# Permutation and Combination- Formula, Difference, Examples, Theorem

SOURCE: Don’t Memorise

**Pe rmutation and Combination** is a mathematical formula used in various ways in which a set of objects are selected and are formed

**into a subset.**

This **arranging of subsets** is known as* permutation** *when the order is the main factor and ** combination **when the

**order is not a factor.**

**Permutation and Combination Formula**

The formula of permutation and combination has **‘r’ as an element **out of **‘n’ number of elements **which is total elements**. ‘P’ and ‘C’ are permutation and combination **of the objects.

Permutation and combination deal with the counting and arrangement of a particular set of data. Factorial (*n*!): It is the product of all positive integers less than or equal to *n. *Example: 4! = 4 × 3 × 2 × 1 = 24.

**Note:** 0! = 1

**Permutation** refers to the act of arranging all the elements of a set into a sequence or order. It is denoted by ^{n}P_{r}. A permutation is the choice of *r* things from a set of *n* things without replacement and where the order matters.** **

^{n}**P**_{r}** = (n!) / (n-r)!**

**Example**: Arrange the given 3 numbers 1, 2, 3, taking two at a time. The numbers can be arranged in 6 ways: (12, 21, 13, 31, 23, 32).**Note:** 12 and 21, 13 and 31 or 23 and 32 do not mean the same, because the order of numbers is important.

**Formulas for ****Permutation and combination**** is as follows:**

#### Permutation formula:^{n }**P **_{r}** = n! / (n – r)!**

^{n }

_{r}

**Combination formula:**^{n}** C **_{r}** = n! / [(r !)(n – r)!]**

**Permutation Formula**

** Permutation **arranges all the objects accordingly while

**subsets**out of it. There is one easy and tricky way to get the number of letters for which the

**permutation**is required and it is

^{n }

**P**

_{n}

**= n!**The required

**formula**is mentioned below

^{n }

**P**

_{r}

**= n! / (n – r)!.**To understand more about

*Permutation,*go through Brett Berry’s post.

**Non-Circular Permutation**

**Arrangement of ‘****n****‘ distinct objects:**

**n**

**Example:** How many different ways can the letters of the word PRISM be arranged? **Solution:** There are 5 distinct letters. The total number of arrangements is 5! = 120

#### Arrangement of ‘*n*‘ objects in which some elements are repeated:

**Example**: How many different ways can the letters of the word INDIA be arranged?**Solution**: There are 5 letters, 2 among them are repeated. To avoid duplication in the counting, divide the total number of arrangements by 2!. The total number of arrangements is 5! / 2! = 120/2 = 60

#### Arrangement of ‘*n*‘ objects on a conditional basis:

**Example**: How many different ways can the letters from the word LONG be re-arranged so that the word starts with a vowel?**Solution**: Out of the 4 distinct letters, only 1 letter (O) is a vowel. The first letter can be arranged in only one way and the remaining 3 places can be arranged in 3! ways. The total number of arrangements is 1 x 3! = 6

**Cyclic Permutation**

Permutation around a circular object is done by fixing one object and permuting the remaining objects.

**Example**: How many different ways can 6 students be seated around a circular table? **Solution**: The number of circular permutations of *n *different items taken all at a time is (n – 1)! = (6 – 1)! = 120

**Combination Formula**

The** combination** is different from

**and it’s formula remains as it is i.e.**

*Permutation*^{n}

**C**

_{r}

**= n! / [(r !)(n – r)!]**A combination is the choice of

*r*things from a set of

*n*things without replacement and where order does not matter.

^{n}**C**_{r}** = n!/ r! × (n-r)!**

**Example**: If we have to select two balls out of 3 balls, X, Y, Z, then find the number of combinations possible.**Solution**:Only two balls are to be selected and arranged. Therefore, this is possible in 3 different ways: (XY, YZ, XZ,).

**Note:** XY and YX are the same combinations.

**Difference between Permutation and Combination**

Permutation | Combination |

The order of elements is taken into consideration | In combination, the order does not matter |

^{n }P _{r} = n! / (n – r)! | ^{n} C _{r} = n! / [(r !)(n – r)!] |

There are different ways in which a collection of elements can be arranged | Whereas in combination wecannot |

**Permutation and Combination Questions**

There are various questions solved in this particular chapter and a **student needs to learn the trickest way to solve **it for various entrances and competitive exams.

**There are some various questions for the students with the solution:**

**Q. There are 7 consonants and 4 vowels. How many words of 3 consonants and 2 vowels can be formed?**

**Ans**. Number of ways of selecting **3 consonants from 7= **^{7 }**C**_{3}

Number of ways of selecting** 2 vowels from 4= **^{4}**C**_{2 }** ****The total number of ways is **^{7 }**C**_{3 }**x **^{4}**C**_{2}** = 210**

Number of ways of arranging **5 letters is 5! =120**

Required number of ways is **210 x 120 = 25200**

**Q. In how many of the ways can a group of 5 men and 2 women be made out of 7 men and 3 women?**

**Ans**. Number of ways of selecting 5 men from** 7 = **^{7}**C**_{5}** **

Number of ways of selecting 2 women from** 3 = **^{3}**C**_{2}

Required number of ways is by multiplying both the solutions using** **^{n}**C**_{r}** = **^{n}**C**_{(n-r)}** = 63**

**Q. How many 3 letter words with and without meaning can be formed out of the letters of the word BACKGROUND if repetition of letters is not allowed?**

**Ans**. The word has** 10 different letters**

The number of 3 letter words with and without meaning formed by using these letters **= **^{10}**P**_{3}** = 10 x 9 x 8 = 720**

## Theorem of Counting

- Rule of Addition: If the first task is performed in
*x*ways and the second task is performed in*y*ways, then either of the two can be performed in**(****x + y****) ways.** - Rule of Multiplication: If the first task is performed in
*x*ways and the second task is performed in*y*ways, then both of the two can be performed in**(****x × y****) ways.**

**Check Out: **

**Profit and Loss Formula****Percentage Formula****Probability Formulas****Mensuration Formulas****Combination Formula**

**FAQ**

**✅**What is the difference between combination and permutation?**Ans. **1. In **permutation**, the order of elements is taken into consideration while in **combination**, the order does not matter.

2. In permutation, there are different ways in which a collection of **elements **can be arranged whereas in **combination** we cannot.**Example**: If your password is 1234 and if you enter 4321 as your password, it won’t open because the order is different i.e permutation.

**✅**When to use permutation and when to use combination?**Ans. Permutation and combination **deal with the counting and arrangement of a particular set of data. Factorial (*n*!): It is the product of all positive integers less than or equal to *n. *Example: 4! = 4 × 3 × 2 × 1 = 24. **Permutation**refers to the act of arranging all the elements of a set into a sequence or order. It is denoted by ^{n}P_{r}. A permutation is the choice of *r* things from a set of *n* things without replacement and where the order matters.

**✅**How do you solve permutations and combinations easily?**Ans. **This **arranging of subsets** is known as** permutation when** the order is the main factor and

**when the**

*combination***order is not a factor.**

See this video so that you can solve permutations and combinations easily: Quick Ways of Doing Permutations and Combinations

**✅**How many unique combinations are there?**Ans. **According to the fundamental counting principle if you want to find the unique combinations of 2 independent events just multiply the number of ways each event can happen together. **Example**: if you have 5 shirts and 7 pants then the unique combination will be 5 * 7 = 35 unique combination.

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