# Permutation and Combination Class 11

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Have you ever imagined how many combinations of your contact number are possible if we interchange one of its digits? Or, In how many unique ways the total number of students in your class can sit together? Class 11 maths syllabus introduces us to a number of new topics, one amongst which is permutation and combination. Know as one of the most interesting topics, permutation and combination chapter for class 11 is a highly scoring concept if you are well-versed with it. Whether you are gearing up for competitive or scholastic exams, our notes on this chapter will help you understand and retain it. So, let’s get started!

## What is Permutation?

Permutation refers to a particular arrangement of a set of objects in a defined order or a process of arranging numbers or letters in a sequence. We can represent permutations by:

Let us understand the term permutation better, consider the numbers 456, and 654. Both of them consist of the same digits 4, 5, 6. But, they are arranged in a different order. So, we can say they are different permutations of the digits 4, 5, 6. We can form many different permutations from a given set of objects taking all of the digits from the set at a time or a particular number of objects at a time. The number of permutations that can be formed is represented by r at a time out of n. We shall denote this number of permutations by P(n, r). The picture given below contains the formula of permutation.

Where, the symbol ‘!’ denotes factorial, which is the product of all the integers less than or even equal to n, but should be greater or equal to 1.

For Example: How many license plates consisting of three different digits can be made out of given integers 3, 4, 5, 6, 7?
Solution:
It is just like arranging 3 objects out of 5 objects. So, we have
P(5,3) = 5!/ (5-3)!
=5 x 4 x 3 x 2 x 1/ 2 x 1
= 5 x 4 x 3 = 60

## Types of Permutation

You may across various types of permutations even in day to day life. Hence, as per permutation and combination class 11 chapter, permutation can be classified into different categories:

### Permutation of Objects When Repetition Not Allowed

Permutations without repetitions are arrangements of n elements in different groups so that they can be differentiated in the order they have been placed.

For Example : Let us consider a set of alphabets: A = { a, b, c, d, e}
Then, the permutation of these 5 alphabets can be arranged in the following ways: abcde, dbeca, bedac, acbde, adcea, cdbae, edabc, etc.

The permutation for a number of n elements is given by:  Pn = n! = n(n-1) (n-2).. 2.1

The total number of elements in the above example is 5 and hence,
P5 = 5! = 5.4.3.2.1 = 120
So, at least 60 permutations can be made with elements A= { a, b, c, d, e}

Let us go through some more examples to be clear with this topic.

For Example: In how many different ways can the letters of the word “MEMBER” be arranged?Solution:
The word member consists of 6 letters, 2 of which are M’s, 2 are E’s and the rest are different. So, writing n = 6, p = 2, and q = 2 the required number of permutations are: n!/ p!q!
n!/ p!q! = 6!/ 2!2! = 180

## Circular Permutation

According to the permutation and combination class 11 chapter, the permutations used when the objects/ elements are placed in a circular order. For Example, the number of circular permutations of “n” different things at all times is taken as P= (n-1)!

For Example: In how many different ways can the numbers on a clock face be arranged?
Solution:
In a clock face, there are 12 numbers. So they can be arranged in:
(12-1)! = 11! ways

Going clockwise they are ABCD, ABDC, ADBC, ADCB, ACBD, ACDB. The arrangements BCDA, CDAB, DABC, are the same as the arrangement ABCD.

## Permutation of Repeated Objects

To find the permutation of n, which is occurring any number of times or repetition of n elements, suppose there are r places, and there are n objects. The first place is filled up by any of the n objects. So, as mentioned in the permutation and combination class 11 chapter, there are ‘n’ choices for filling up the first place.

Let us consider this example that from a lock you have to choose 10 numbers (0,1,2..9) and you picked 3 of them.
So, 10 * 10 * ( 3times) = 1000 Permutations

The permutation of repetition is denoted by PR(n,k)

From a given set on n numbers, the numbers are formed, so that there are n1 identical numbers of type1, n2 identical numbers for type 2…. and nk numbers of type k. Generally, repetitions are done by dividing permutation by the fractional numbers that are identical.

Let us clear this out through an example, how many three-digit numbers can be formed from the numbers 0,1,2,3,4,5
Solution:
Here, n = 3 and K= 6

First, the numbers will be separated into two blocks. The first set of blocks will consist of only one from the 5 digits because a number will not begin from zero.

So, n=5 and K = 1
The second block will consist of two numbers of any digit
N=6 and K= 2
P(5,1) x Pr (6,2) = 5 x 6^2 = 180

## Properties Of Permutations

The permutation and combination class 11 chapter emphasis on a variety of properties, understanding which you will be through with an essential part of the concept.

### Principle of Multiplication

The principle states “If one operation can be done in m ways, there are also n ways to perform a second operation as well, then the number of ways to perform two operations together is m*n

In an event of A and B, if the events do not occur simultaneously, i.e., first event A occurs in m different ways and the second event B occurs in n different ways, then one and the other will occur in (m+n ways). That is A will occur in m+n ways and B will also occur in m+n ways, according to permutations and combinations class 11 topic.

## Combinations

A combination is known as selecting items from a combination or selection of all or part of a set of objects, without giving regard to the order in which the objects are selected.  To calculate combination, we will use the formula:

Where, n represents the total number of items, and r represents the number of items being chosen at a time. An example of combination form class 11 NCERT book is given below-

Example: Calculate the value of C(12,9).
Solution:
C(12,9) = C(12,12-9)
= C(12,3) = 12!/9!3!
= 10.11.12/ 1.2.3
= 220

## Permutation and Combination Class 11 Formulas

• nPr = n!/(n-r)!
• nPn = n!
• nP1 = n
• nCr = n!/(r! (n-r)!)
• nC1 = n
• nC0 = 1 = nCn
• nCr = nCn-r
• nCr = nPr/r!
• Number of diagonals in a geometric figure of n sides = nC2-n

## Solved Examples for Permutation and Combinations

Question 1: How many different combinations do you get if you have 4 items and choose 2?

Answer: Insert the given numbers into the combinations equation and solve. “n” is the number of items that are in the set (4 in this example); “r” is the number of items you’re choosing (2 in this example):

C(n,r) = n! / r! (n – r)!
= 4! / 2! (4 – 2)!
= 4! /2! * 2!
= 4 x 3 x 2 x 1 / 2 x 1 * 2 x 1
= 24 / 4= 6

Question 2:  A bag contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the bag if at least one black ball is to be included in the draw?

Answer: From 2 white balls, 3 black balls and 4 red balls, 3 balls are to be selected such that at least one black ball should be there.

Hence we have 3 choices as given below.
Choice 1: We can select 3 black balls.
Choice 2: We can select 2 black balls and 1 non-black ball.
Choice 3: We can select 1 black ball and 2 non-black balls.

Number of ways to select 3 black balls
= 3C3
Number of ways to select 2 black balls and 1 non-black ball
= 3C2 × 6C1
Number of ways to select 1 black ball and 2 non-black balls
= 3C1 × 6C2

Total number of ways
= 3C3 + 3C2 × 6C1 + 3C1 × 6C2
= 3C3 + 3C1 × 6C1 + 3C1 × 6C2 [ nCr = nC(n-r)]= 1 + (3×6) + [3 x (6×5)/(2×1)]= 1 + 18 + 45
= 64

Question 3: How many words with or without meaning can be formed by using all the letters of the word, ‘DELHI’, using each letter exactly once?

Solution: The word ‘DELHI’, contains 5 different letters.

Required number of words
= Number of arrangements of 5 letters, taken all at time
= 5P5 = 5 ! = (5×4×3×2×1)=120.

Question 4: How many words can be formed from the letters of the word ‘SIGNATURE’ so that the vowels always come together?

Solution: The word ‘SIGNATURE’ contains 9 different letters.
When the vowels IAUE are taken together, they can be supposed to form an entity, treated as one letter.
Then, the letter to be arranged are SGNTR (IAUE).
These 6 letters can be arranged in 6P6 = 6! = 720 ways.

The vowels in the group (IAUE) can be arranged amongst themselves in 4P4 = 4! = 24 ways.
Required number of words = (720×24) = 17280.

## Permutation and Combination Class 11: Practice Questions

• Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
•  In how many different ways can the letters of the word ‘OPTICAL’ be arranged so that the vowels always come together?
• How many 4-letter words with or without meaning, can be formed out of the letters of the word, ‘ALGORITHMS’, if repetition of letters is not allowed?
• There are 18 stations between Hyderabad and Bangalore. How many second class tickets have to be printed, so that a passenger can travel from any station to any other station?
• In how many different ways can the letters of the word ‘CORPORATION’ be arranged so that the vowels always come together?
• In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
•  How many 3 digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9 which are divisible by 5 and none of the digits is repeated?
• How many words with or without meaning, can be formed by using all the letters of the word, ‘DELHI’ using each letter exactly once?
• Six points are marked on a straight line and five points are marked on another line which is parallel to the first line. How many straight lines, including the first two, can be formed with these points?
• A selection is to be made for one post of principal and two posts of vice-principal amongst the six candidates called for the interview only two are eligible for the post of principal while they all are eligible for the post of vice-principal. The number of possible combinations of selectees is?

So, through this article about permutation and combination class 11, we hope that you have understood this important concept. Reach out to our experts at Leverage Edu and they will guide you in choosing the right career path as per future goals. Hurry up! Book an-meeting.

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