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

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NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations

Free PDF download of NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3 (Ex 7.3) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 11 Maths Chapter 7 Permutations and Combinations Exercise 7.3 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.


Class:

NCERT Solutions for Class 11

Subject:

Class 11 Maths

Chapter Name:

Chapter 7 - Permutations and Combinations

Exercise:

Exercise - 7.3

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

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

Exercise 7.3

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

Ans:

We have to form a $3$-digit numbers using digits $1-9$. That means, the order of the digits matter.

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

So, required number of $3$-digit numbers 

$={{\ }^{9}}{{P}_{3}}$

 $\Rightarrow \frac{9!}{(9-3)!}=\frac{9!}{6!}$

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

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

Hence, there are $504$  $3$-digit numbers formed.


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

Ans:

The thousands place of the $4$-digit number is to be filled with any of the digits from $1$ to $9$ as the digit $0$ cannot be included. Therefore, the number of ways in which thousands place can be filled is $9$. The hundreds, tens, and units place can be filled by any of the digits from $0$ to $9$. However, the digits cannot be repeated in the $4$-digit numbers and thousands place is already occupied with a digit. The hundreds, tens, and units place is to be filled by the remaining $9$-digits. Therefore, there will be as many such $3$-digit numbers as there are permutations of $9$ different digits taken $3$ at a time.

Number of such $3$-digit numbers $={{\ }^{9}}{{P}_{3}}$

$\Rightarrow \frac{9!}{(9-3)!}=\frac{9!}{6!}$

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

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

Thus, by multiplication principle, the required number of $4$-digit numbers are $9\times 504=4536$.


3. How many $3$-digit even numbers can be made using the digits \[\mathbf{1},\text{ }\mathbf{2},\text{ }\mathbf{3},\text{ }\mathbf{4},\text{ }\mathbf{6},\text{ }\mathbf{7},\] if no digit is repeated?

Ans:

We have to form $3$-digit even numbers using the given six digits, \[1,\text{ }2,\text{ }3,\text{ }4,\text{ }6,\] ,and $7$, without repeating the digits. Then, units can be filled in $3$ ways by any of the digits, $2$, $4$ or $6$. Since the digits cannot be repeated in the $3$-digit numbers and units place is already occupied with a digit (which is even), the hundreds and tens place is to be filled by the remaining $5$-digits. Therefore, the number of ways in which hundreds and tens place can be filled with the remaining $5$-digits is the permutation of $5$ different digits taken $2$ at a time.

So, the number of ways of filling hundreds and tens place

${{\Rightarrow }^{5}}{{P}_{2}}$

$\Rightarrow \frac{5!}{(5-2)!}=\frac{5!}{3!}$

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

$\Rightarrow 5\times 4=20$

Thus, by multiplication principle, the required number of $3$-digit numbers are $3\times 20=60$.


4. Find the number of $4$-digit numbers that can be formed using the digits \[\mathbf{1},\text{ }\mathbf{2},\text{ }\mathbf{3},\text{ }\mathbf{4},\text{ }\mathbf{5}\] if no digit is repeated. How many of these will be even?

Ans:

We have to form $4$-digit numbers using the digits, $1,\ 2,\ 3,\ 4,$ and $5$. 

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

That is, required number of $4$-digit numbers

${{\Rightarrow }^{5}}{{P}_{4}}$

$\Rightarrow \frac{5!}{(5-4)!}=\frac{5!}{1!}$

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

$\Rightarrow 120$

Among the $4$-digit numbers formed by using the digits, $1,\ 2,\ 3,\ 4,\ 5$ even numbers end with either $2$ or $4$. The number of ways in which units place is filled with digits is $2$. Since the digits are not repeated and the units place is already occupied with a digit (which is even), the remaining places are to be filled by the remaining $4$-digits. Therefore, the number of ways in which the remaining places can be filled is the permutation of $4$ different digits taken $3$ at time. 

Number of ways of filling the remaining places

${{\Rightarrow }^{4}}{{P}_{3}}=\frac{4!}{(4-3)!}$

$\Rightarrow 4!$

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

Thus, by multiplication principle, the required number of even numbers is $24\times 2=48$.                                                                                   


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?

Ans:

We have to choose a chairman and a vice chairman from a committee of $8$ persons in such a way that one person cannot hold more than one position. Here, the number of ways of choosing a chairman and a vice chairman is the permutation of $8$ different objects taken $2$ at a time. 

Thus, required number of ways

${{\Rightarrow }^{8}}{{P}_{2}}=\frac{8!}{(8-2)!}$

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

$\Rightarrow 56$


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

Ans:

Here,

$^{n-1}{{P}_{3}}{{:}^{n}}{{P}_{4}}=1:9$

$\Rightarrow \frac{^{n-1}{{P}_{3}}}{^{n}{{P}_{4}}}=\frac{1}{9}$

$\Rightarrow \frac{\left[ \frac{(n-1)!}{(n-1-3)!} \right]}{\left[ \frac{n!}{(n-4)!} \right]}=\frac{1}{9}$

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

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

\[\Rightarrow \frac{1}{n}=\frac{1}{9}\]

Hence, $n=9$.


7. Find $r$ if:

(i). $^{5}{{P}_{r}}={{2}^{6}}{{P}_{r-1}}$

Ans:

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

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

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

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

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

$\Rightarrow r=3\text{ or }r=10$

It is known that $^{n}{{P}_{r}}=\frac{n!}{(n-r)!}$, where $0\le r\le n$.

$\Rightarrow 0\le r\le 5$

Hence, $r=3$.

(ii). $^{5}{{P}_{r}}{{=}^{6}}{{P}_{r-1}}$

Ans: 

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

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

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

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

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

It is known that $^{n}{{P}_{r}}=\frac{n!}{(n-r)!}$, where $0\le r\le n$.

$\Rightarrow 0\le r\le 5$

Hence, $r=4$.


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:

There are $8$ different letters in the word EQUATION. Therefore, the number of words that can be formed using all the letters of the word EQUATION, using each letter exactly once, is the number of permutations of $8$ different objects taken $8$ at a time, which is $^{8}{{P}_{8}}=8!$.

Thus, required number of words that can be formed is

$8!=40320$


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,

Ans:

There are 6 different letters in the word MONDAY.

Number of $4$-letter words that can be formed from the letters of the word MONDAY, without repetition of letters, is the number of permutations of $6$ different objects taken $4$ at a time, which is $^{6}{{P}_{4}}$.

Thus, required number of words that can be formed using $4$ letters at a time are $^{6}{{P}_{4}}=\frac{6!}{(6-4)!}$

\[\Rightarrow \frac{6\times 5\times 4\times 3\times 2!}{2!}=6\times 5\times 4\times 3\]

$\Rightarrow 360$

(ii). all letters are used at a time,

Ans:

Number of words that can be formed by using all the letters of the word MONDAY at a time is the number of permutations of $6$ different objects taken $6$ at a time, which is $^{6}{{P}_{6}}=6!$.

Thus, required number of words that can be formed when all letters are used at a time

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

$\Rightarrow 720$

(iii). all letters are used but first letter is a vowel.

Ans:

In the given word, there are $2$ different vowels, which have to occupy the rightmost place of the words formed. This can be done only in $2$ ways. Since the letters cannot be repeated and the rightmost place is already occupied with a letter (which is a vowel), the remaining five places are to be filled by the remaining $5$ letters. This can be done in \[5!\] ways.

Thus, in this case, required number of words that can be formed is \[5!\times 2=120\times 2\]

$\Rightarrow 240$


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

Ans: 

In the given word MISSISSIPPI, I appear $4$ times, S appears $4$ times, P appears $2$ times, and M appears just once. Therefore, number of distinct permutations of the letters in the given word

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

$\Rightarrow \frac{11\times 10\times 9\times 8\times 7\times 6\times 5\times 4!}{4!\times 4\times 3\times 2\times 1\times 2\times 1}$

$\Rightarrow \frac{11\times 10\times 9\times 8\times 7\times 6\times 5}{4\times 3\times 2\times 1\times 2\times 1}$

$\Rightarrow 34650$

There are $4$ I’s in the given word. When they occur together, they are treated as single object [IIII] for the time being. This single object together with the remaining $7$ objects will account for $8$ objects.

These $8$ objects in which there are $4$ S’s, and $2$ P’s can be arranged in $\frac{8!}{4!2!}$ ways that means, $840$ ways.

Number of arrangements where all I’s occurred together is in $840$ ways.

Thus, number of distinct permutations of the letters in MISSISSIPPI in which four IS do not come together \[34650-840=33810\].


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,

Ans:

In the word PERMUTATIONS, there are $2$ T’s, and all the other letters appear only once.

If P and S are fixed at the extreme ends (P at the left end and S at the right end), then $10$ letters are left.

Hence, in this case, required number of arrangements

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

(ii). Vowels are all together,

Ans:

There are $5$ vowels in the given word, each appearing only once.

Since they have to always occur together, they are treated as a single object for the time being. This single object together with the remaining $7$ objects will account for $8$ objects, in total. 

These $8$ objects in which there are $2$ T’s can be arranged in $\frac{8!}{2!}$ ways.

Corresponding to each of these arrangements, the $5$ different vowels can be arranged in $5!$ ways. Therefore, by multiplication principle, required number of arrangements in this case

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

(iii). There are always $4$ letters between P and S?

Ans:

The letters have to be arranged in such a way that there are always $4$ letters between P and S. Therefore, in a way, the places of P and S are fixed. The remaining $10$ letters in which there are $2$ T’s can be arranged in $\frac{10!}{2!}$ ways.

Since,Possible places of P & S are 1 & 6, 2 & 7, 3 & 8, 4 & 9, 5 & 10, 6 & 11, 7 & 12. Then there can be $4$ letters between P & S.Also P & S can be interchanged as it won’t effect the number of letters between them. 

So, the letters P and S can be placed such that there are $4$ letters between them in $2\times 7=14$ ways.Therefore, by multiplication principle, required number of arrangements in this case

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


NCERT Solutions for Class 11 Maths Chapters


List of Topics Covered Under Exercise 7.3

Exercise 7.3 is based on the following topics:

1. Permutations

The permutation is a different way of arranging objects in a definite order. It can also be defined as the rearrangement of objects in a linear order of an already ordered set. nPr is the symbol used to denote the no. of permutations.

Here, 

n = total no of objects

r = no of objects selected


2. Derivation of the formula for nPr

As we know a permutation involves the selection of r distinct objects without replacing from n items. Here, the order is important. 

By the fundamental counting principle, 

P (n, r) = n . (n-1) . (n-2) . (n-3)…… (n-(r-1)) ways.

This can be expressed as:

P (n, r) = n.(n-1).(n-2). (n-3) …. (n-r+1)

To simplify, let’s use factorial notations. 

Lets Multiply and divide the above expression by (n - r) ... 3 · 2 · 1

We have,

nPr = [n (n - 1) (n - 2) ...(n - r + 1) (n - r) ... 3 · 2 · 1] / [(n - r) ... 3 · 2 · 1]

nPr = n! / (n - r)!


3. In this exercise we will also learn permutations when all the objects are not distinct objects. 


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

Opting for the NCERT solutions for Ex 7.3 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 7.3 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 11 students who are thorough with all the concepts from the Subject Permutations and Combinations textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 11 Maths Chapter 7 Exercise 7.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 11 Maths Chapter 7 Exercise 7.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 11 Maths Chapter 7 Exercise 7.3 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.


Benefits of choosing Vedantu for NCERT Solutions Permutations and Combinations Exercise 7.3

  • NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3 is a well-written resource that provides a thorough understanding of each topic related to Permutations and Combinations. 

  • Students will be able to easily grasp any complex concepts with these properly structured solutions.

  • In this exercise, there are solutions available that are sufficient to deliver concepts in an easier way. They elaborate all steps clearly with examples to impart an accurate knowledge of each concept.

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

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

To find the four-digit number (digits does not repeat)


Now we will have 4 places where 4 digits are to be put.


So, in thousand’s position = There are 9 ways as 0 will not be at thousand’s position = 9 ways


In hundredth’s position = There are 9 digits which are to be filled as 1 digit is already taken = 9 ways


In ten’s position = There are now 8 digits that are to be filled as 2 digits are previously taken = 8 ways


At unit’s place = There are 7 digits that can be filled = 7 ways


Total Number of methods to fill the four places = 9 × 9 × 8 × 7 = 4536 ways.


So a total of 4536 four-digit figures can be there with no numbers repeated.

2. What are permutations and combinations?

In mathematics, permutation relates to the act of organizing all the members of a set into some succession or order, or if the set is already arranged, rearranging its components, a process called permuting. Permutations happen, in more or less noticeable ways, in almost all areas of math. They usually arise when various orderings on specific finite sets are recognised.

 

The combination is a way of choosing items from a group, such that (unlike permutations) the order of preference does not matter. In smaller events, it is possible to figure the number of combinations. Combination introduces the combination of n items taken k at a time without recurrence. To refer to combinations in which recurrence is allowed, the terms k-selection or k-combination with the return are often used.

3. What is the difference between permutations and combinations?

A permutation is used for the list of data (where the order of the data matters) and the combination is used for a collection of data (where the order of data doesn’t matter).


Permutations:

  • Arranging people, digits, numbers, alphabets, letters, and colours.

  • Picking a team captain, player or pitcher, and shortstop from a collection.

  • Picking two favourite colours, in order, from a colour brochure.

  • Picking first, second and third place winners.

Combinations:

  • Selection of menu, food, clothes, subjects, team.

  • Picking three team members from a group.

  • Picking two colours from a colour brochure.

  • Picking three winners.

4. How do NCERT answers of Vedantu help me in scoring higher marks?

Our innovative gaining knowledge of methodology in conjunction with the smart and easy study strategies will make your studying methods extra fun, interesting, interactive and in a nicely deliberate way. Our NCERT Solutions for Class eleven Maths have been carefully designed that will help you increase your information base which will, in the end, improve your retention rate.

 

All the essential and necessary concepts have been included through us in an effort to make your studying much less complicated for examination preparation. They have been crafted from the exam point of view. Our solutions are framed by our experts and they have included each and every part of the concept additionally the exercise questions that are covered at the end of the chapter.

5. What do you understand by factorial notation mentioned in Chapter 7 of Class 11 Maths?

Factorial notation (n!) is the number that you will get after multiplying the first n natural numbers. The product that you will get multiplying 1 x 2 x 3 x 4 x 5..(n - 1)n is known as n! Or factorial notation. For example, 4! will be calculated as 1x2x3x4 = 24. Likewise if you have to calculate 5! - 2!, then you first calculate 5! = 1x2x3x4x5 = 120, then you calculate 2! = 1 x 2 = 2. Now, you subtract 120-2 = 118. 

6. Is 3! + 4! = 7!? Determine using the concept used in Chapter 7 of Class 11 Maths.

Let’s have a look at the answer step wise. 3! = 1 x 2 x 3, which is equal to 6; 4! = 1 x 2 x 3 x 4 = 24; 7! = 1 x 2 x 3 x 4 x 5 x 6 x 7, which is equal to 5040. 3! + 4! = 6 + 24 = 30. We can clearly see that 3! + 4! is 30, while 7! is 5040. Therefore,  we can establish here that 3! + 4! is not equal to 7!. 

7. Which is the best NCERT Solution for Exercise 7.3 of Chapter 7 of Class 11 Maths?

The NCERT Solutions for Exercise 7.3 of Chapter 7 of Class 11 Maths, that you find at Vedantu is the best NCERT Solution manual that you can lay your hands on for solving all the questions based on the respective topic. This solution manual provides answers to all questions from the exercise. Students who solve this exercise from Vedantu’s NCERT Solutions will find the exercise so easy and will be able to finish it in no time. The PDF’s of the Solutions or any study material provided by Vedantu can be downloaded absolutely free of cost.

8. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6, if the digits can be repeated? Solve by the method used in  Chapter 7 of Class 11 Maths.

To get 3-digit even numbers, we know that in the one place of these digits, there should be even numbers, which can be filled in 3 ways (either 2, 4 or 6). The tens and hundreds place can be filled in 6 ways each as the digits can be repeated. So going by the multiplication rule, the number of 3-digit even numbers formed from digits 1, 2, 3, 4, 5, and 6 would be 3 x 6 x 6 = 108. 

9. What is the formula to calculate combinations as mentioned in Chapter 7 of Class 11 Maths?

The formula that you use to calculate the total number of combinations of ‘n’ number of different things taken ‘r’ at a time, indicated by nCr, is given by nCr = n!r! (n - r)!, where 0 r n. By using this simple formula, you can calculate combinations. If you are looking for more solutions from this chapter, check out the NCERT Solutions for Chapter 7 of Class 11 Maths on Vedantu.