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# NCERT Solutions for Class 11 Maths Chapter 7: Permutations and Combinations - Exercise 7.4

Last updated date: 23rd May 2024
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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 7.4 (Ex 7.4) 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 7.4 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 - 7.4 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes

### Exercise 7.4 of Chapter 7- Permutations and Combinations is based on the following topic.

Combination

Combinations are defined as the selection of things from a given set of things. Here we just select the thing but not intend to arrange things like we do in permutations. The no. of unique combinations or r-selections out of a group of n objects by nCr.

Combinations Formula

The combinations formula is used to easily determine the no. of possible different groups of ‘r’ objects each, which can be formed by the given ‘n’ different objects. The Combinations formula is expressed as the n factorial, divided by the product of the factorial of r factorial, and the factorial of the difference of n and r.

Combinations Formula: nCr = n!(n – r)! r!

Note: A permutation is a process of arranging the objects in order. Combinations are just a way of selecting the objects from a group of objects where the order of the selection of objects does not matter. We can say permutations are ordered combinations.

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

Exercise-7.4

1. If $^{n}{{C}_{8}}{{=}^{n}}{{C}_{2}}$, find $^{n}{{C}_{2}}$.

Ans: Since, we know that $^{n}{{C}_{a}}{{=}^{n}}{{C}_{b}}$

$\Rightarrow a=b$

Or, $n=a+b$

Here n=8+2=10

Therefore,

$^{10}{{C}_{2}}=\dfrac{10!}{2!(10-2)!}$

$\Rightarrow \dfrac{10!}{2!8!}=\dfrac{10\times 9\times 8!}{2\times 1\times 8!}$

Hence, $^{n}{{C}_{2}}=45$.

2. Determine $n$ if

(i) $^{2n}{{C}_{3}}:{{\text{ }}^{n}}{{C}_{3}}=12:1$

Ans: $\dfrac{^{2n}{{C}_{3}}}{^{n}{{C}_{3}}}=\dfrac{12}{1}$

$\Rightarrow \dfrac{(2n)!}{3!(2n-3)!}\times \dfrac{3!(n-3)!}{n!}=\dfrac{12}{1}$

$\Rightarrow \dfrac{(2n)(2n-1)(2n-2)(2n-3)!}{(2n-3)!}\times \dfrac{(n-3)!}{n(n-1)(n-2)(n-3)!}=12$

Therefore,

$\Rightarrow \dfrac{2(2n-1)(2n-2)}{(n-1)(n-2)}=12$

$\Rightarrow \dfrac{4(2n-1)(n-1)}{(n-1)(n-2)}=12$

$\Rightarrow 2n-1=3(n-2)$

$\Rightarrow 3n-2n=-1+6$

Hence, $n=5$.

(ii) $^{2n}{{C}_{3}}:{{\text{ }}^{n}}{{C}_{3}}=11:1$

Ans: $\Rightarrow \dfrac{(2n)!}{3!(2n-3)!}\times \dfrac{3!(n-3)!}{n!}=\dfrac{11}{1}$

$\Rightarrow \dfrac{(2n)(2n-1)(2n-2)(2n-3)!}{(2n-3)!}\times \dfrac{(n-3)!}{n(n-1)(n-2)(n-3)!}=11$

$\Rightarrow \dfrac{2(2n-1)(2n-2)}{(n-1)(n-2)}=11$

Therefore,

$\Rightarrow \dfrac{(2n-1)}{(n-2)}=\dfrac{11}{4}$

$\Rightarrow 8n-4=11n-22$

$\Rightarrow 3n=18$

Hence, $n=6$.

3. How many chords can be drawn through $21$ points on a circle?

Ans: For drawing one chord a circle, only $2$ points are required.

To know the number of chords that can be drawn through the given $21$ points on a circle, the number of combinations have to be counted.

Therefore, the chords can be drawn through $21$ points taken $2$ as equal to each chord.

Thus, required number of chords

$^{21}{{C}_{2}}=\dfrac{21!}{2!(21-2)!}$

$\Rightarrow \dfrac{21!}{2!19!}=\dfrac{21\times 20}{2}$

$\Rightarrow 210$

4. In how many ways can a team of $3$ boys and $3$ girls be selected from $5$ boys and $4$ girls?

Ans: The team which is selected is of $3$ boys and $3$ girls from $5$ boys and $4$ girls.

The team of $3$ boys can be selected from $5$ boys in $^{5}{{C}_{3}}$ different ways.

The team of $3$ girls can be selected from $4$ girls in $^{4}{{C}_{3}}$ different ways.

Therefore, by multiplication principle, number of ways in which a team of $3$ boys and $3$ girls can be selected

${{\Rightarrow }^{5}}{{C}_{3}}{{\times }^{4}}{{C}_{3}}=\dfrac{5!}{3!2!}\times \dfrac{4!}{3!1!}$

$\Rightarrow \dfrac{5\times 4\times 3!}{3!\times 2}\times \dfrac{4\times 3!}{3!}=10\times 4$

$\Rightarrow 40$

5. Find the number of ways of selecting $9$ balls from $6$ red balls, $5$ white balls and $5$ blue balls if each selection consists of $3$ balls each colour.

Ans: The number of balls from which we have to select $6$ red balls, $5$ white balls, and $4$ blue balls is $9$ balls. These balls have to be selected in such a manner that each selection of balls consists of $9$ balls from each of the colour.

Here, given we have :

$3$ balls which can be opted from the $6$ red balls in $^{6}{{C}_{3}}$ different ways.

$3$ balls which can be opted from the $5$ white balls in $^{5}{{C}_{3}}$ different ways.

$3$ balls which can be opted from the $5$ blue balls in $^{5}{{C}_{3}}$ different ways.

Thus, after applying the multiplication principle, required number of ways of selecting the $9$ balls will be –

${{\Rightarrow }^{6}}{{C}_{3}}{{\times }^{5}}{{C}_{3}}{{\times }^{5}}{{C}_{3}}=\dfrac{6!}{3!3!}\times \dfrac{5!}{3!2!}\times \dfrac{5!}{3!2!}$

$\Rightarrow \dfrac{6\times 5\times 4\times 3!}{3!3\times 2}\times \dfrac{5\times 4\times 3!}{3!2\times 1}\times \dfrac{5\times 4\times 3!}{3!2\times 1}$

$\Rightarrow 20\times 10\times 10=2000$

6. Determine the number of $5$ card combinations out of a deck of $52$ cards if there is exactly one ace in each combination.

Ans: We have a deck of $52$ cards, which contains $4$ aces. When we make a combination, $5$ cards should be made in such a manner that there is exactly one ace.

Now, $1$ ace can be opted in $^{4}{{C}_{3}}$ different ways and the cards which are left, out of them $4$ cards can be opted out of the $48$ cards in $^{48}{{C}_{4}}$ different ways.

Therefore, after applying multiplication principle, the required number of $5$ card combinations will be –

${{\Rightarrow }^{48}}{{C}_{4}}{{\times }^{4}}{{C}_{1}}=\dfrac{48!}{4!44!}\times \dfrac{4!}{1!3!}$

$\Rightarrow \dfrac{48\times 47\times 46\times 45\times 44!}{44!4!\times 3\times 2\times 1}\times 4!$

$\Rightarrow \dfrac{48\times 47\times 46\times 45}{3\times 2\times 1}$

$\Rightarrow 778320$

7. In how many ways can one select a cricket team of eleven from $17$ players in which only $5$ players can bowl if each cricket team of $11$ must include exactly $4$ bowlers?

Ans: The number of players out of which we have to select is $17$ players and only $5$ players from them are bowlers.

A cricket team of $11$ players need to be opted in such a way that there remains exactly $4$ players which are bowlers.

$4$ bowlers can be opted in $^{5}{{C}_{4}}$ different ways whereas out of the remaining $7$ players can be opted out from the $12$ players in $^{12}{{C}_{7}}$ different ways.

Therefore, after applying multiplication principle, the required number of ways for selecting the cricket team will be –

$^{5}{{C}_{4}}{{\times }^{12}}{{C}_{7}}=\dfrac{5!}{4!1!}\times \dfrac{12!}{7!5!}$

$\Rightarrow 5!\times \dfrac{12\times 11\times 10\times 9\times 8}{5!\times 4\times 3\times 2\times 1}=3960$

8. A bag contains $5$ black and $6$ red balls. Determine the number of ways in which $2$ black and $3$ red balls can be selected.

Ans: There are $5$ black and $6$ red balls in the bag. $2$ black balls can be selected out of $5$ black balls in $^{5}{{C}_{2}}$ ways and $3$ red balls can be selected out of $6$ red balls in $^{6}{{C}_{3}}$ ways.

Thus, by multiplication principle, required number of ways of selecting $2$ black and $3$ red balls

$^{5}{{C}_{2}}{{\times }^{6}}{{C}_{3}}=\dfrac{5!}{2!3!}\times \dfrac{6!}{3!3!}$

$\Rightarrow \dfrac{5\times 4}{2}\times \dfrac{6\times 5\times 4}{3\times 2\times 1}=10\times 20$

$\Rightarrow 200$

9. In how many ways can a student choose a programme of $5$ courses if $9$ courses are available and $2$ specific courses are compulsory for every student?

Ans: There are $9$ courses available out of which, $2$ specific courses are compulsory for every student.

Therefore, every student has to choose $3$ courses out of the remaining $7$ courses. This can be chosen in $^{7}{{C}_{3}}$ ways.

Thus, required number of ways of choosing the programme is

$^{7}{{C}_{3}}=\dfrac{7!}{3!4!}$

$\Rightarrow \dfrac{7\times 6\times 5\times 4!}{3\times 2\times 1\times 4!}=35$

### NCERT Solution Class 11 Maths of Chapter 7 Exercise

 Chapter 7 - Permutations and Combinations Exercises in PDF Format Exercise 7.1 6 Questions & Solutions Exercise 7.2 5 Questions & Solutions Exercise 7.3 11 Questions & Solutions Exercise 7.4 9 Questions & Solutions Miscellaneous Exercise 11 Questions & Solutions

## NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Exercise 7.4

Opting for the NCERT solutions for Ex 7.4 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 7.4 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 Maths 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 7.4 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 7.4, 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.

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## FAQs on NCERT Solutions for Class 11 Maths Chapter 7: Permutations and Combinations - Exercise 7.4

1. Which topic Class 11 Maths Chapter 7 Exercise 7.4 is based upon?

Class 11 Maths Chapter 7 of NCERT book deals with Permutations and Combinations. Among that, Exercise 7.4 of Class 11 Maths Chapter 7 is based on the topic Combinations. Class 11 students can gain in-depth knowledge about the topic by referring to the problems given in the exercise. These exercises are extremely helpful while doing the homework and practising. A student can gain thorough knowledge regarding this topic by solving the problems from these exercises of every chapter of NCERT Class 11 Maths textbook.

2. How many questions are there in the Class 11 Maths Chapter 7 Exercise 7.4 of NCERT book?

The Class 11 Maths Chapter 7 Exercise 7.4 of NCERT book consists of nine questions in total. The chapter deals with Permutations and Combinations. All the solutions to the nine questions have been provided in the solutions PDF provided by Vedantu. The solutions are provided in a step-wise manner so that it can be grasped easily and created by the top-notch Maths experts so that the solutions are accurate.

3. Where can I find the best quality NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.4 online for free?

If you are looking for the best quality NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.4 online, then Vedantu, a leading ed-tech portal in India, is the right place for you. Visit the official website of Vedantu i.e. vedantu.com to access the Class 11 Maths Chapter 7 Exercise 7.4 NCERT Solutions online for free of cost. You can also access the same study materials from Vedantu app as well. For that, you need to download the app on your mobile or smart device.

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You can download Vedantu’s NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.4 by visiting the official website of Vedantu i.e. vedantu.com or by downloading Vedantu app on your mobile/ smart device. All you have to do is visit the site of Vedantu or open the mobile app and then register for yourself. Followingly, you will find the specific study material under the ‘Study Material’ tab.

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Vedantu follows an interesting way to design the NCERT solutions which is suitable for all the students. The subject experts have typically tried to break down problems into smaller chunks and solve them in parts. And also, our faculty focuses on making students understand the concepts thoroughly as well as grasping the methods and formulas to solve the questions. These solutions will improve the students' problem-solving skills and give them the confidence to solve any type of questions in the exam.

6. Do I need to practice all the questions provided in the Class 11 Maths Chapter 7 Exercise 7.4?

Yes. Students should practice all the questions from the Class 11 Maths Chapter 7 Exercise 7.4 provided by Vedantu. This will help the students to understand the concepts and methods clearly. Practising once or twice will give a quick revision of all the concepts and students will know if they have difficulty in understanding any problem or steps to solve. It is recommended for the students not to skip any questions.  For the board exam, everything is important.

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Vedantu provides students with the most accurate NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.4 in a stepwise explanation making it easier for the students to understand the problem-solving method easily and effortlessly. These solutions can be viewed online and can also be downloaded in PDF format from the official website of Vedantu. Students can also access all the material free of cost on the Vedantu app also. The concepts for this chapter are explained in detail very clearly by the Maths experts.

8. Are the NCERT solutions for Class 11 Maths Chapter 7 Exercise 7.4 important from the exam point of view?

Yes. To get high grades in the board exams, all the questions from the class 11 Maths Chapter 7 Exercise 7.4 are important. So, students must practise all the questions from the exercise without skipping any questions to score really good marks. These stepwise solutions provided by the Vedantu are very accurate and are detailed. It will help the students to understand the concepts, methods and analyse the types of questions that would appear in the exam.

9. Is there any need to practice extra questions for Class 11 Maths Chapter 7 Exercise 7.4?

There is no need to practice questions for the Class 11 Maths Chapter 7 Exercise 7.4 from resources other than the NCERT Solutions by Vedantu. The NCERT textbook questions, exemplar questions and examples from the book are highly sufficient to practice. It depends on the student to practice extra questions. Although it is not compulsory. Students can refer to NCERT solutions to clarify their doubts regarding the methods or formulas.