Examples  Permutations as Arrangements
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. Permute means to position. Permutation locks, schedules of buses, trains or flights, allocation of zip codes and phone numbers are a few situations where permutations are used.
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, for getting 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. On analyzing, we find that this could be done in 5! × 4 ! × 3! = 60 ways.
On generalization, 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 arrangements of r people in n chairs can be expressed as: ^{n}P_{r} = \(\dfrac{{n!}}{{(n  r)!}}\)
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}n_{2 } ways, if we don't have repetitions. ^{n}P_{r} = \(\dfrac{10!}{(10  2)!}\) = \(\dfrac{10!}{8!}\) = 90 ways. If we have repetitions, we always have n arrangements every time. We have 10^{2 }ways = 100 ways.
Another important concept which should be understood in permutations is the factorial of the 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
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: 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 \(10 \times 10 \times 10 \times 10 \times 10 = {10^5}\).
(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 possibilities of^{ 50}P_{3} ways.
^{6}P_{3 }= \(\dfrac{{50!}}{{(50  3)!}}\) = 117,600
Important Notes
 Permutations are ordered combinations of objects that can be done with or without repetitions.
 Permutations are calculated by the formula: ^{n}P_{r} = \(\dfrac{{n!}}{{(n  r)!}}\), where n different things are taken r at at time.
 When the items are to be arranged or ordered or positioned, then we consider permutations.
Related topics on Permutations
Solved Examples on Permutations

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 = 4! words = 4 × 3 × 2 × 1 = 24 words 
2. How many 6digit codes could be generated from the digits 0 to 9 if repetition is allowed and if repetition is not allowed?
Solution:
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.
If repetition is not allowed, we have ^{10}n_{6 } codes. ^{10}n_{6 } = \(\dfrac{{10!}}{{(10  6)!}}\) = \(\dfrac{{10!}}{{4!}}\) = 151,200 codes 
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.
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 letters 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
Frequently Asked Questions (FAQs)
What is the Permutation Formula?
The formula used for permutation is: nPr = \(\dfrac{{n!}}{{(n  r)!}}\) , where the number of objects are taken at a time for the arrangements required.
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 multi sets.
In How Many Ways Can you Order 3 things?
3 things can be ordered or arranged in 6 ways. (3! = 3 × 2 × 1)
What is the Permutation of 6?
The permutation of 6 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.