NCERT Solutions for Exercise 7.3 Class 11 Maths Chapter 7 - Permutations and Combinations

# NCERT Solutions for Exercise 7.3 Class 11 Maths Chapter 7 - Permutations and Combinations

Edited By Vishal kumar | Updated on Nov 07, 2023 01:07 PM IST

## NCERT Solutions for Class 11 Maths Chapter 7 Permutations And Combinations Exercise 7.3- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 7: Permutations and Combinations Exercise 7.3- In the previous exercise, you have already learned about the fundamental principle of counting, factorial notations, permutations when all the objects are distinct, etc. In the NCERT solutions for Class 11 Maths chapter 7 exercise 7.3, you will learn about the permutations when all the objects are not distinct objects. If you have understood the previous exercise of this chapter you will understand the Class 11 Maths chapter 7 exercise 7.3 easily.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics WallahAakash Unacademy

Scholarship Test: Vidyamandir Intellect Quest (VIQ)

Class 11th Maths chapter 7 exercise 7.3 is an extension of the previous exercise where you have learned about the permutations when all the objects are distinct. Class 11th Maths chapter 7 exercise 7.3 contains the permutations problems related to our daily life. Exercise 7.3 Class 11 Maths is a good exercise to practice real-life permutations problems to get command on this chapter. If you are looking for click on the above-given link where you will get NCERT solutions from Classes 6 to 12 for Science and Maths in a detailed manner. All the solutions are crafted by highly experienced subject experts in a detailed and easily understandable manner. Additionally, PDF versions of the solutions are available to students at no cost, ensuring they can access and utilize them at any time.

Also, see

**As per the CBSE Syllabus for 2023-24, this chapter has been renumbered as Chapter 6.

## Question:1 How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?

3-digit numbers have to be formed by using the digits 1 to 9.

Here, the order of digits matters.

Therefore, there will be as many 3-digit numbers as there are permutations of 9 different digits taken 3 at a time.

Therefore, the required number of 3-digit numbers $=^{9}P_3$

$=\frac{9!}{(9-3)!}$

$=\frac{9!}{6!}$

$=\frac{9\times 8\times 7\times 6!}{6!}$

$=9\times 8\times 7=504$

The thousands place of 4-digit numbers has to be formed by using the digits 1 to 9(0 cannot be included).

Therefore, the number of ways in which thousands place can be filled is 9.

Hundreds,tens, unit place can be filled by any digits from 0 to 9.

The digit cannot be repeated in 4-digit numbers and thousand places is occupied with a digit.

Hundreds, tens, unit place can be filled by remaining any 9 digits.

Therefore, there will be as many 3-digit numbers as there are permutations of 9 different digits taken 3 at a time.

Therefore, the required number of 3-digit numbers $=^{9}P_3$

$=\frac{9!}{(9-3)!}$

$=\frac{9!}{6!}$

$=\frac{9\times 8\times 7\times 6!}{6!}$

$=9\times 8\times 7=504$

Thus, by multiplication principle, required 4 -digit numbers is $9\times 504=4536$

3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated.

The unit place can be filled in 3 ways by any digits from 2,4 or 6.

The digit cannot be repeated in 3-digit numbers and the unit place is occupied with a digit(2,4 or 6).

Hundreds, tens place can be filled by remaining any 5 digits.

Therefore, there will be as many 2-digit numbers as there are permutations of 5 different digits taken 2 at a time.

Therefore, the required number of 2-digit numbers $=^{5}P_2$

$=\frac{5!}{(5-2)!}$

$=\frac{5!}{3!}$

$=\frac{ 5\times 4\times 3!}{3!}$

$=5\times 4=20$

Thus, by multiplication principle, required 3 -digit numbers is $3\times 20=60$

4-digit numbers that can be formed using the digits 1, 2, 3, 4,5.

Therefore, there will be as many 4-digit numbers as there are permutations of 5 different digits taken 4 at a time.

Therefore, the required number of 4-digit numbers $=^{5}P_4$

$=\frac{5!}{(5-4)!}$

$=\frac{5!}{1!}$

$= 5\times 4\times 3\times 2\times 1=120$

4-digit even numbers can be made using the digits 1, 2, 3, 4, 5 if no digit is repeated.

The unit place can be filled in 2 ways by any digits from 2 or 4.

The digit cannot be repeated in 4-digit numbers and the unit place is occupied with a digit(2 or 4).

Thousands, hundreds, tens place can be filled by remaining any 4 digits.

Therefore, there will be as many 3-digit numbers as there are permutations of 4 different digits taken 3 at a time.

Therefore, the required number of 3-digit numbers $=^{4}P_3$

$=\frac{4!}{(4-3)!}$

$=\frac{4!}{1!}$

$=4\times 3\times 2\times 1=24$

Thus, by multiplication principle, required 4 -digit numbers is $2\times 24=48$

From a committee of 8 persons, chairman and a vice chairman are to be chosen assuming one person can not hold more than one position.

Therefore,number of ways of choosing a chairman and a vice chairman is permutations of 8 different objects taken 2 at a time.

Therefore, required number of ways $=^{8}P_2$

$=\frac{8!}{(8-2)!}$

$=\frac{8!}{6!}$

$=\frac{ 8\times 7\times 6!}{6!}$

$=8\times 7=56$

Given : $^{n-1}P_{3}:^{n}P_{4}=1:9.$

To find the value of n

$^{n-1}P_{3}:^{n}P_{4}=1:9.$

$\Rightarrow \frac{^{n-1}P_3}{^nP_4}=\frac{1}{9}$

$\Rightarrow \, \, \frac{\frac{(n-1)!}{(n-1-3)!}}{\frac{n!}{n-4!}}=\frac{1}{9}$

$\Rightarrow \, \, \frac{\frac{(n-1)!}{(n-4)!}}{\frac{n!}{(n-4)!}}=\frac{1}{9}$

$\Rightarrow \, \, \frac{(n-1)!}{n!}=\frac{1}{9}$

$\Rightarrow \, \, \frac{(n-1)!}{(n-1)!\times n}=\frac{1}{9}$

$\Rightarrow \, \, \frac{1}{ n}=\frac{1}{9}$

$\Rightarrow \, \, n=9.$

Question:7(i) Find r if

$^{5}P_{r}=2\; ^{6}\! P_{r -1}$

Given : $^{5}P_{r}=2\; ^{6}\! P_{r -1}$

To find the value of r.

$^{5}P_{r}=2\; ^{6}\! P_{r -1}$

$\Rightarrow \, \, \frac{5!}{(5-r)!}=2\times \frac{6!}{(6-(r-1))!}$

$\Rightarrow \, \, \frac{5!}{(5-r)!}=2\times \frac{6!}{(6-r+1)!}$

$\Rightarrow \, \, \frac{5!}{(5-r)!}=2\times \frac{6\times 5!}{(6-r+1)!}$

$\Rightarrow \, \, \frac{1}{(5-r)!}=2\times \frac{6}{(7-r)!}$

$\Rightarrow \, \, \frac{1}{(5-r)!}= \frac{12}{(7-r)\times (6-r)\times (5-r)!}$

$\Rightarrow \, \, \frac{1}{1}= \frac{12}{(7-r)\times (6-r)}$

$\Rightarrow \, \, (7-r)\times (6-r)=12$

$\Rightarrow \, \, 42-6r-7r+r^2=12$

$\Rightarrow \, \, r^2-13r+30=0$

$\Rightarrow \, \, r^2-3r-10r+30=0$

$\Rightarrow \, \, r(r-3)-10(r-3)=0$

$\Rightarrow \, \, (r-3)(r-10)=0$

$\Rightarrow \, \, r=3,10$

We know that

$^nP_r=\frac{n!}{(n-r)!}$

where $0\leq r\leq n$

$\therefore 0\leq r\leq 5$

Thus the value of, $r=3$

Question:7(ii) Find r if

$^{5}P_{r}=^{6\! \! }P_{r-1}$

Given : $^{5}P_{r}=^{6\! \! }P_{r-1}$

To find the value of r.

$^{5}P_{r}=^{6\! \! }P_{r-1}$

$\Rightarrow \, \, \frac{5!}{(5-r)!}= \frac{6!}{(6-(r-1))!}$

$\Rightarrow \, \, \frac{5!}{(5-r)!}=\frac{6!}{(6-r+1)!}$

$\Rightarrow \, \, \frac{5!}{(5-r)!}=\frac{6\times 5!}{(6-r+1)!}$

$\Rightarrow \, \, \frac{1}{(5-r)!}= \frac{6}{(7-r)!}$

$\Rightarrow \, \, \frac{1}{(5-r)!}= \frac{6}{(7-r)\times (6-r)\times (5-r)!}$

$\Rightarrow \, \, \frac{1}{1}= \frac{6}{(7-r)\times (6-r)}$

$\Rightarrow \, \, (7-r)\times (6-r)=6$

$\Rightarrow \, \, 42-6r-7r+r^2=6$

$\Rightarrow \, \, r^2-13r+36=0$

$\Rightarrow \, \, r^2-4r-9r+36=0$

$\Rightarrow \, \, r(r-4)-9(r-4)=0$

$\Rightarrow \, \, (r-4)(r-9)=0$

$\Rightarrow \, \, r=4,9$

We know that

$^nP_r=\frac{n!}{(n-r)!}$

where $0\leq r\leq n$

$\therefore 0\leq r\leq 5$

Thus, $r=4$

There are 8 different letters in word EQUATION.

Therefore, words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once

is permutations of 8 different letters taken 8 at a time, which is $^8P_8=8!$

Hence, the required number of words formed $=8!=40320$

4 letters are used at a time

There are 6 letters in word MONDAY.

Therefore, words that can be formed using 4 letters of the word MONDAY.

Hence, the required number of words formed using 4 letters $=^6P_4$

$=\frac{6!}{(6-4)!}$

$=\frac{6!}{(2)!}$

$=\frac{6\times 5\times 4\times 3\times 2!}{(2)!}$

$=6\times 5\times 4\times 3$

$=360$

all letters are used at a time

There are 6 letters in word MONDAY.

Therefore, words that can be formed using all 6 letters of the word MONDAY.

Hence, the required number of words formed using 6 letters at a time $=^6P_6$

$=\frac{6!}{(6-6)!}$

$=\frac{6!}{(0)!}$

$=6\times 5\times 4\times 3\times 2\times 1$

$=720$

all letters are used but first letter is a vowel?

There are 6 letters and 2 vowels in word MONDAY.

Therefore, the right most position can be filled by any of these 2 vowels in 2 ways.

Remaining 5 places of the word can be filled using any of rest 5 letters of the word MONDAY.

Hence, the required number of words formed using 5 letters at a time $=^5P_5$

$=5!$

$=5\times 4\times 3\times 2\times 1$

$=120$

Words formed starting from vowel using 6 letters $=2\times 120=240$

In the given word MISSISSIPPI, I appears 4 times, S appears 4 times, M appears 1 time and P appear 2 times.

Therefore, the number of distinct permutations of letters of the given word is

$=\frac{11!}{4!4!2!}$

$=\frac{11\times 10\times 9\times 8\times 7\times 6\times 5\times 4!}{4!4!2!}$

$=\frac{11\times 10\times 9\times 8\times 7\times 6\times 5}{4\times 3\times 2\times 2}$

$=34650$

There are 4 I's in the given word. When they occur together they are treated as a single object for the time being. This single object with the remaining 7 objects will be 8 objects.

These 8 objects in which there are 4Ss and 2Ps can be arranged in $\frac{8!}{4!2!}=840$ ways.

The number of arrangement where all I's occur together = 840.

Hence, the distinct permutations of the letters in MISSISSIPPI in which the four I's do not come together$=34650-840=33810$

There are 2 T's in word PERMUTATIONS, all other letters appear at once only.

If words start with P and ending with S i.e. P and S are fixed, then 10 letters are left.

The required number of arrangements are

$=\frac{10!}{2!}=1814400$

vowels are all together?

There are 5 vowels in word PERMUTATIONS and each appears once.

Since all 5 vowels are to occur together so can be treated as 1 object.

The single object with the remaining 7 objects will be 8 objects.

The 8 objects in which 2 T's repeat can be arranged as

$=\frac{8!}{2!}$ ways.

These 5 vowels can also be arranged in $5!$ ways.

Hence, using the multiplication principle, the required number of arrangements are

$=\frac{8!}{2!}\times 5!=2419200$ ways.

there are always 4 letters between P and S?

The letters of the word PERMUTATIONS be arranged in such a way that there are always 4 letters between P and S.

Therefore, in a way P and S are fixed. The remaining 10 letters in which 2 T's are present can be arranged in

$=\frac{10!}{2!}$ ways.

Also, P and S can be placed such that there are 4 letters between them in $2\times 7=14$ ways.

Therefore, using the multiplication principle required arrangements

$=\frac{10!}{2!}14=25401600$

## More About NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3:-

Class 11 Maths chapter 7 exercise 7.3 consists of questions related to finding the different arrangements with the given conditions. There are eight examples and some theorems given before Class 11 Maths chapter 7 exercise 7.3. You must go through these examples and theorems to get conceptual clarity. There are 11 problems given in the Class 11 Maths chapter 7 exercise 7.3 which you must try solve by yourself.

## Benefits of NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3:-

• NCERT problems given in the Class 11th Maths chapter 7 exercise 7.3 are based on real-life problems.
• Exercise 7.3 Class 11 Maths is considered to be tougher than the previous exercise of this chapter but if you have conceptual clarity, you won't get any difficulty while the NCERT problems of this exercise.
• You are advised to solve the NCERT problems on your own before going through the NCERT solutions.
JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%

## Key Features of NCERT 11th Class Maths Exercise 7.3 Answers

1. Comprehensive Coverage: The ex 7.3 class 11 solution encompass all exercises and problems presented in Chapter 7.3 of the NCERT 11th Class Mathematics textbook.

2. Step-by-Step Solutions: The class 11 maths ex 7.3 solutions are provided in a step-by-step format, making it easier for students to follow and understand the solution process.

3. Clarity and Precision: The class 11 ex 7.3 answers are written with clarity and precision, ensuring that students can grasp the mathematical concepts and problem-solving methods.

4. Proper Mathematical Notation: The 11th class maths exercise 7.3 answers use appropriate mathematical notations and terminology, helping students become familiar with the language of mathematics.

5. Free Access: Typically, these ex 7.3 class 11 answers are available free of charge, allowing students to access them without any cost, making it a valuable resource for self-study.

6. Homework and Practice: Students can use these answers to check their work, practice problem-solving, and enhance their overall performance in mathematics.

Also see-

## Subject Wise NCERT Exampler Solutions

Happy learning!!!

1. How many 3-digit numbers are there with no digit repeated?

The number of 3-digit numbers with no digit repeated = 9 x 9 x 8 = 648

2. How many 4-digit numbers can be formed by numbers 1,2,3,4,5,6,7 if no digit is repeated?

The number of 4-digit numbers can be formed by numbers 1,2,3,4,5,6,7 if no digit is repeated = 7 x 6 x 5 x 4 = 840

3. How many permutations of word APPLE could be are there ?

APPLE  has 5 letters but two letters PP are the same.

The total number of permutations = 5!/2! = 60

4. How many different words can be formed using all letters of word MATHS.

MATHS  has 5 letters.

The total number of permutations = 5!= 120

5. What is weightage of Determinants in the JEE Main Maths ?

Determinants have 6.6% weightage in the JEE Main Maths.

6. What is weightage of 3-D Geometry in the JEE Main Maths ?

3-D Geometry has 6.6% weightage in the JEE Main Maths.

## Upcoming School Exams

#### National Means Cum-Merit Scholarship

Application Date:05 August,2024 - 25 September,2024

#### National Rural Talent Scholarship Examination

Application Date:05 September,2024 - 20 September,2024

#### National Institute of Open Schooling 12th Examination

Admit Card Date:13 September,2024 - 07 October,2024

#### National Institute of Open Schooling 10th examination

Admit Card Date:13 September,2024 - 07 October,2024

#### Nagaland Board High School Leaving Certificate Examination

Application Date:17 September,2024 - 30 September,2024

Get answers from students and experts

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9