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Permutation and combination represent a way in which a group of objects can be represented by selecting them in sets and then forming subsets. It also defines ways in order to arrange a group of data. In the case where we select the data from a certain group, it is known to be permutation whereas the order in which they are represented in known as combination. A permutation is a list of data and the combination is used for a group of data.

Questions from Permutation and combination are frequently asked in various competitive exams.

The list of exams where such type of questions are asked includes bank exams such as IBPS PO, SBI PO, IBPS RRB, IBPS Clerk, IBPS SO as well as exams like LIC AAO, SSC CGL etc.

## Understanding what is Permutation

Permutation refers to the arrangement of all elements or objects of a set in a sequence or if a sequence is already ordered, rearranging it is called permuting. They occur in almost every topic of maths and often arise when different order on certain finite set is considered.

## Understanding what is combination

It refers to the way of selecting items from a group such that the order of the selection doesn’t matter. In smaller cases, it is also possible to count the number of combinations. It refers to combination of ‘n’ number of things, taken at ‘k’ time and does not include repetition. In the case of referring to a combination where repetition is allowed, the terms k-combination is often used.

## Understanding Factorial

It refers to the product of the numbers which start from 1 and go up to a number ‘n’. In this case, it is known as the factorial of the number ‘n’.

n! = 1 × 2 × 3 × 4 × 5 × 6…….. × (n – 2) × (n – 1) × n

For example:

4! = 1 × 2 × 3 × 4 = 24

5! = 1 × 2 × 3 × 4 × 5 = 120

6! = 1 × 2 × 3 × 4 × 5 × 6 = 720

Key points to consider in case of Factorial:

0! & 1! Is equal to 1.

Also, we can’t find the factorial of a negative number.

## Factorial application

It is one of the most common applications in case of arrangement. Let us try to understand how can factorial help to arrange things.

For example, when we have 5 persons and we need to arrange them in 5 vacant places. We can start with the first place when we choose 1 person out of 5 for the first place. So, we can do that in 5 ways. Now with 4 vacant seats, we have 4 persons left. We can choose 1 person out of 4 for the second place. In this manner, the number of seats, as well as people, keep on decreasing.

As all the people are required to do these activities, we multiply them to get the final answer and find out the different ways of arrangement.

Hence, the total ways are = 5 × 4 × 3 × 2 × 1 which is 5! = 120

Or we can also say that wherever we need to arrange ‘n’ things at ‘n’ places, the total arrangement can be equal to n!

Whenever we need to choose things from a group without arraignment, the combination comes into the picture.

## Concept of combination

It is applicable whenever we need to choose things from a group and no arrangement is required.

In this case, we select the things at random & then check out the different ways of selection. This is step one in the process. It is also known as a collection. The formula, in this case, is nCr

nCr = n! / [r! × (n – r)!]

5C2 = [5 × 4] / [1 × 2] = 10

nCr = nC(n – r)

For example:

10C7 = 10C3 = [10 × 9 × 8] / [1 × 2 × 3] = 120

Whenever we are required to choose things from a group and arrangement is required to be made, in that case permutation comes into the picture.

## Concept of permutation

The formula which is used for permutation is nPr = n! / (n – r)! S

For example,

Five people out of whom only two can drive are to be seated in a five seater car with two seats in front and three in the rear. The people who know driving don’t sit together. Only someone who knows driving can sit on the driver’s seat. Find the number of ways the five people can be seated.

A. 40
B .60
C. 48
D. 36
E. None of these

Ans. D

Number of people who can drive = 2

Number of ways of selecting driver = 2C1

The other person who knows driving can be seated only in the rear three seats in 3 ways

Total number of ways of seating the two persons = 2C1 × 3

Number of ways of seating remaining = 3!

Total number of all five can be seated = 2C1 × 3 × 3! = 36