Arrangement of Different Objects - Theorems, Formulas and Examples for JEE

The arrangement of different objects in mathematics essentially refers to how you can arrange a set of numbers, integers, or values. The formal term used to refer to this phenomenon is ‘permutation.’ Normally, permutation refers to a pattern or an order given to an arrangement, especially in cases where demand exists. In instances where integers or whole numbers are already in ordered pairs, the permutation refers to a rearrangement of sorts. 

Some real time examples of permutation would be the rearrangements of the letters of an anagram, where the same letters are repositioned to form a completely different word with a different meaning. For example, if we consider the set {1,2,3}, the numbers in the set can be arranged in 6 different ways: {1,2,3}, {1,3,2}, {2,1,3}, {2,3,1}, {3,1,2} and { 3,2,1}. For ‘n’ distinct objects, the number of permutations possible is n!

Theorems Related to the Concept of Permutation in Mathematics

The critical point you should remember when it comes to permutation is the order of the arrangement mentioned. There are four fundamental theorems in the permutation field that are widely implemented and extensively used. Read on to find out the theorems and their respective permutation formulas. 

Theorem 1: For n distinct, the number of arrangements possible is n!.

Example 1: If there are 5 different objects, in how many ways can they be arranged?

Solution: If there are 5 distinct objects, they can be arranged in 5 ways! Therefore, the required number of arrangements of 5 different objects = 5! =120.

Theorem 2: The number of permutations of n distinct objects, when taken r at a time, is denoted by the expression $P_{r}^{n}$( in cases where there are no repetitions in the arrangement). In some textbooks, $P_{r}^{n}$ is also denoted as nPr. Despite the difference in notations, it means one and the same thing.

The mathematical representation is : $P_{r}^{n}=n(n-1)(n-2) \ldots(n-r+1)=\dfrac{n !}{(n-r) !}$

Example 2: If there are 6 items to be arranged into 3 different bags, in how many permutations can it be done?

Solution: As per the theorem, if n different objects have to be arranged in r different ways, then, the number of possible arrangements = $P_{r}^{n}=\dfrac{n !}{(n-r) !}$

Here n = 6 and r = 3. 

Therefore, the required number of arrangements = $P_{3}^{6}=\dfrac{6 !}{(6-3) !}=\dfrac{6.5 \cdot 4 \cdot 3 !}{3 !}=120$

Thus, 6 items can be arranged in 3 bags in 120 different ways. 

Theorem 3: The number of possible permutations or arrangements when each distinct object is of a different kind i.e. for nk items , n₁ is of one type, n₂ is of the seconds type. . . nk is of the k th type, is given by the formula $\dfrac{n !}{n_{1} ! \times n_{2}! \times \dots \times n_{k} !}$.

Example 3: Let there be 5 balls: two red, one black and one white. In how many possible numbers of permutations can these five balls be arranged?

Solution: As you can see from the example, there are 5 distinct objects that have 3 different types.

Therefore, the number of ways in which these five objects can be arranged is $=\dfrac{5 !}{2 ! \times 1 ! \times 1 !}=60$

Theorem 4: The fourth fundamental theorem deals with the concept of cyclic permutation and has two clauses:

1. If the clockwise and the anti-clockwise orders are different, then the total number of possible circular permutations is given by $P_{n}=(n-1) !$

2. When the clockwise and the anticlockwise orders of arrangement are not considered to be different, the total number of circular permutations is given by $P_{n}=\dfrac{(n-1) !}{2 !}$

Circular Permutation

As mentioned before, the fourth theorem of permutation deals with cyclic permutation. Just like linear permutation, circular permutation deals with the number of ways certain objects can be arranged. If they are already in an ordered pair, then they are rearranged around a circle.

There can be two distinct cases of circular permutation. When the clockwise and the anticlockwise orders are different, the number of ways in which they can be arranged in a circle is given by $P_{n}=(n-1) !$. When the arrangement of items in the clockwise and anticlockwise directions are the same, then the circular arrangement formula is given by  $P_{n}=\dfrac{(n-1) !}{2 !}$. 

In the case of cyclic permutation, there is no beginning point, just like there is no ending point. One item remains fixed in this type of arrangement, while the others are arranged or rearranged in relation to this fixed object. There can be many circular permutation examples, one of which is given below. 

Example 4: In a playground, 3 sisters and 5 other girls are seated. In how many ways can the girls be seated so that the 3 sisters are not seated together?

Solution: 

Number of sisters = 3

Number of other girls = 5

Total number of girls in the playground= 3 + 5 = 8

Therefore, the number of ways to arrange the girls in a circular manner = (8-1)!=7!

The number of ways the three sisters can rearrange themselves= 3!

Therefore, the number of ways the sisters are always sitting together = 3!$\cdot$5!

Thus, the required number of ways in which the arrangement occurs where none of the sisters is sitting together= 7! - ( 3!$\cdot$5!)= 4320 

Conclusion

A permutation is a mathematical concept of arranging a bunch of different and distinct objects in a particular pattern or if they are already in order, then rearranging that order.  We can arrange n distinct objects into n! ways. If n objects are to be arranged in r different ways then it is represented as  $P_{r}^{n}$. A cyclic permutation is the field of mathematics that deals with arranging a number of different objects, distinctly around a circle under two conditions, the arrangement of both the clockwise and anticlockwise directions are the same or different.