NCERT Solutions for Exercise 7.3 Class 11 Maths Chapter 7

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.

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NCERT Solutions for Class 11 Maths Chapter 7 Permutations And Combinations Exercise 7.3- Download Free PDF
Download the PDF of NCERT Solutions for Class 11 Maths Chapter 7 – Permutations and Combinations Exercise 7.3
Access Permutation And Combinations Class 11 Chapter 7 Exercise 7.3
Question:1 How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
More About NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3:-
Benefits of NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3:-
Key Features of NCERT 11th Class Maths Exercise 7.3 Answers
NCERT Solutions of Class 11 Subject Wise
Subject Wise NCERT Exampler Solutions

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

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 NCERT solutions, 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.

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

Download the PDF of NCERT Solutions for Class 11 Maths Chapter 7 – Permutations and Combinations Exercise 7.3

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Access Permutation And Combinations Class 11 Chapter 7 Exercise 7.3

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

Answer:

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 $=^{9}P_3$

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

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

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

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

Question:2 How many 4-digit numbers are there with no digit repeated?

Answer:

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 $9\times 504=4536$

Question:3 How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?

Answer:

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 $3\times 20=60$

Question:4 Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4,5 if no digit is repeated. How many of these will be even?

Answer:

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 $2\times 24=48$

Question:5 From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?

Answer:

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 $=^{8}P_2$

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

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

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

$=8\times 7=56$

Question:6 Find n if $^{n-1}P_{3}:^{n}P_{4}=1:9.$

Answer:

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

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

Answer:

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

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

Answer:

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, $r=4$

Question:8 How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?

Answer:

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 $=8!=40320$

Question:9(i) How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if.

4 letters are used at a time

Answer:

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 $=^6P_4$

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

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

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

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

$=360$

Question:9(ii) How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if.

all letters are used at a time

Answer:

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 $=^6P_6$

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

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

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

$=720$

Question:9(iii) How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if.

all letters are used but first letter is a vowel?

Answer:

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 $=2\times 120=240$

Question:10 In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?

Answer:

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 $=34650-840=33810$

Question:11(i) In how many ways can the letters of the word PERMUTATIONS be arranged if the

words start with P and end with S?

Answer:

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

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

Question:11(ii) In how many ways can the letters of the word PERMUTATIONS be arranged if the

vowels are all together?

Answer:

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.

Question:11(iii) In how many ways can the letters of the word PERMUTATIONS be arranged if the

there are always 4 letters between P and S?

Answer:

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

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

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.

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Key Features of NCERT 11th Class Maths Exercise 7.3 Answers

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

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Frequently Asked Questions (FAQs)

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.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

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

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.

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

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

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

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

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

Download the PDF of NCERT Solutions for Class 11 Maths Chapter 7 – Permutations and Combinations Exercise 7.3

Access Permutation And Combinations Class 11 Chapter 7 Exercise 7.3

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

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