

In general, the number of circular permutations that can be achieved with all the n elements of a set is:

(n – 1) P (n – 1) = (n – 1)! = (n – 1) (n – 2)… (2) (1)

Note that (n – 1)! It is known as n factorial and abbreviates the product of all numbers from the number (n – 1) to the number one, inclusive.

__Examples__

__Examples__

**Example 1**

How many different ways do 6 people have to sit at a circular table?

You want to find the number of different ways that 6 people can sit around a round table.

N ° of ways to sit = (6 – 1) P (6 – 1) = (6 – 1)!

No. of ways to sit = 5 * 4 * 3 * 2 * 1 = 120 different ways

**Example 2**

How many different ways do 5 people have to locate themselves at the vertices of a pentagon?

The number of ways in which 5 people can be located at each of the vertices of a pentagon is sought.

N ° of ways to be located = (5 – 1) P (5 – 1) = (5 – 1)!

N ° of ways to be located = 4 * 3 * 2 * 1 = 24 different ways

__Solved exercises__

__Solved exercises__

**– Exercise 1**

A jeweler acquires 12 different precious stones to place them in the points of the hours of a clock that he is preparing on behalf of the royal house of a European country.

a) How many different ways does he have to arrange the stones on the clock?

b) How many different shapes does it have if the stone that goes to 12 o’clock is unique?

c) How many different shapes if the 12 o’clock stone is unique and the stones of the other three cardinal points, 3, 6 and 9 o’clock; Are there three particular stones, which can be exchanged, and the rest of the hours are assigned from the rest of the stones?

**Solutions**

a) The number of ways to arrange all the stones on the circumference of the clock is requested; that is, the number of circular arrangements involving all available stones.

Number of arrangements on the clock = (12 – 1) P (12 – 1) = (12 – 1)!

Number of fixes on the clock = 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

Number of arrangements on the clock = 39976800 different shapes

b) He wonders how many different ways of ordering exist knowing that the stone of the 12 o’clock handle is unique and fixed; that is, the number of circular arrangements involving the remaining 11 stones.

Number of arrangements on the clock = (11 – 1) P (11 – 1) = (11 – 1)!

Number of fixes on the clock = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

Number of arrangements on the clock = 3,628,800 different shapes

c) Finally, the number of ways to order all the stones is sought except for the 12 o’clock stone which is fixed, the 3, 6 and 9 stones that have 3 stones to be assigned to each other; that is, 3! arrangement possibilities, and the number of circular arrangements involving the remaining 8 stones.

Number of arrangements in the clock = 3! * [(8–1) P (8-1)] = 3! * (8–1)!

Number of arrangements in the clock = (3 * 2 * 1) (8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)

Number of arrangements on the clock = 241920 different shapes

**– Exercise 2**

The steering committee of a company consists of 8 members and they meet at an oval table.

a) How many different forms of arrangement around the table does the committee have?

b) Suppose the chairman sits at the head of the table in any committee arrangement, how many different forms of arrangement does the rest of the committee have?

c) Suppose the vice president and the secretary sit on either side of the president in any committee arrangement. How many different forms of arrangement does the rest of the committee have?

**Solutions**

a) We want to find the number of different ways to arrange the 12 members of the committee around the oval table.

N ° of committee arrangements = (12 – 1) P (12 – 1) = (12 – 1)!

N ° of committee arrangements = 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

N ° of committee arrangements = 39976800 different forms

b) Since the committee chair is located in a fixed position, the number of ways to order the remaining 11 committee members around the oval table is sought.

N ° of committee arrangements = (11 – 1) P (11 – 1) = (11 – 1)!

N ° of committee arrangements = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

N ° of committee arrangements = 3,628,800 different forms

c) The president is located in a fixed position and to the sides are the vice president and the secretary with two possibilities of arrangement: vice president on the right and secretary on the left or vice president on the left and secretary on the right. Then you want to find the number of different ways of ordering the remaining 9 members of the committee around the oval table and multiply by the 2 forms of arrangements that the vice president and the secretary have.

N ° of committee arrangements = 2 * [(9–1) P (9-1)] = 2 * [(9–1)!]

N ° of committee arrangements = 2 * (8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)

N ° of committee arrangements = 80640 different forms

__References__

__References__

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