Permutation: Definition, Formula, Types & Examples

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Permutation is a method of elements or objects in a defined sequence or series. Permutation and Combinations are integral concepts in Mathematics.

It refers to the rearrangement of items in a linear order of an Ordered Set.
Permutation is denoted by the symbol ⁿP_r.
The sequence of arrangement matters in permutation, i.e. the elements should be arranged in a particular order.

Example: Consider a Set A = {2, 6}. Here, the Permutations will be 2 as the elements of the given set can be arranged in only two ways which are {2, 6}, and {6, 2}. Apart from these two arrangements, there are no other ways to arrange the elements of set A.

Table of Content

What is Permutation?
Permutation Formula
Types of Permutation
Permutation Solved Examples
Difference Between Permutation and Combination
Things to Remember
Previous Years’ Questions
Sample Questions

Key Terms: Permutation, Combination, Factorial, Permutation Formula, Arrangement, Set, Elements, Permutations and Combinations

What is Permutation?

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Permutation is an arrangement of objects or items in a specific way or order. Permutation involves arranging the elements of a given set into all possible arrangements.

It rearranges the elements to find the number of all possible arrangements.
It is used for data sets where the sequence or order of data matters.
Permutation involves both selections as well as arrangement.
Thus, it can be referred to as an Ordered Combination.
It is generally denoted by the symbol ⁿP_r.

Example: Assume Set B = {1, 5}. Now, the Permutation will be 2 as there are only two possible arrangements of the elements of the given set: {5, 1}, and {1, 5}.

What is Permutation?

Representation of Permutation

Permutation can be represented in numerous ways as follows:

\(\begin{array}{l}{P(n,k)}\end{array}\)
\(\begin{array}{l} {P^{n}_{k}}\end{array}\)
\(\begin{array}{l}{_{n}P_{k}}\end{array}\)
\(\begin{array}{l}{^{n}P_{k}}\end{array}\)
\({P _{n}{,k}}\)

Permutations and Combinations Detailed Video Explanation

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Relevant Concepts
Combinatorics	Factorial Formula	Calculus
Set Theory in Maths	Fundamental Theorem of Arithmetic	Sets Formula

Permutation Formula

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Permutation Formula is used when the given data sets cannot be arranged manually. Permutation Formula is used to arrange data in sequences.

Permutation Formula is given as

\({^nP_r} = \frac{n!}{(n-r)!} \)

Where

n: Total number of objects in a Set.
r: Number of objects selected from the Set.

Solved Example

Example: What will be the number of 3-letter words that can be formed from the letters a, b, c, d, and e in which the letters should not be repeated?

Solution: Given that,

Number of Letters, n = 5
Number of Letters in Each Word is, r = 3

Using the Permutation Formula,

\({^nP_r} = \frac{n!}{(n-r)!} \)

Substituting the values, we get

⁵P₃ = (5!) / (5 - 3)! = 5! / 2! = 120/2 = 60

Thus, there are 60 three-letter words that can be formed from the letters a, b, c, d, and e with no repetition of letters.

Types of Permutation

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Permutation can be categorized into three major types:

Permutation of n Different Objects (No Repetition)
Permutation of n Different Objects (Repetition Allowed)
Permutation when Objects are not Distinct (Permutation of Multisets)

Permutation of n Different Objects (No Repetition)

Assume that 'n' is a positive integer, and ‘r’ is a whole number, where r is less than n. Then, Permutation or arrangements of 'n' different things when taken 'r' at a time is represented by P(n,r). In this case, it is without repetition, so it can be derived as follows:

P(n, r) = n(n-1)(n-2)(n-3)……..up to r factors

This gives us P(n, r) = n(n-1) (n-2) (n-3)……..(n – r +1)

\( P(n,r) = \frac{n!}{(n-r)!}\)

Where “ⁿP_r” represents the “n” objects to be chosen from “r” objects without repetition, in which the sequence or order matters.

Solved Example

Example: How many three-letter words with or without meaning can be formed out of the letters of the word SWING with no repetition of letters?

Solution: Given that,

Number of Letters in Word SWING, n = 5
Number of Letters in Each Word is, r = 3

Using the Permutation Formula,

\({^nP_r} = \frac{n!}{(n-r)!} \)

Substituting the values, we get

⁵P₃ = (5!) / (5 - 3)! = 5! / 2! = 120/2 = 60

So, there is a possibility of 60 different permutations to form a 3-letter word from the word SWING.

Permutation of n Different Objects (Repetition Allowed)

Permutation can easily be calculated with repetition. Permutation with repetition of items may be written using the exponent form. If the number of objects is “n,” and “r” is the selection of an object, then, the object can be chosen in n different ways (each time).

Therefore, the permutation of objects when repetition is allowed will be equal to

n x n x n x ……(r times) = ( n^r)

This is the formula to calculate the number of permutations for the selection of “r” objects from the “n” objects when repetition is allowed.

Solved Example

Example: How many three-letter words with or without meaning can be formed out of the letters of the word SWING with while the repetition of letters is allowed?

Solution: Given that,

Number of Letters in Word SWING, n = 5
Number of Letters in Each Word is, r = 3

We need to create three-letter phrases with or without meaning with repetition being allowed, thus,

P (n, r) = n^r

P (5, 3) = 5³ = 5 x 5 x 5 = 125

Thus, there are 125 three-letter words that can be formed using the given five-lettered words with repetitions allowed.

Permutation of Multisets

Permutation of n different things when P₁ objects among ‘n’ things are similar, P₂ things of the second type are similar, P₃things of the third type are similar ……… and so on, P_k things of the kth kind are similar and the remaining of all are of a different kind, it forms a multiset.

Permutation of Multisets is given as

\( \frac{n!}{p_{1}!\; p_{2}!\; p_{3}…..p_{n}!}\)

Permutation Solved Examples

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Here are a few solved examples on Permutations for better concept clarity:

Example 1: A lock will open only if the right choice of 3 numbers (from 1 to 50) is selected. How many permutations can be made assuming no number is repeated?

Solution: It is given that there are 50 digits out of which 3 digits need to be arranged. We have the possibility of⁵⁰P₃ ways.

Using the Permutation Formula,

\({^nP_r} = \frac{n!}{(n-r)!} \)

Substituting the values,

⁵⁰P₃= 50! / (50-3)!
= 50! / (47!)
= (50 × 49 × 48 × 47!) / 47!
= 50 × 49 × 48
= 117,600

Thus, there is 117,600 possible permutations for the password of the lock.

Example 2: Determine the number of 3-letter words that can be formed from the letters a, b, c, d, and e in which the letters are allowed to be repeated.

Solution: Given that,

Number of Letters in Given, n = 5
Number of Letters in Each Word is, r = 3

As the repetition of letters is allowed, we will use the formula:

P (n, r) = n^r

5³ = 125

Thus, there are 125 three-letter words that can be created from the letters a, b, c, d, and e in which the letters are allowed to be repeated.

Fundamental Counting Principle

Fundamental Counting Principle states that

“If one operation can be performed in ‘m’ ways and there are n number of ways of performing a second operation, then the number of ways of performing the two operations together will be m x n.”

It can be extended to the case in which the different operations are be performed in m, n, p, . . . . . . ways.
In such scenario, the number of ways of performing all the operations one after the other is m x n x p x . . . . . . . . and so on

Difference between Permutation and Combination

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Permutations and Combinations are two integral concepts in Mathematics that are frequently used in the principle of counting.

Permutation is used to calculate the number of all possible arrangements of given elements.
Combination is a used to calculate the number of subgroups that may be created from a larger set.
Sequence of arrangement of elements matters in Permutations while it doesn’t matter in Combinations.

Given below is the difference between permutation and combination:

Permutation	Combination
Permutation is a defined as a method of arranging members of a set in a defined sequence or series.	Combination is a method of combining data from other larger sets to form a subset.
Sequence or order of arrangement of elements matters in Permutation.	Sequence or order of arrangement of elements doesn’t matter in Permutation.
Permutation is denoted by the symbol ⁿP_r.	Combination is denoted by the symbol ⁿC_r.
Permutation Formula is \({^nP_r} = \frac{n!}{(n-r)!} \).	Combination Formula is \({^nC_r}= \frac{n!}{r!(n-r)!} \).
Example: Arranging things, people, digits, alphabets, colors, etc.	Example: Selecting menu items, food, clothes, subjects, etc.

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Related Topics
Set Theory Symbols	Probability and Statistics	Set Operations
Geometry	Types of Sets	Number Systems

Things to Remember

Permutation is the arrangement of elements of a set in particular order or sequence.
It is denoted by ⁿP_rand is used for a list of data where the order of data matters.
Permutation of n Different Objects where Repeating is not allowed is \({^nP_r} = \frac{n!}{(n-r)!} \).
Permutation of n Different Objects where Repeating is allowed is P (n, r) = n^r
Permutation of Multisets is \( \frac{n!}{p_{1}!\; p_{2}!\; p_{3}…..p_{n}!}\).
Combination is the method of combining data from other larger sets to form a subset.
Sequence or Order of arrangement matters in Permutations, unlike Combination.

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Previous Years’ Questions

Sample Questions

Ques. What is Permutation? (3 Marks)

Ans. Permutation is a method to determine how many times can data be arranged in different sequences.

The arrangements are made in sequence or order in permutations.
It is represented by the symbol ⁿP_r.
Examples include arranging alphabets, words, numbers, etc.

Ques. How many words can be formed by using 3 letters from the word “DELHI”? (3 Marks)

Ans. The word “DELHI” has 5 different letters.

Therefore, the required number of words will be ⁵P₃.

Using the Permutation Formula,

\({^nP_r} = \frac{n!}{(n-r)!} \)

⁵P₃ = 5! / (5 – 3)!

= 5! / 2! = 120 / 2 = 60 Words

Ques. Determine how many 4-digit numbers can be formed with the digits 1 to 9 when no repetition is allowed. (1 Mark)

Ans. To fill nine digits in the place of a four-digit number, the order must be relevant. Here, the 4-digit numbers can be as many as there are permutations of 9 digits picked 4 at a time. Therefore, the four-digit number will be 9! / 4! x 2! = 7560.

Ques. What is Permutation Formula? (3 Marks)

Ans. Permutation Formula is given as

\({^nP_r} = \frac{n!}{(n-r)!} \)

Where

n is the total number of objects in a given set.
r is the number of objects to be chosen from the set.

Ques. How many words can be formed by using the letters from the word “DRIVER” such that all the vowels are always together? (3 Marks)

Ans. In these types of questions, we assume all the vowels to be a single character, i.e., “IE” is a single character.

So, now we have a total of 5 characters in the word, namely, D, R, V, R, I, E.

But, R occurs 2 times.

Number of Possible Arrangements = 5! / 2! = 60

Now, the two vowels can be arranged in 2! = 2 ways.

Total number of possible words such that the vowels are always together= 60 x 2 = 120

Ques. In how many ways, can we select a team of 4 students from a given choice of 15? (2 Marks)

Ans. Number of Possible Ways of Selection = ¹⁵C₄ = 15! / [(4!) x (11!)]

= (15 x 14 x 13 x 12) / (4 x 3 x 2 x 1) = 1365

Thus, there are 1365 possible ways of selection.

Ques. The word 'OPTICAL' needs to be arranged such that the vowels always come together. Determine in how many ways can this be done. (3 Marks)

Ans. Given word, ‘OPTICAL’ has 3 vowels: O, I, and A. These vowels can be grouped together to form a single letter, something like CLTP(OIA).

Assuming this to be 5 different letters (with each letter being unique),

Number of Ways = 5! = 5 × 4 × 3 × 2 × 1 = 120

The remaining 3 vowels (OIA) can be arranged among themselves as well.

Number of ways to arrange these vowels among themselves = 3! = 3 × 2 × 1 = 6

Hence, the number of ways of arranging the word according to the question = 120 × 6 = 720.

Ques. In how many ways can a group of 5 members be formed by selecting 3 boys out of 6 and 2 girls out of 5? (3 Marks)

Ans. Number of Ways 3 boys can be selected Out of 6 = ⁶C₃ = 6! / [(3!) x (3!)] = (6 x 5 x 4) / (3 x 2 x 1) = 20

Number of Ways 2 girls can be selected Out of 5 = ⁵C₂ = 5! / [(2!) x (3!)] = (5 x 4) / (2 x 1) = 10

Therefore, Total Number of Ways of forming the Group = 20 x 10 = 200

Ques. How many words can be formed by using the letters from the word “DRIVER” such that all the vowels are never together? (3 Marks)

Ans. We assume all the vowels to be a single character, i.e. “IE” is a single character.

So, now we have a total of 5 characters in the word, namely, D, R, V, R, and IE.

But, R occurs 2 times.

Number of Possible Arrangements = 5! / 2! = 60

Now, the two vowels can be arranged in 2! = 2 ways.

Total number of possible words such that the vowels are always together = 60 x 2 = 120

Also, total number of possible words = 6! / 2! = 720 / 2 = 360

Therefore, the total number of possible words such that the vowels are never together = 360 – 120 = 240

Ques. Determine the number of ways in which the word 'DETAIL' can be arranged such that vowels only occupy the odd positions. (3 Marks)

Ans. 'DETAIL' has 6 letters, out of which there are 3 vowels (EAI) and 3 consonants (DTL).

The 3 vowels(EAI) must occupy only the odd positions.

Let the positions be 1, 2, 3, 4, 5, 6.

The remaining 3 vowels should only occupy positions marked 1/3/5 only.

Hence, number of ways to arrange these vowels = ³P₃ = 3! = 3 × 2 × 1 = 6

There are 3 consonants (DTL) that can be arranged in any order.

Hence, number of ways to arrange these consonants = ³P₃= 3! = 3 × 2 × 1 = 6

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Permutation: Definition, Formula, Types & Examples