6. How do I decide whether to use a permutation (nPr) or a combination (nCr) for a given problem in Chapter 6?

The choice depends entirely on whether the order of items is important for the outcome.Use Permutation (nPr) when the problem is about arrangements, sequences, or assigning distinct roles (e.g., arranging letters in a word, assigning 1st/2nd/3rd prize). The order creates a new outcome.Use Combination (nCr) when the problem is about selection, forming a group, or choosing a team where the order of selection does not matter (e.g., selecting 3 students from a class of 10).

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NCERT Solutions

Class 11

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Chapter 6 Permutations And Combinations Exercise 6.2

CBSE Class 11 Mathematics Chapter 6 Permutations and Combinations – NCERT Solutions 2025-26

Q: 5. How can I avoid common mistakes when solving permutation and combination questions?

To improve accuracy, watch out for these common errors:Confusing Permutations and Combinations: Always ask if the order of selection matters. If yes, it is a permutation. If no, it is a combination.Factorial Calculation Errors: When dealing with fractions, always cancel common factorial terms first (e.g., 8!/6! = 8×7) before multiplying to avoid large number errors.Mishandling Word Problems: For words with repeated letters, always use the formula n! / (p1! p2! ...), where p represents the frequency of each repeating letter.

Q: 9. Do the problem-solving skills from NCERT Solutions for Class 11 Maths Chapter 6 help in preparing for competitive exams like JEE Main?

Absolutely. Mastering the NCERT concepts is the most critical first step for competitive exams. These solutions build a strong foundation in core logic, such as differentiating permutations from combinations, handling complex restrictions, and applying formulas correctly. This conceptual clarity is essential for tackling the advanced, multi-topic questions that appear in JEE Main and other engineering entrance exams.

Download Free PDF of Permutations and Combinations Exercise 6.2 for Class 11 Maths

You’re now working through Exercise 6.2 from Class 11 Mathematics, focused on permutations and combinations—one of the most practical and scoring chapters in your CBSE syllabus. Here, you’ll tackle arrangements, factorial notation, and the skill of applying nPr in real situations, using the exact stepwise approach expected in board exams.

If you’ve ever typed “exercise 6.2 class 11 permutation and combination” while searching for instant answers, you’re not alone. These NCERT solutions guide you through each question methodically, breaking down arrangement problems with or without repetition, just like actual exam patterns.

With chapter weightage helping fill a key part of Algebra’s 25-mark contribution, a solid grasp here can boost your overall score. Each solution on Vedantu is designed for clarity and accuracy—so you can master permutations confidently and meet every CBSE requirement this year.

Access NCERT Solutions for Class 11 Maths Chapter 6 – Permutations and Combinations

Exercise 6.2

Refer to pages 3-4 for exercise 6.2 in the PDF

1. Evaluate.

(i) $8!$

Ans: A number's factorial is the function that multiplies it by each natural number below it. The product of the first $n$ natural numbers is $n$ factorial, which is written as $n!$ and is given by, $n! = n \times \left( {n - 1} \right)!$.

Now, the given expression is $8!$.

According to the definition of the factorial,

$8! = 8 \times \left( {8 - 1} \right)!$

$8! = 8 \times 7!$

This is further evaluated as,

$ 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$

$ 8! = 40320 $

Therefore, the value of $8!$ is found to be 40320.

(ii) $4! - 3!$

Ans: The product of the first $n$ natural numbers is $n$ factorial, which is written as $n!$ and is given by, $n! = n \times \left( {n - 1} \right)!$.

Now, the given expression is $4! - 3!$.

First, consider and evaluate $4!$.

According to the definition of the factorial,

$ 4! = 4 \times \left( {4 - 1} \right)!$

$ 4! = 4 \times 3!$

This is further evaluated as,

$ 4! = 4 \times 3 \times 2 \times 1$

$ 4! = 24$

Thus, the value of $4!$ is found to be 24.

Now, consider and evaluate for $3!$.

According to the definition of the factorial,

$3! = 3 \times \left( {3 - 1} \right)!$

$3! = 3 \times 2! $

This is further evaluated as,

$ 3! = 3 \times 2 \times 1$

$3! = 6$

Thus, the value of $3!$ is found to be 6.

Finally, consider the expression, $4! - 3!$ and evaluate.

$ 4! - 3! = 24 - 6$

$4! - 3! = 18$

Therefore, the value of the expression on evaluation is found to be 18.

2. Is $4! + 3! = 7!?$

Ans: The product of the first $n$ natural numbers is $n$ factorial, which is written as $n!$ and is given by, $n! = n \times \left( {n - 1} \right)!$.

First, consider and evaluate $4!$.

According to the definition of the factorial,

$ 4! = 4 \times \left( {4 - 1} \right)!$

$4! = 4 \times 3!$

This is further evaluated as,

$ 4! = 4 \times 3 \times 2 \times 1$

$ 4! = 24 $

Thus, the value of $4!$ is found to be 24.

Now, consider and evaluate for $3!$.

According to the definition of the factorial,

$ 3! = 3 \times \left( {3 - 1} \right)!$

$3! = 3 \times 2!$

This is further evaluated as,

$3! = 3 \times 2 \times 1$

$ 3! = 6$

Thus, the value of $3!$ is found to be 6.

Now, consider the expression, $4! + 3!$ and evaluate.

$ 4! + 3! = 24 + 6$

$ 4! + 3! = 30$

Further, consider $7!$.

According to the definition of the factorial,

$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$

$7! = 5040$

Therefore, it is observed that $4! + 3! \ne 7!$.

3. Compute $\dfrac{{8!}}{{6! \times 2!}}$.

Ans: The product of the first $n$ natural numbers is $n$ factorial, which is written as $n!$ and is given by, $n! = n \times \left( {n - 1} \right)!$.

First consider the expression in the numerator, that is, $8!$.

According to the definition of the factorial,

$ 8! = 8 \times \left( {8 - 1} \right)!$

$ 8! = 8 \times 7!$

This is further evaluated as,

$8! = 8 \times 7 \times 6!$

Now, consider the expression in the denominator, that is, $6! \times 2!$.

This expression can also be written in a way that is, $6! \times 2 \times 1$.

Consider the expression that is to be determined and evaluated.

$\dfrac{{8!}}{{6! \times 2!}} = \dfrac{{8 \times 7 \times 6!}}{{6! \times 2 \times 1}}$

$\dfrac{{8!}}{{6! \times 2!}} = \dfrac{{8 \times 7}}{2}$

$\dfrac{{8!}}{{6! \times 2!}} = 28$

Therefore, the value of the expression given on evaluation is found to be 28.

4. $\dfrac{1}{{6!}} + \dfrac{1}{{7!}} = \dfrac{x}{{8!}}$

Ans: The product of the first $n$ natural numbers is $n$ factorial, which is written as $n!$ and is given by, $n! = n \times \left( {n - 1} \right)!$.

The given expression is $\dfrac{1}{{6!}} + \dfrac{1}{{7!}} = \dfrac{x}{{8!}}$.

Consider making the denominator of the left-hand side of the given expression equal.

$\dfrac{1}{{6!}} + \dfrac{1}{{7 \times 6!}} = \dfrac{x}{{8!}}$

Take out the common terms aside,

$\dfrac{1}{{6!}}\left( {1 + \dfrac{1}{7}} \right) = \dfrac{x}{{8!}}$

$\dfrac{1}{{6!}}\left( {\dfrac{8}{7}} \right) = \dfrac{x}{{8!}}$

Now, rearrange the above equation and solve for $x$.

$x = \dfrac{{8!}}{{6!}}\left( {\dfrac{8}{7}} \right)$

$x = \dfrac{{8 \times 8 \times 7 \times 6!}}{{7 \times 6!}}$

$x = 8 \times 8$

$x = 64$

Therefore, the value of $x$ of the expression on evaluation is found to be 64.

5. Evaluate $\dfrac{{n!}}{{\left( {n - r} \right)!}}$, when

(i) $n = 6,r = 2$

Ans: The expression given is $\dfrac{{n!}}{{\left( {n - r} \right)!}}$.

Substitute 6 for $n$ and 2 for $r$ in the given expression, the expression obtained is,

$\dfrac{{6!}}{{\left( {6 - 2} \right)!}}$

On evaluating the denominator, the expression becomes, $\dfrac{{6!}}{{4!}}$.

Now, consider the expression in the numerator, that is $6!$.

According to the definition of the factorial,

$6! = 6 \times \left( {6 - 1} \right)!$

$6! = 6 \times 5!$

$6! = 6 \times 5 \times 4!$

Further, consider the expression $\dfrac{{6!}}{{4!}}$.

$\dfrac{{6!}}{{4!}} = \dfrac{{6 \times 5 \times 4!}}{{4!}}$

$\dfrac{{6!}}{{4!}} = 6 \times 5$

$\dfrac{{6!}}{{4!}} = 30$

Therefore, the value of the expression $\dfrac{{n!}}{{\left( {n - r} \right)!}}$, when $n = 6,r = 2$ is found to be 30.

(ii) $n = 9,r = 5$

Ans: The expression given is $\dfrac{{n!}}{{\left( {n - r} \right)!}}$.

Substitute 9 for $n$ and 5 for $r$ in the given expression, the expression obtained is,

$\dfrac{{9!}}{{\left( {9 - 5} \right)!}}$

On evaluating the denominator, the expression becomes, $\dfrac{{9!}}{{4!}}$.

Now, consider the expression in the numerator, that is $9!$.

According to the definition of the factorial,

$9! = 9 \times \left( {9 - 1} \right)! $

$9! = 9 \times 8!$

$9! = 9 \times 8 \times 7 \times 6 \times 5 \times$

Further, consider the expression $\dfrac{{9!}}{{4!}}$.

$\dfrac{{9!}}{{4!}} = \dfrac{{9 \times 8 \times 7 \times 6 \times 5 \times 4!}}{{4!}}$

$\dfrac{{9!}}{{4!}} = 9 \times 8 \times 7 \times 6 \times 5$

$\dfrac{{9!}}{{4!}} = 15120$

Therefore, the value of the expression $\dfrac{{n!}}{{\left( {n - r} \right)!}}$, when $n = 9,r = 5$ is found to be 15120.

NCERT Solution Class 11 Maths of Chapter 1 All Exercises

Exercise	Number of Questions
Exercise 6.1	6 Questions and Solutions
Exercise 6.3	11 Questions and Solutions
Exercise 6.4	9 Questions and Solutions
Miscellaneous Exercise	11 Questions and Solutions

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Exercise 6.2

Opting for the NCERT solutions for Ex 6.2 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 6.2 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 6.2 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 6.2, 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 6.2 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.

CBSE Class 11 Maths Chapter 6 Other Study Materials

S. No	Important Links for Chapter 6 Permutations and Combinations
1	Class 11 Permutations and Combinations Important Questions
2	Class 11 Permutations and Combinations Revision Notes
3	Class 11 Permutations and Combinations Important Formulas
4	Class 11 Permutations and Combinations NCERT Exemplar Solution
5	Class 11 Permutations RD Sharma Solutions
6	Class 11 Combinations RD Sharma Solutions
7	Class 11 Permutations RS Aggarwal Solutions
8	Class 11 Combinations RS Aggarwal Solutions

NCERT Class 11 Maths Solutions Chapter-wise Links - Download the FREE PDF

S. No	NCERT Solutions Class 11 Maths All Chapters
1	Chapter 1 - Sets Solutions
2	Chapter 2 - Relations and Functions Solutions
3	Chapter 3 - Trigonometric Functions Solutions
4	Chapter 4 - Complex Numbers and Quadratic Equations Solutions
5	Chapter 5 - Linear Inequalities Solutions
6	Chapter 6 - Permutations and Combinations Solutions
7	Chapter 7 - Binomial Theorem Solutions
8	Chapter 8 - Sequences and Series Solutions
9	Chapter 9 - Straight Lines Solutions
10	Chapter 10 - Conic Sections Solutions
11	Chapter 11 - Introduction to Three Dimensional Geometry Solutions
12	Chapter 12 - Limits and Derivatives Solutions
13	Chapter 13 - Statistics Solutions
14	Chapter 14 - Probability Solutions

Important Related Links for CBSE Class 11 Maths

S.No.	Important Study Material for Maths Class 11
1	CBSE Class 11 Maths NCERT Books
2	CBSE Class 11 Maths Revision Notes
3	CBSE Class 11 Maths Important Questions
4	CBSE Class 11 Maths NCERT Solutions
5	CBSE Class 11 Maths NCERT Exemplar Solution
6	CBSE Class 11 Maths RS Aggarwal Solutions
7	CBSE Class 11 Maths RD Sharma Solutions
8	CBSE Class 11 Maths Formulas
9	CBSE Class 11 Maths Syllabus
10	CBSE Class 11 Maths Sample Papers

FAQs on CBSE Class 11 Mathematics Chapter 6 Permutations and Combinations – NCERT Solutions 2025-26

1. What is the core concept of factorial notation (n!) used throughout NCERT Class 11 Maths Chapter 6?

Factorial notation, shown as n!, represents the product of all positive integers up to n. It is a fundamental building block for solving permutation and combination problems. The formula is n! = n × (n-1) × (n-2) × ... × 3 × 2 × 1. For example, 5! equals 120. A crucial rule to remember is that 0! is defined as 1, not 0.

2. What is the correct step-by-step method to solve permutation problems where repetition is not allowed?

To solve permutation problems without repetition as per the NCERT methodology, you should follow these steps:

Step 1: Identify 'n' and 'r'. Determine the total number of distinct objects available (n) and the number of objects you need to arrange (r).
Step 2: Confirm Order Matters. Ensure the problem requires an 'arrangement' or 'sequence', which confirms the use of permutations.
Step 3: Apply the Formula. Use the standard permutation formula ⁿP_r = n! / (n-r)!.
Step 4: Calculate the Result. Substitute the values and simplify the factorials to find the final answer.

3. How should I approach permutation problems from Chapter 6 that include specific restrictions?

When solving permutation problems with restrictions, you must adjust the standard approach:

For items that must be together: Treat the restricted group of items as a single unit. First, calculate the permutations for this new set of items. Then, multiply the result by the internal arrangements of the items within the unit.
For items that must be separate: First, calculate the total possible arrangements without any restrictions. From this, subtract the number of arrangements where the specific items are together. This difference gives the number of arrangements where they are always separate.

4. Which key formulas are essential for solving all exercises in NCERT Class 11 Maths Chapter 6?

To confidently solve problems from Chapter 6, you must master these four essential formulas:

Factorial Notation: n! = n × (n-1) × ... × 1
Permutations (Arrangements): ⁿP_r = n! / (n-r)!
Combinations (Selections): ⁿC_r = n! / (r! × (n-r)!)
Permutations with Identical Objects: n! / (p₁! × p₂! × ... × p_k!)

5. How can I avoid common mistakes when solving permutation and combination questions?

To improve accuracy, watch out for these common errors:

Confusing Permutations and Combinations: Always ask if the order of selection matters. If yes, it is a permutation. If no, it is a combination.
Factorial Calculation Errors: When dealing with fractions, always cancel common factorial terms first (e.g., 8!/6! = 8×7) before multiplying to avoid large number errors.
Mishandling Word Problems: For words with repeated letters, always use the formula n! / (p₁! p₂! ...), where p represents the frequency of each repeating letter.

6. How do I decide whether to use a permutation (ⁿP_r) or a combination (ⁿC_r) for a given problem in Chapter 6?

The choice depends entirely on whether the order of items is important for the outcome.

Use Permutation (ⁿP_r) when the problem is about arrangements, sequences, or assigning distinct roles (e.g., arranging letters in a word, assigning 1st/2nd/3rd prize). The order creates a new outcome.
Use Combination (ⁿC_r) when the problem is about selection, forming a group, or choosing a team where the order of selection does not matter (e.g., selecting 3 students from a class of 10).

7. How are NCERT problems involving permutations with repeated letters, like in the word 'ASSASSINATION', solved?

To find the number of unique arrangements for a word with repeating letters, you use a specific formula. First, count the total letters (n). Then, count the frequency of each repeating letter (p₁, p₂, etc.). The formula is: Total Arrangements = n! / (p₁! × p₂! × ... × pₖ!). For 'ASSASSINATION', n=13. The letter 'A' repeats 3 times, 'S' repeats 4 times, 'I' repeats 2 times, and 'N' repeats 2 times. The solution is 13! / (3! × 4! × 2! × 2!).

8. Are these NCERT Solutions for Chapter 6 on Permutations and Combinations fully aligned with the CBSE 2025-26 syllabus?

Yes, the NCERT Solutions for Class 11 Maths Chapter 6 are fully updated and strictly follow the latest CBSE 2025-26 curriculum. All solved examples and exercise questions adhere to the prescribed NCERT methodology, ensuring they are perfectly suitable for your board exam preparation.

9. Do the problem-solving skills from NCERT Solutions for Class 11 Maths Chapter 6 help in preparing for competitive exams like JEE Main?

Absolutely. Mastering the NCERT concepts is the most critical first step for competitive exams. These solutions build a strong foundation in core logic, such as differentiating permutations from combinations, handling complex restrictions, and applying formulas correctly. This conceptual clarity is essential for tackling the advanced, multi-topic questions that appear in JEE Main and other engineering entrance exams.