The chapter, Permutations and Combinations, is part of Algebra, accounting for 30 of the total 80 points. This chapter includes four problems and a miscellaneous activity to assist students in grasping the concepts of permutations and combinations. Some important principles driving the concepts are as follows:

The number of permutations of n things when selected r at a time, where repetition is not allowed, is denoted by nPr

n! = 1  x  2  x  3  x  …  x  n

n! = n  x  (n – 1) !

The number of permutations of n different things, taken r at a time, where repetition is allowed, is nr.

 

Therefore, nr is the number of permutations of n different items, taken r at a time, with repetition permitted.

 

As a result, we can conclude that studying NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations enables students to comprehend the Fundamental Principle of Counting, Factorial (n! ), Permutations and Combinations, Formulae for nPr and nCr and their connections, and simple applications of permutations and combinations. Permutations and combinations formulas are important, and therefore, you should remember its derivation and usage.

The fundamental principle of counting states that if an event can happen in m different ways, then another event can happen in n different ways. The total number of times the events can happen in the given order is m  x  n.

 

If an event can happen in m different ways, then another event can happen in n different ways, then a third event can happen in p different ways. The total number of times the events happen in the given order is m  x  n  x  p.

A permutation is an arrangement of several objects in a specific order, taken one at a time or all at once.

 

3.1 Permutations when all the objects are distinct

 

n(n–1)(n–2)...(n–r+1), which nPr indicates, is the number of permutations of n different objects taken r at a time, where rn and the items do not repeat.

Proof: There will be as many variants as ways to fill in r vacant spaces... by r vacant places the n objects. The first spot can be filled in n ways; the second place can be filled in (n – 1) ways; the third place can be filled in (n – 2) ways; and so on until the rth place is filled in [n – (r – 1)] ways. As a result, the number of methods to fill in r unoccupied spaces in a sequential order is n(n–1)(n–2)... (n–(r–1)) or n (n–1)(n–2)... (n – r + 1). You can refer to the NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations  to understand how the permutation questions are solved when the objects are distinct.

 

3.2 Factorial Notation

The symbol n! denotes the product of the first n natural numbers, i.e., 1  x  2  x  3. ... (n – 1) n is symbolized by the symbol n! This symbol denotes 'n factorial'. Thus,1 x 2 x 3 x 4. ..  x (n–1)  x n=n!

1=1!

1 x 2=2!

1 x 2 x 3 = 3!

1 2 3 4 equals 4! and so forth.

0! = 1 is how we define it.

We can write a total of five! =5 x 4!=5 x 4 x 3!=5 x 4 x 3 x 2! = 5  x  4  x  3  x  2  x  1! Clearly, n n! =n(n – 1) for a natural number!

and so forth.

(n – 1)(n – 2)! =n(n – 1)(n – 2)!

=n(n-1)(n-2)(n-3)!

[provided (n≥2)] [provided (n≥3)]

 

3.3 Derivation of the formula for nPr

 

Assume we have some objects that can be arranged in  r ways:

The first object can be selected in n ways

Second object can be selected in n-1 ways

Third object can be selected in n-2 ways

Similarly rth object can be selected in 

𝑛−(𝑟−1)=𝑛−𝑟+1

n−(r−1)=n−r+1ways

So all r objects can be selected in 

𝑛(𝑛−1)(𝑛−2)..(𝑛−𝑟+1)

n(n−1)(n−2)..(n−r+1)ways

𝑛𝑃𝑟=𝑛(𝑛−1)(𝑛−2)..(𝑛−𝑟+1)

nPr=n(n−1)(n−2)..(n−r+1)

For the above, divide and multiply by (n-r)!

𝑛𝑃𝑟=(𝑛(𝑛−1)(𝑛−2)..(𝑛−𝑟+1)(𝑛−𝑟)!)/(𝑛−𝑟)!

nPr=(n(n−1)(n−2)..(n−r+1)(n−r)!)/(n−r)!

𝑛𝑃𝑟=𝑛!/(𝑛−𝑟)!

1. 1. 1. nPr=n!/(n−r)!

 

3.4 Permutations when all the objects are not distinct

 

Assume we need to determine the number of different ways to rearrange the letters in the word, ROOT. The letters of the word are not all different in this situation. There are two Os that are of the same kind. Let's pretend the two Os are different and call them O1 and O2. In this situation, the number of permutations of four different letters taken all at once is 4! Consider RO1O2T as an e x ample of one of these permutations. We have 2! permutations that correspond to this permutation: RO1O2T and RO2O1T, identical if O1 and O2 are not treated as different, i.e., if O1 and O2 are the same as O in both places.

Assume a group of three lawn tennis players,  x , Y, and Z. It is necessary to construct a two-player team. How many different ways can we do it? Is the  x  and Y squad different from the Y and  x  teams? Order isn't important here. There are only three viable methods to put together the team.

 

These are the letters  x Y, YZ, and Z x .

 

Each selection is a mi x ture of three separate objects that are taken two at a time. The order of the elements in a combination is irrelevant.

You can refer to NCERT Solutions for Class 11 Permutations and Combinations for a detailed understanding of how problems can be solved. Some e x amples are listed below:

 

1.  Find the number of four-letter words, with or without meaning that can be made from the letters of the word ROSE, with no duplication of letters.

 

Solution:  There are as many words as there are methods to fill in four empty spaces with four letters, e x cept that repetition is not permitted. Any of the four letters R, O, S and  E can fill the first spot in four different ways. Following that, the second spot can be filled in three different ways with any of the remaining three letters, the third place can be filled in two different ways, and the fourth-place can be filled in one way. As a result, there are a variety of approaches to fill the four positions. Here, using the multiplication principle: 4  x  3  x  2 x  1=24 ways.

 

2.  How many different signals can be made with four different coloured flags if each signal requires two flags, one above the other?

 

Solution: There will be as many signals as methods to fill in two vacant spots in a row with four flags of various colours. The upper vacant spot can be filled in four different ways by any of the four flags, and the lower vacant spot can be filled in three different ways by any of the three remaining flags. As a result, the required number of signals = 4  x  3 = 12, according to the multiplication principle.

1. How can I score full marks in Class 11 Maths Chapter 7?

 

You can easily score full marks by practising the fundamental concept of counting and the multiplication principle. Once you have understood these concepts, you can then easily solve questions on permutations and combinations.

 

2. What are the important topics covered in the NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations?

 

The combination is a precise arrangement of multiple elements, whereas permutation is the various ways in which a number or set of objects can be organized. The first is the fundamental theorem of counting, which is based on some of the fundamental notions. For instance, if one event occurs in 'n' different ways while another occurs in 'm' various ways, the net occurrences can be e x pressed in the order n  x  m.

The second e x plains that the number of permutations 'n' separate items at 'r' time when no repetition is written as nPr, and when there is repetition is e x pressed as nr.

 

3. What are the fundamental topics of permutations and combinations?

 

The NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations cover the following topics:

1. Introduction

2. Fundamental Principle of Counting

3. Permutations

4. Combinations