Permutations

What are Permutations?

A permutation is an ordered sequence of outcomes. For example, there are 5 seats available, and 3 persons are expected to occupy them. We have five sitting options for the first person, four seating options for the second person, and three seating options for the third person. Thus, we multiply the available choices to get the number of ways to place three individuals on five chairs. We do it in five various ways: five, four, three, and two. That is, there are 60 different ways to do it. It’s worth mentioning that 543 can also be written as (5!) / (2!) or (5!) / (5 – 3)!.

When we generalize this, we obtain n possibilities for the first chair, n-1 options for the second, and n-2 options for the third chair. Thus, the total number of permutations of r individuals in n chairs is nP_r = n! / (n – r)!. This is known as the permutation formula.

nP_r=n!/(n-r)!

Representation of Permutation

Permutation can be represented in a variety of ways, including:

• nP_r

• nP_r

• Pk_n

• P_n,_k

• P_(n,k)

Permutation Formulas

The different permutations formulas are as follows:

• The formula: nPr = n! / (n – r)! gives the number of permutations of ‘n’ different items, when ‘r’ things are selected at a time and repetition is not allowed.

• Using the preceding formula, the total number of ways to arrange n different items is n!.

• According to the circular permutation formula, the number of ways to arrange ‘n’ things in a circular shape is (n-1)!

• The number of permutations of ‘n’ things (where all are the same) in which ‘r1’ items are of one type, ‘r2’ objects are of the second type,…, ‘rn’ objects of the nth type is n! / (r₁! r₂!… r_n!).

• The formula: nr gives the number of permutations of ‘n’ items from which ‘r’ things are chosen and when repetition is permitted.

Steps to Calculate Permutations

Permutations can be computed both with and without repetitions. If there are 10 pairs of socks and you choose two of them, then you do it 10P2 times.

• Without repetition: nP_r= 10! / (10-2)! = 10! / 8! = 90 ways.

• With repetition: 10²=100 ways.

The factorial of numbers is another key topic to understand in permutations. The product of the first n positive integers is a factorial. For instance, the factorial of 8 can be written as 8! = 87654321.

Types of Permutations

1. Repetition

The permutation with repetition is simple to compute. The exponent form can be used to write the permutation with object repetition. At the point when the quantity of articles is “n,” and “r” is the item determination, then, at that point, choosing an item should to be sure be possible in an assortment of ways. Thus, when recurrence is permitted, the permutation of items will be equal to,

 nnn…… = nr

When repetition is allowed, this is the permutation formula for calculating the number of permutations possible for the selection of “r” items from the “n” objects.

Example: There are ten numbers to choose from (0,1,2,3,4,5,6,7,8,9), and we pick three.

Solution: 10 × 10 × … (3 times) = 10³ = 1,000 permutations

nr= when n is the number of objects to choose from and we select r of them, repetition is permitted, and order is important.

2. Without Repetition

P(n, r) indicates the number of all possible arrangements or permutations of n unique items taken r at a time if n is a positive integer and r is a whole number. In the circumstance of permutation without repetition, the number of possible options decreases with each iteration. It can also be written as nPr

P(n,r)=n!/(n-r)!

Example: Count of 3 out of 16 pool balls.

Solution: 16P₃=16!16-3!=16!13!=20,922,789,888,0006,227,20,800=3,360

nPr= n objects to be selected from r objects, repetition is not permitted, and order is important.

3. Multi-sets

Permutations of n various articles when p1 objects among ‘n’ objects are comparable, p2 objects of the subsequent kind are comparable, p3 objects of the third kind are comparable … … …, etc, pk objects of the kth kind are comparable.

As a result, multiset is formed, with the permutation mentioned as

n!p1!p2!p3……pn!

Permutation Examples

1. A class consists of ten individuals. Jack and Daniel are two of these students that do not get along. How many different ways can the teacher arrange the pupils in a row so that Jack and Daniel are not in the same row?

Solution:

The total count of arrangements made by the ten children is 10P10 = 10! Count the number of configurations in which Jack and Daniel are present. If we treat Jack and Daniel as a single entity (let’s call it JD), we have a total of nine entities that we can permute: the eight students and JD. These nine entities can be permuted in 9P9= 9! different ways. J and D, on the other hand, can be permuted among themselves in two! or two ways for each of the nine! permutations: JD or DJ. As a result, the total number of permutations in which Jack and Daniel are present is 2 9!

We infer that there are 10 permutations in which Jack and Daniel are not together 10! – (2 9!)

2. If the correct combination of three numbers (from 1 to 50) is chosen, a permutation lock will be opened. How many lock variations are possible if no number is repeated?

Solution:

We have 50 digits, of which we arrange three. We have the option of 50P3 methods.

6P3=50! / (50-3)!

= 50! / (47!)

= (50 × 49 × 48 × 47!) / 47!

= 50 × 49 × 48

= 117,600

Conclusion

Permutations are several ways of arranging objects in a specific order. It can alternatively be defined as the rearrangement of items in a previously ordered set in a linear order. It secures bus, train, and airplane timetables, as well as the assignment of zip codes and phone numbers. Permutations are employed in the following situations.