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

After the students have completed their Class 10 and passed out successfully, they are required to choose subjects that they want to study for two years in Class 11 and Class 12. Students must decide whether they want to study arts, science, or commerce in Class 11 based on that they are allotted. Generally, students who choose science or commerce are the ones who study mathematics in Class 11 and Class 12. Students who have chosen to study mathematics know how challenging it can be. So it is required for every student to focus on their studies and health to be as productive and concentrated as they can be. This concentration will help them gain marks in exams and achieve whatever goal they set for themselves. Students can choose to study math further in undergraduate programmes if they wish to.

Students must remember that mathematics requires practising in order to be able to solve advanced problems. Students must practice from NCERT, and if they feel as if they need help, they can refer to the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3. Students who need help with mathematics should take a look at the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3, available on Extramarks’ website and app. The permutations and combinations could come off as a new and challenging topic for students, but once they practice and understand the chapter, they’ll be able to figure out how to manage all these new topics and how to answer them properly with no mistakes. Extramarks NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3 will help students to manage their questions and figure out how to answer them efficiently.

Students must remember that, however challenging it may seem, there is help available for them. Students can ask their teachers for explanations about a topic, or they can also ask their peers to help them out with a problem. There is also the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3, if students do not understand the method of answering or if they need to check their answers. They can get their answers corrected or identify their mistakes from the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3. This way, students can solve their answers faster during the examination, whether it is a board exam or an entrance examination.

## NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations (Ex 7.3) Exercise 7.3

Chapter 7 of Class 11 is called Permutations and Combinations. This chapter can be mainly divided into Permutation and Combination. The topics included in this chapter are the fundamental principle of counting, permutation, permutations when all the objects are distinct, factorial notation, derivation of the formula for ‘nPr’, permutations when all the objects are not distinct objects, properties of permutation, theorems based on the formula of permutation, combinations, the formula of combination, theorems based on the formula of combinations, and theorems for both the formula of combination and permutation.

This chapter has four exercises in addition to a miscellaneous exercise. The four exercises have been divided so, students can learn each topic in the chapter with much concentration practice and precision. The exercises have been divided in a way that provides students with practice for each topic equally and according to its importance. This helps students set up a strong foundation for concepts of permutations and combinations. Students can struggle to gain knowledge of permutations and combinations, but it can be made easier for them to understand this with the help of the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3.

It is important for students to understand each concept and practice it as much as they can to achieve perfection, as it is required of them to solve further problems and is also of use while solving other related problems and chapters. If students have not studied it with focus, they might have trouble understanding the concepts further in the book. It could also cause difficulty for them while solving advanced problems. Hence, students must practise all the topics in depth to avoid having trouble with this chapter in the future. They can get help to solve each exercise from Extramarks like the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3.

Students remember that support for academics is available. They can have doubt solving sessions regarding their academic problems on Extramarks. They can also get other study materials for any subject they want on the Extramarks website and app. Students must remember to study with the help of study notes and study material provided and use the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3 only for doubts. Students must not depend on Extramarks study material for NCERT solutions like the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3.

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

It is known to students that NCERT is an excellent book for anybody who is starting out with a new chapter. NCERT is considered a great book for making the basics of a concept or topic strong. Starting out with NCERT will help students form a strong foundation of topics and ultimately make them get better with every exercise that they practise. NCERT covers all topics, from basic to advanced, and helps students solve all the problems. If students find themselves struggling, they can always take help from the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3. Students must remember that NCERT is a very important book for students who are taking CBSE board exams. As CBSE affiliated schools, they have NCERT as a part of their curriculum to help students gain knowledge of each topic and concept.

NCERT is an excellent book for anyone. Students who are preparing for CBSE board exams pay special attention to NCERT. Teachers have recommended practising NCERT as a need for students preparing for any exam, an entrance exam, or a board exam, especially the CBSE board exam. This is because NCERT provides knowledge about a chapter from the start, which helps students form a better understanding of each topic in a specified manner. It also provides examples of questions, so students can understand how the answer must be written and how it should be solved. It also contains the answer key for each exercise at the end of the book. This helps students know if they have the correct answer. If students unfortunately do not get the right answer, they can always come to the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3 to check where they might have made a mistake or to see how the question should be done.

The purpose of the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3 is to help students practise NCERT and to prepare for exams so that they can perform well in their exam. This practice is required so that students learn from it and can correctly write the answer during examination using the correct method as the method of solution is very important for students giving board examinations, whether it is CBSE or ICSE or State Board. Whichever board students are giving exams from, it is important to solve the NCERT and students can take help from the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3 if faced with any challenge while practising.

### NCERT Solutions for Class 11 Maths Chapters

The NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3, is just one of the many chapters in the NCERT Mathematics book for Class 11. The NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3, only provide solutions for one exercise, but Extramarks has provided students with solutions for other exercises and other chapters on Extramarks website and app. Extramarks provides NCERT solutions for all exercises including the last exercise, i.e., miscellaneous exercises. The miscellaneous exercise at the end of each chapter has questions of an advanced level and includes questions about every topic students have studied in the chapter. Some students might have difficulty solving NCERT exercises, Extramarks has provided the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3 to help those students. That’s why, Extramarks also provides a solution to the miscellaneous exercise.

Some students may also want to solve more questions and practice for exams. Extramarks provide study material for each class and also preparation for entrance exams like JEE, NEET, CUET, etc. Extramarks also has study material for international exams like the SATs, PSATs, and TOEFL. Students preparing for any exam can get help from Extramarks. Students must remember to use all the resources available to them on Extramarks website and app, for example the NCERT solutions provided like the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3.

Extramarks provided NCERT solutions are with no mistakes as it can be seen from the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3. Therefore, students must use the Extramarks NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3 for studying this chapter, and they can look for any chapter they have a doubt in through the Extramarks portal on website and app. The study material is provided by Extramarks experts and is made sure to have no mistakes. Students can learn a lot from the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3 and the NCERT solutions provided for other chapters. These NCERT solutions are helpful not only for solving NCERT but also in advanced questions as NCERT form the basis of solution and foundation of knowledge for students. This will help them solve any question however challenging it may seem. The NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3 will guide students to solve any question from this chapter easily and methodically. The NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3 will work as a guide for students to perform better for the tests based on the chapter Permutations and Combinations and especially the topic permutations as this exercise is based on it.

### List of Topics Covered Under Exercise 7.3

Exercise 7.3 of Chapter 7 Permutations and combinations covers some very important topics for the permutations part of the chapter. In the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3, an overview of the topics covered in Exercise 7.3 Class 11th Maths can be seen. The topics covered in Exercise 7.3 Class 11, include the derivation of the formula for ‘nPr’ and permutation when all the objects are not distinct objects. This exercise will help students solve advanced problems in permutations. There are advanced and mixed questions in the Miscellaneous Exercise of Chapter 7. These questions could cause trouble for students if they have not studied properly in the previous exercises.

In this exercise, the formula of permutation is used to solve questions, and questions are also made to teach students how to solve permutations when all the objects are not distinct objects. The method of answering can be seen in the examples given in NCERT book for Maths Class 11 and also the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3. Understanding how the question should be solved and the answer written is important, as board examinations require students to solve whole questions whereas entrance exams like JEE require students to mark the right answer or to write the correct answer. As we can deduce, the method of solution is quite important for board examinations, though it might not be that important for the entrance exams.

The topics covered in the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3, will help students when they move on to solving advanced questions and when they give tests. It will help them correctly interpret the question and figure out which formula should be used in the questions. This will save their time during the exam. This way, they will be able to avoid getting confused and properly answer the question. Using the proper method will help them obtain the correct answers easily. Students will gain an idea about these types of questions from the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3, and it will help them score well in exams. Hence, students must use the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3 to get to know the chapter, the pattern of answers, and the core knowledge of topics. This will help them to score good marks.

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

Students tend to overlook some concepts while studying for their exams. It could be because they do not get enough time to study the whole chapter before the exam, so when they skim through their book, some concepts might get lost without understanding. This could cause trouble for students if the concept turns out to be an important one. Then students might face problems during the exam if a question from that concept is asked. To avoid this, students must always read notes and look through the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3 as it provides both an overview of all the concepts in the exercise and also questions that have a chance of being asked in the exams. To avoid getting stuck during the exam, students must practise using the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3 It is an important part of their curriculum to study from NCERT, and they should not take it lightly. It is important to solve NCERT for many reasons. Students preparing for any entrance exam like JEE or CUET must remember that NCERT is important to be solved as it provides basics to the advanced questions of entrance. Students will not be able to solve advanced questions if they haven’t studied the fundamentals of the chapter thoroughly. Hence, students must solve the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3. Extramarks is an educational portal that can help with any doubts students might have. Students can take help solving NCERT through the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3.

### Benefits of choosing Extramarks for NCERT Solutions Permutations and Combinations Exercise 7.3

There are a number of benefits for choosing Extramarks for NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3, some of which are as follows:

• Extramarks study material is provided to students by Extramarks Experts. Hence, these solutions are checked for any mistakes before being put up online.
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• The Extramarks NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3 is a great way for students to learn and understand about permutations and also to understand its applications.
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Extramarks also has other study material ready for other classes.in addition to the solutions provided for Class 11 and Class 12, Extramarks also has NCERT Solutions Class 10, NCERT Solutions of Class 9, NCERT Solutions Class 8, NCERT Solutions Class 7, NCERT Solutions Class 6, NCERT Solutions Class 5, NCERT Solutions Class 4, NCERT Solutions Class 3, NCERT Solutions Class 2 and NCERT Solutions Class 1 which is available to students through the Extramarks website and app in both Hindi and English.

The NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3 are beneficial to students who are just starting to solve the chapter. It will help them resolve any doubts that arise later. This will also keep them from making careless mistakes during the exam. Practising from the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3 will surely help students improve themselves academically. Students must make use of the resources available to them to make the best out of their time and gain the chances of cracking their entrance as well as  board exams with the practice from the NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3.

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

Ans.
Number of digits from 1 to 9 = 9
Number of digits in each number = 3
If no digit is repeated, then
Number of 3-digit numbers = 9P3
= 9! /(9 – 3)!
= 9 × 8 × 7
= 504
Thus, there are 504, 3-digit numbers which can be formed by using the digits 1 to 9.

Q.2 How many 4-digit numbers are there with no digit repeated?

Ans.
Number of digits from 0 to 9 = 10
Number of digits in each number = 4
If no digit is repeated, then
Number of 4-digit numbers = 9 × 9P3
= 9 × 9 × 8 × 7
= 4536
Thus, there are 4536, 4-digit numbers which can be formed by using the digits 0 to 9.

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

Ans.
Number of given digits = 6
Number of digits in each number = 3
Since, each number is even, so last digit may be 2, 4 or 6.
If no digit is repeated, then Number of ways to fill remaining two places of 3-digit numbers = 5P2
= 5! /(5 – 2)!
= 5! / 3!
= 20
Number of ways to arrange 3 even numbers at one’s place = 3
Total formed 3-digit even numbers = 20 × 3 = 60
Thus, there are 60, 3-digit even numbers.

Q.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?

Ans.
Number of given digits = 5
Number of digits in each number = 4
Number of 4-digit numbers = 5P4
= 5!
= 120
Now, we find 4 digit even numbers.
Only 2 numbers are possible at units place (2, 4) as we need even number.
If no digit is repeated, then number of ways to fill remaining three places of 4-digit numbers = 4P3
= 4! /(4 – 3)!
= 4! / 1!
= 24
Number of ways to arrange 2 even numbers at one’s place = 2
Total formed 4-digit even numbers = 24 x 2
= 48
Thus, there are 48, 4 digit even numbers.

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

Ans.

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Number}\text{of persons in committee}=8\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Number}\text{of chairman to be selected}=\text{1}\\ \mathrm{Number}\text{of vice chairman to be selected}=\text{1}\\ \text{Number of persons to be selected for both posts}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{2}\\ \text{Number of ways to select a chairman and vice chairman}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\text{\hspace{0.17em}}}^{8}{\mathrm{P}}_{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{8!}{\left(8-2\right)!}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{8×7×6!}{6!}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=56\\ \mathrm{Thus},\text{​ the number of ways to choose both persons is 56.}\end{array}$

Q.6

$\mathbf{Find}\mathbf{}\mathbf{n}\mathbf{}\mathbf{if}{\mathbf{}}^{\mathbf{n}\mathbf{–}\mathbf{1}}{\mathbf{P}}_{\mathbf{3}}\mathbf{}\mathbf{:}{\mathbf{}}^{\mathbf{n}}{\mathbf{P}}_{\mathbf{4}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{}\mathbf{:}\mathbf{}\mathbf{9}$

Ans.

$\begin{array}{l}\mathrm{We}\text{ are given:}\\ {\text{ }}^{\text{n}-1}{\text{P}}_{3}\text{}:{}^{\text{n}}{\text{P}}_{4}=1:9\\ ⇒\text{ }\frac{{}^{\text{n}-1}{\text{P}}_{3}}{{}^{\text{n}}{\text{P}}_{4}}=\frac{1}{9}\\ ⇒\text{ }\frac{\left\{\frac{\left(\text{n}-1\right)!}{\left(\text{n}-1-3\right)!}\right\}}{\left\{\frac{\text{n}!}{\left(\text{n}-4\right)!}\right\}}=\frac{1}{9}\\ ⇒\text{ }\frac{\left(\text{n}-1\right)!}{\left(\text{n}-1-3\right)!}×\frac{\left(\text{n}-4\right)!}{\text{n}!}=\frac{1}{9}\\ ⇒\text{ }\frac{\left(\text{n}-1\right)!}{\left(\text{n}-4\right)!}×\frac{\left(\text{n}-4\right)!}{\text{n}\left(\text{n}-1\right)!}=\frac{1}{9}\\ ⇒\text{ }\frac{1}{\text{n}}=\frac{1}{9}\\ ⇒\text{ n}=9\end{array}$

Q.7

$\begin{array}{l}\mathbf{\text{Find r if}}\\ \left(\text{i}\right){\mathbf{\text{ }}}^{\mathbf{\text{5}}}{\mathbf{\text{P}}}_{\mathbf{\text{r}}}{\mathbf{\text{= 2 }}}^{\mathbf{\text{6}}}{\mathbf{\text{P}}}_{\mathbf{\text{r}}\mathbf{–}\mathbf{\text{1}}}\\ \left(\text{ii}\right){\mathbf{\text{ }}}^{\mathbf{\text{5}}}{\mathbf{\text{P}}}_{\mathbf{\text{r}}}{\mathbf{\text{= }}}^{\mathbf{\text{6}}}{\mathbf{\text{P}}}_{\mathbf{\text{r–1}}}\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right){\text{ }}^{5}{\mathrm{P}}_{\mathrm{r}}={2}^{6}{\mathrm{P}}_{\mathrm{r}-1}\\ ⇒\text{ }\frac{5!}{\left(5-\mathrm{r}\right)!}=2\cdot \frac{6!}{\left(6-\mathrm{r}+1\right)!}\\ ⇒\text{ }\frac{5!}{\left(5-\mathrm{r}\right)!}=2\cdot \frac{6!}{\left(7-\mathrm{r}\right)!}\\ ⇒\text{ }\frac{5!}{\left(5-\mathrm{r}\right)!}=2\cdot \frac{6.5!}{\left(7-\mathrm{r}\right)\left(6-\mathrm{r}\right)\left(5-\mathrm{r}\right)!}\\ ⇒\text{ }\frac{1}{1}=2\cdot \frac{6}{\left(7-\mathrm{r}\right)\left(6-\mathrm{r}\right)}\\ ⇒\text{ }\left(7-\mathrm{r}\right)\left(6-\mathrm{r}\right)=12\\ ⇒\text{ }42-13\mathrm{r}+{\mathrm{r}}^{2}=12\\ ⇒\text{ }30-13\mathrm{r}+{\mathrm{r}}^{2}=0\\ ⇒\text{ }\left(10-\mathrm{r}\right)\left(3-\mathrm{r}\right)=0\\ ⇒\text{ }\left(10-\mathrm{r}\right)\left(3-\mathrm{r}\right)=0\\ ⇒\text{ }\mathrm{r}=10,3\\ ⇒\text{ }\mathrm{r}=3\left[\mathrm{Neglecting}\text{ }10,\mathrm{because}\text{ }0<\mathrm{r}\le \mathrm{n}.\right]\\ \left(\mathrm{ii}\right){\text{ }}^{5}{\mathrm{P}}_{\mathrm{r}}{=}^{6}{\mathrm{P}}_{\mathrm{r}-1}\\ ⇒\text{ }\frac{5!}{\left(5-\mathrm{r}\right)!}=\frac{6!}{\left(6-\mathrm{r}+1\right)!}\\ ⇒\text{ }\frac{5!}{\left(5-\mathrm{r}\right)!}=\frac{6!}{\left(7-\mathrm{r}\right)!}\\ ⇒\text{ }\frac{5!}{\left(5-\mathrm{r}\right)!}=\frac{6.5!}{\left(7-\mathrm{r}\right)\cdot \left(6-\mathrm{r}\right)\cdot \left(5-\mathrm{r}\right)!}\\ ⇒\text{ }1=\frac{6}{\left(7-\mathrm{r}\right)\left(6-\mathrm{r}\right)}\\ ⇒\text{ }\left(7-\mathrm{r}\right)\left(6-\mathrm{r}\right)=6\\ ⇒\text{ }42-13\mathrm{r}+{\mathrm{r}}^{2}=6\\ ⇒\text{ }36-13\mathrm{r}+{\mathrm{r}}^{2}=0\\ ⇒\text{ }\left(9-\mathrm{r}\right)\left(4-\mathrm{r}\right)=0\\ ⇒\text{ }\mathrm{r}=9,4\\ ⇒\text{ }\mathrm{r}=4\text{ }\left[\text{Neglecting}9,\text{ because}0<\mathrm{r}\le \mathrm{n}.\right]\end{array}$

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

Ans.

$\begin{array}{l}\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Number}\text{of letters in ‘EQUATION’}=8\\ \mathrm{Number}\text{\hspace{0.17em}of words using each letter once}={\text{\hspace{0.17em}}}^{8}{\mathrm{P}}_{8}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}$

$\begin{array}{l}=\frac{8!}{\left(8-8\right)!}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{8!}{0!}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8×7×6×5×4×3×2×1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=40320\\ \mathrm{Thus},\text{number of words formed by using letters of ‘Equation’}\\ \text{once is 40320.}\end{array}$

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

(i) 4 letters are used at a time,
(ii) all letters are used at a time,
(iii) all letters are used but first letter is a vowel?

Ans.

$\begin{array}{l}\mathrm{Number}\text{of letters in ‘MONDAY’}=6\\ \left(\mathrm{i}\right)\text{If 4 letters are used at a time.}\\ \mathrm{Number}\text{of 4-letter words that can be formed from the letters}\\ \text{of ‘MONDAY’, then}\\ \text{Number of words}={\text{\hspace{0.17em}}}^{6}{\mathrm{P}}_{4}\\ =\frac{6!}{\left(6-4\right)!}\\ =\frac{6!}{2!}\\ =6×5×4×3\\ =360\end{array}$

$\begin{array}{l}\mathrm{Thus},\text{number of words formed by using 4 letters is 360.}\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}}\mathrm{all}\text{letters are used at a time.}\\ \text{Number of words}={\text{\hspace{0.17em}}}^{6}{\mathrm{P}}_{6}\\ \text{\hspace{0.17em}}=\frac{6!}{\left(6-6\right)!}\\ \text{\hspace{0.17em}}=\frac{6!}{0!}\\ \text{\hspace{0.17em}}=6×5×4×3×2×1\\ \text{\hspace{0.17em}}=720\\ \left(\mathrm{iii}\right)\mathrm{Number}\text{of vowels}=2\\ \mathrm{Number}\text{of ways to select first letter}\\ ={\text{\hspace{0.17em}}}^{2}{\mathrm{P}}_{1}\\ =\frac{2!}{\left(2-1\right)!}\\ =2\\ \mathrm{Number}\text{​ of ways to select remaining 5 letters}\\ ={\text{\hspace{0.17em}}}^{5}{\mathrm{P}}_{5}\\ =\frac{5!}{\left(5-5\right)!}\\ =5×4×3×2×1\\ =120\\ \mathrm{So},\text{\hspace{0.17em}}\mathrm{total}\text{number of words}=2×120\\ =240\end{array}$

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

Ans.

$\begin{array}{l}\mathrm{Number}\text{of letters in ‘MISSISSIPPI’}=11\\ \mathrm{Number}\text{of I’s}=4\\ \mathrm{Number}\text{of S’s}=4\\ \mathrm{Number}\text{of P’s}=2\end{array}$

$\begin{array}{l}\mathrm{Number}\text{of permutations}=\frac{11!}{4!4!2!}\\ =\frac{11×10×9×8×7×6×5×4!}{4!4!2!}\\ =\frac{11×10×9×8×7×6×5}{4×3×2×1×2×1}\\ =34650\\ \mathrm{When}\text{4 I’s come together.}\\ \mathrm{Then},\text{\hspace{0.17em}number of words}=\frac{8!}{4!2!}\\ =\frac{8×7×6×5×4!}{4!×2×1}\\ =840\\ \mathrm{So},\text{required distinct permutations}=34650-840\\ =33810\end{array}$

Q.11 In how many ways can the letters of the word PERMUTATIONS be arranged if the

(i) words start with P and end with S,
(ii) vowels are all together,
(iii) there are always 4 letters between P and S?

Ans.

$\begin{array}{l}\mathrm{Number}\text{of letters in ‘PERMUTATIONS’}=12\\ \text{ }\mathrm{Number}\text{of T’s}=\text{2}\\ \left(\text{i}\right)\text{ }\mathrm{When}\text{words start with P and end with S,}\\ \text{ then number of the words formed}=\frac{10!}{2!}\\ \text{ }=1814400\\ \left(\mathrm{ii}\right)\text{ }\mathrm{Vowels}\text{}\mathrm{are}\text{all together,}\\ \text{ Number of vowels in PERMUTATIONS}=5\\ \text{ }\mathrm{If}\text{all vowels are considered as a single letter, then}\\ \text{ number of letters to be arranged}=12-5+1\\ \text{ }=8\\ \text{ then number of the words formed}=\frac{8!}{2!}×{\text{ }}^{5}{\text{P}}_{5}\\ \text{ }=20160×120\\ \text{ }=2419200\\ \left(\mathrm{iii}\right)\text{ If there are always 4 letters between P and S, then}\\ \text{ }\mathrm{number}\text{of letters to be arranged}=12-2\\ \text{ }=10\\ \text{ }\mathrm{Number}\text{repeated letter T}=\text{2}\\ \text{ Number of ways to arrange 10 letters}=\frac{10!}{2!}\\ \text{ }=1814400\\ \text{ }\begin{array}{cccccccccccc}\text{P}& & & & & \text{S}& & & & & & \end{array}\\ \text{ }\begin{array}{ccccccc}1& 2& 3& 4& 5& 6& 7\end{array}\\ \text{ }\mathrm{Total}\text{number of places to shift P and S}=7\\ \text{ P and S can be interchange, so total number}\\ \text{ of ways of arrangement of P and S}=2×7\\ \therefore \text{ }\mathrm{Number}\text{of ways to arrange letters}=14×1814400\\ \text{ }=25401600\end{array}$

## FAQs (Frequently Asked Questions)

### 1. How can Extramarks NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3 help students score better?

Extramarks NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3 can help students score better as it has the correct answers available to students anywhere, anytime on their phone or laptop through Extramarks app and website. The NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3 is perfect for students who need to work on their problem-solving time and who need solutions. NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3 also helps students figure out where they make mistakes and this will help them during the exam. The NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3 also help them with the last minute quick revision. This way, they will be able to save time for chapters and topics they find difficult to understand and answer, and it will help them find a way to answer quickly.

### 2. How can students study productively?

Studying productively means that more work is done in less time, hence saving time for other work and leisure. Students can study effectively and productively if all of their focus is on solving the questions and understanding the questions. Students must first understand the concepts and the questions and learn their method of answering so that they understand the concept holistically. It is important as this will help them to figure out how the questions should be approached and the correct method that should be used to obtain the answer. This will affect their scores positively. Moreover, students must prioritise their health- both mental and physical health. Having hobbies and doing extracurricular activities will help students maintain their health. It will keep them motivated to study and keep their interest alive. It should be noted that students who take breaks sometimes get more work done in less time. Studying for long hours might affect their physical health badly, which is why it is advised to take breaks. This will affect their academic performance in a positive way. And it will also affect their health positively and lead to personal growth and skill development.