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NCERT Solutions for Class 11 Maths Chapter 6 Permutations And Combinations Ex 6.4

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NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

Free PDF download of NCERT Solutions for Class 11 Maths Chapter 6 Exercise 6.4 (Ex 6.4) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 11 Maths Chapter 6 Permutations and Combinations Exercise 6.4 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.


Exercise 6.4 of Chapter 6- 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 6 - Permutations and Combinations

Exercise-6.4

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 1 All Exercises

Exercise

Number of Questions

Exercise 6.1

6 Questions and Solutions

Exercise 6.2

5 Questions and Solutions

Exercise 6.3

11 Questions and Solutions

Miscellaneous Exercise

11 Questions and Solutions

CBSE Class 11 Maths Chapter 6 Other Study Materials

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Exercise 6.4

Opting for the NCERT solutions for Ex 6.4 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 6.4 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 6 Exercise 6.4 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 6 Exercise 6.4, 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. 

Do not delay any more. Download the NCERT solutions for Class 11 Maths Chapter 6 Exercise 6.4 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.


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FAQs on NCERT Solutions for Class 11 Maths Chapter 6 Permutations And Combinations Ex 6.4

1. Where can I find stepwise NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations as per CBSE 2025–26 pattern?

You can access the stepwise NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations on this page, aligned with the latest CBSE 2025–26 syllabus. All answers are provided in the official NCERT format, covering each exercise such as 6.1, 6.2, 6.3, and the Miscellaneous Exercise in a clear, textbook-based manner.

2. How do I download the NCERT Solutions for Class 11 Maths Chapter 6 PDF for all exercises?

To download the complete NCERT Solutions PDF for Class 11 Maths Chapter 6, use the provided link on this page. The PDF includes all questions and correct answers for Exercises 6.1, 6.2, 6.3, and the Miscellaneous set, following the stepwise CBSE-approved method for each problem.

3. What is the correct method to solve Exercise 6.1 of Class 11 Maths Chapter 6 as per the NCERT format?

The correct method to solve Exercise 6.1 of Class 11 Maths Chapter 6 involves identifying permutations and solving each question step-by-step as per the NCERT pattern. Ensure you apply the appropriate formula for permutation calculations and write each answer with proper justification according to CBSE-approved NCERT answer format.

4. Are the NCERT Solutions for Class 11 Maths Chapter 6 Miscellaneous Exercise provided here fully solved and CBSE approved?

Yes, the NCERT Solutions for the Miscellaneous Exercise of Class 11 Maths Chapter 6 presented here are fully solved, with step-by-step reasoning for each question. All solutions strictly follow the latest CBSE and NCERT guidelines for the 2025–26 academic year.

5. Do the provided NCERT Solutions help in answering intext and back questions from Class 11 Maths Chapter 6, including tricky applications?

Absolutely. These NCERT Solutions address all intext, back, and application-based questions from Class 11 Maths Chapter 6 with detailed stepwise explanations. The answers use NCERT’s official answer structure, ensuring clarity for both conceptual and calculation-based questions.

6. Are the answers given for Permutations and Combinations in Chapter 6 suitable for CBSE board exam preparation?

Yes, every answer for Permutations and Combinations in Chapter 6 is prepared using the NCERT stepwise answer method, which is the recommended structure for CBSE 2025–26 board exams. The solutions help you score maximum marks when reproduced as per the given format.

7. Is every question from Chapter 6, including Ex 6.2 and Ex 6.3, solved as per latest NCERT guidelines?

All questions from Chapter 6, including Exercise 6.2 and Exercise 6.3, are solved using detailed steps strictly based on the CBSE and NCERT guidelines. The explanations clarify the logic for each calculation and provide correct answers in the official textbook answer style.

8. Why is stepwise explanation important in NCERT Solutions for Class 11 Maths Chapter 6?

Stepwise explanations in NCERT Solutions help students understand each stage of solving a question in Permutations and Combinations. This method matches the CBSE answer marking scheme, ensures conceptual clarity, and enables students to secure full marks in both school and board-level assessments.

9. Can I use these NCERT answer keys for clarifying doubts while solving miscellaneous and higher-order questions from Chapter 6?

Yes, the NCERT answer keys provided here offer clear, step-by-step clarification for every question, including those in the Miscellaneous section and higher-order application questions in Chapter 6. Students are guided through each type of permutation and combination problem with textbook-accurate reasoning.

10. Are these NCERT Solutions for Class 11 Maths Chapter 6 valid for the upcoming 2025–26 CBSE examination?

All NCERT Solutions for Class 11 Maths Chapter 6 here are completely up to date, following the current NCERT textbook and CBSE syllabus for the 2025–26 session. This ensures that you practice and answer according to the latest examination requirements.

11. What should I do if two answers to a permutation question in Exercise 6.3 seem different?

If you find different answers for the same permutation problem in Exercise 6.3, carefully check the method used and ensure it matches the NCERT-recommended approach. Always follow the stepwise explanation provided here, which uses the correct formula and logic as per NCERT and CBSE guidelines for Class 11 Maths Chapter 6.

12. How should I write my answers in the exam to match the NCERT solutions for Permutations and Combinations?

When writing answers in your CBSE exam for Permutations and Combinations, follow the stepwise structure used in the NCERT solutions. Present your calculation, state the formulas used, show substitution of values, and justify your answer with a concluding statement, just as shown in the provided correct CBSE answer format.