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Permutations
Permutations are different ways of arranging objects in a definite order. It can also be expressed as the rearrangement of items in a linear order of an already ordered set. The symbol \(^n{P_r}\) is used to denote the number of permutations of n distinct objects, taken r at a time. It locks schedules of buses, trains or flights, allocation of zip codes and phone numbers. These are a few situations where permutations are used.
Permute means to position. Let us learn more about permutations along with a few solved examples.
What are Permutations?
A permutation is an ordered arrangement of outcomes and an ordered combination. For example, there are 5 chairs and 3 persons are to be seated. We have 5 ways to seat the first person; 4 ways to seat the next person and 3 ways to seat the third person. Thus, to find the number of ways for arranging 3 persons in 5 chairs, we multiply the options available to us. We do it in 5 × 4 × 3 ways. i.e., it can be done in 60 ways. Observe that 5 × 4 × 3 can be written as (5!) / (2!) (or) (5!) / (5  3)!.
Generalizing this, we get n options to fill the first chair, n1 options to fill the second and n2 options to fill the third chair. Thus, the total number of permutations (arrangements) of r people in n chairs can be expressed as: ^{n}P_{r} = n! / (n  r)!.
This is called the permutations formula. We can see the detailed proof of this formula by clicking here.
Permutations Formulas
We have already seen the basic permutations formula in the previous section. Here are different permutations formulas that are used in different scenarios.
 The number of permutations (arrangements) of 'n' different things out of which 'r' things are taken at a time and where the repetition is not allowed is given by the formula: ^{n}P_{r} = n! / (n  r)!.
 Using the above formula, the total number of ways of arranging n different things (taking all at a time) is n! (this is because ^{n}P_{n} = n! / (n  n)! = n!/0! = n!/1 = n!).
 The circular permutation formula says, the number of ways of arranging 'n' things in the circular shape is (n1)!.
 The number of permutations of 'n' things (where all are not different), among which there are 'r_{1}' objects are of one type, 'r_{2}' objects belong to the second type, ... , 'r_{n}' objects belong to the n^{th} type is n! / (r_{1}! × r_{2}! × ... × r_{n}!).
 The number of permutations of 'n' things out of which 'r' things are taken and where the repetition is allowed is given by the formula: n^{r}.
How to Calculate Permutations?
Permutations can be calculated with or without repetitions. If there are 10 pairs of socks and you choose 2 pairs out of them, then you do it in ^{10}P_{2 } ways, then
 if we don't have repetitions. ^{n}P_{r} = 10! / (102)! = 10! / 8! = 90 ways.
 if we have repetitions, we always have n arrangements every time. We have 10^{2 }ways = 100 ways.
Another important concept that should be understood in permutations is the factorial of numbers. Factorials are the product of the first n positive integers. For example, the factorial of 8 can be expressed as: 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1.
Examples on Permutations As Arrangements
Here are some applications of permutations in real life scenarios.
Example 1: (a) How many words can be formed using the letters of the word TRIANGLES? (b) How many of these words start with T and end with S?
Solution:
(a) There are 9 distinct letters in the given word. Thus, the number of different permutations (or arrangements) of the letters of this word is ^{9}P_{9} = 9!.
(b) If we fix T at the start and S at the end of the word, we have to permute 7 distinct letters in 7 places. This can be done in ^{7}P_{7} = 7! ways. Thus, the number of such words is 7!
Example 2: How many different 5letter words can be formed using the letters from A to J (total ten letters) such that each word has at least one letter repeated?
Solution:
The number of letters from A to J is 10.
We need to count:
(i) The total number of 5letter words (with or without repetition) which can be formed using 10 distinct letters. This number is 10^{5}, since each place in the 5letter word can be filled in 10 different ways, and so, the required number is 10 × 10 × 10 × 10 × 10 = 10^{5}.
(ii) The number of 5letter words which can be formed using 10 distinct digits, such that no digit is repeated. This number is ^{10}P_{5}.
The answer to the original question is the difference of these two numbers. Now, from the set of all the words, if we take out those words which have no repetition of letters, then you get the set of words which have at least one letter repeated. Thus, the final answer is 10^{5}  ^{10}P_{5}.
Example 3: There are 10 students in a class. Two of these students are Jack and Daniel, who don’t get along very well. In how many ways can the teacher arrange the students in a row, so that Jack and Daniel are not together?
Solution:
The total number of arrangements of the 10 students is ^{10}P_{10} = 10!.
Now, let us count the number of arrangements in which Jack and Daniel are together. Treating Jack + Daniel as a single unit (call it JD), we have a total of 9 entities which we can permute: the 8 students, and JD. These 9 entities can be permuted in ^{9}P_{9}= 9! ways. But for each of 9! permutations, J and D can be permuted among themselves in 2! or 2 ways: JD or DJ. Thus, the total number of permutations in which Jack and Daniel are together, is 2 × 9!
We conclude that the number of permutations in which Jack and Daniel are not together is 10!  (2 × 9!)
Example 4: A permutation lock will open if the right choice of 3 numbers (from 1 to 50) is selected. How many lock permutations can be made assuming no number is repeated?
Solution:
We have 50 digits out of which we arrange 3 digits. We have the possibility of^{ 50}P_{3} ways.
^{6}P_{3 }= 50! / (503)!
= 50! / (47!)
= (50 × 49 × 48 × 47!) / 47!
= 50 × 49 × 48
= 117,600
Important Notes on Permutations
 Permutations are ordered combinations of objects that can be done with or without repetitions.
 They are calculated by the formula: ^{n}P_{r} = n! / (n  r)!, where n different things are taken r at a time.
 When the items are to be arranged or ordered or positioned, then we consider permutations.
 The combinations formula is also referred as ncr formula.
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Solved Examples on Permutations

Example 1: Find the number of words, with or without meaning, that can be formed with the letters of the word "PARK".
Solution:The number of letters in the word PARK is 4.
Thus, the number of words that can be arranged with these 4 letters
= The number of permutations with 4 words
= 4! words
= 4 × 3 × 2 × 1
= 24 words
Answer: 24 words 
Example 2: How many 6digit codes could be generated from the digits 0 to 9 (a) if repetition is allowed, and (b) if repetition is not allowed?
Solution:
Since the arrangement matters, we use permutations.
(a) If repetition is allowed we always have the choice of 10 digits every time. Thus, we could have 10^{6 }codes generated. We could have 1,000,000 codes.
(b) If repetition is not allowed, we have ^{10}P_{6 } codes.
^{10}P_{6 } = 10! / (10  6)!
= 10! / 4!
= (10 × 9 × 8 × 7 × 6 × 5 × 4!) / 4!
= 10 × 9 × 8 × 7 × 6 × 5
= 151,200 codes
Answer: (a) 1,000,000 codes (b) 151,200 codes 
Example 3: Find the number of different words that can be formed with the letters of the word ‘TREAT’ so that the vowels are always together using permutations.
Solution:
The number of letters in the word is 7. The vowels E and A should occur together. Thus have EA as a unit.
Now we should arrange T R T (EA). Now we have 4 units to be arranged, which can be done in 4! ways.
The letter T is repeated twice, We have 2! ways in which T can be arranged. We get 4! / 2! ways in arranging T R T (EA) = 12 ways. The arrangement of the letters EA can be done in 2! ways. Hence, the total number of ways in which the letters of the ‘TREAT’ can be arranged such that vowels are always together are (12 × 2!) = 24 ways
Answer: 24 ways
Frequently Asked Questions (FAQs)
What is the Meaning of Permutations?
The meaning of the word "permute" is "to position". Hence, permutations mean the arrangements.
What is the Mathematical Definition of Permutations?
The number of ways of selecting and arranging 'r' things out of 'n' things is called the number of permutations. Wherever "arrangement" has importance, we have to use the permutations there. For example:
 The number of ways of forming 5 letter words in which repetitions are allowed from the letters a, t, y, u, r, c, and p.
 The number of 3 digit numbers in which repetition of digits is not allowed using the digits 1, 2, and 3.
What is the Permutation Without Repetition?
The formula used for permutation without repetition is: nPr = \(\dfrac{{n!}}{{(n  r)!}}\) , where 'r' different things are taken from 'n' different things and repetition is not allowed. Here "n!" means "the factorial of n". We can see how to calculate factorials by clicking here.
What is the Permutation With Repetition?
The formula used for permutation with repetition is: n^{r}, where 'r' different things are taken from 'n' different things and repetition is allowed.
What are the Different Types of Permutations?
There are 3 types of permutations.
 Arrangement of n different objects where repetition is not allowed,
 Arrangement of n different objects where repetition is allowed.
 Permutation of multisets.
What is the Difference Between Permutation and Combination?
Permutation means "both selection and arrangement" whereas combination means just the "selection". Wherever the ordering of objects does not matter, we use combinations there. For a more detailed understanding of the difference between permutation and combination, click here.
What is the Permutation of 6?
The permutation of 6 (which meaning arranging 6 different things among themselves) is 6 P 6 = 6! = 6 × 5 × 4 × 3 × 2 × 1= 720.
Is nPr and nCr the Same?
nPr is calculating the permutations as arrangements where the order matters, whereas, nCr is calculating the combinations, where the order doesn't matter.
What is Circular Permutation Formula?
The circular permutation formula gives the number of ways of arranging 'n' different things around a circle to be (n  1)!.
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