RD Sharma Solutions Class 11 Maths Chapter 16 - Permutations


The RD Sharma Solutions for Class 11 Maths Chapter 16: Permutations are essential resources for students aiming to excel in their final exams. This chapter introduces the concept of selecting and arranging ‘r’ items from a group of ‘n’ distinct objects — a concept known as permutations, where the order of selection is important. In contrast, combinations are used when the order does not matter.

In this chapter, students learn foundational concepts such as factorials, counting principles, and different types of permutations, including those under special conditions and involving non-distinct objects. The RD Sharma solutions make use of illustrative examples and shortcut techniques to help students grasp these concepts more effectively and solve problems accurately.

RD Sharma Solutions Class 11 Chapter 16 – Overview

This chapter is structured into five exercises, each focusing on a specific aspect of permutations. Topics include:

  • Factorials and their notation

  • Fundamental principles of counting

  • Basic permutations and their applications

  • Permutations with specific conditions

  • Permutations involving identical items

Each question in these exercises is answered in a step-by-step manner, making it easier for students to understand even the trickiest problems. The detailed breakdown of concepts also helps in improving analytical thinking and problem-solving speed — crucial for competitive exams.

RD Sharma Solutions for Class 11 Maths Permutations Question with solutions

Q1. How many different 3-digit numbers can be formed using the digits 1 to 5 without repetition?
Digits available = 5
1st digit: 5 choices
2nd digit: 4 choices
3rd digit: 3 choices
Total = 5 × 4 × 3 = 60
 
Q2. In how many ways can the letters of the word "MATH" be arranged?
All letters are different → 4! = 24
 
Q3. How many permutations can be made with all letters of “LEVEL”?
Letters: L(2), E(2), V(1)
Total = 5! / (2! × 2!) = 120 / 4 = 30
 
Q4. How many ways can 4 boys be seated in a row of 4 chairs?
Total permutations = 4! = 24
 
Q5. How many 4-letter words can be formed from the letters of the word “EQUATION” without repetition?
Total letters = 8
Choose and arrange 4 = 8P4 = 8! / 4! = 1680
 
Q6. How many ways can 6 people be arranged around a circular table?
Circular permutations = (6 − 1)! = 120
 
Q7. In how many ways can the letters of “BANANA” be arranged?
Letters: A(3), N(2), B(1)
Total = 6! / (3! × 2!) = 720 / 12 = 60
 
Q8. How many 3-digit numbers can be formed using 2, 3, 5, 7, 9 if repetition is allowed?
Each digit has 5 options → 5 × 5 × 5 = 125
 
Q9. How many 5-letter words can be formed using the letters of the word “CLASS” if all letters are used?
Letters: S repeated twice → Total = 5! / 2! = 60
 
Q10. How many arrangements of the word "DELHI" start with a vowel?
Vowels = E, I (2 options)
Remaining letters = 4! = 24
Total = 2 × 24 = 48
 
Q11. How many permutations of 6 items taken all at a time?
6! = 720
 
Q12. How many ways can 3 boys and 2 girls be seated in a row such that girls sit together?
Treat 2 girls as 1 unit → 4 units
Ways = 4! × 2! = 24 × 2 = 48
 
Q13. How many 4-digit numbers can be formed using digits 2, 4, 6, 8, 0 without repetition if the first digit is not 0?
Valid 1st digit = 4 choices (not 0)
Remaining = 4 digits → 4 × 3 × 2 = 96
 
Q14. How many permutations of the word “SUCCESS” are possible?
Letters: S(3), C(2), U(1), E(1)
Total = 7! / (3! × 2!) = 5040 / 12 = 420
 
Q15. How many ways can 5 people be seated in a circular table if 2 persons must sit together?
Treat 2 people as 1 block → 4 units
Circular permutations = 3!
Internal arrangements = 2!
Total = 6 × 2 = 12
 
Q16. Number of permutations of 7 different books on a shelf taking 4 at a time.
^7P4 = 7! / 3! = 5040 / 6 = 840
 
Q17. In how many ways can the word “BOOKKEEPER” be arranged?
O(2), K(2), E(2) → 10! / (2! × 2! × 2!) = 3628800 / 8 = 453600
 
Q18. Find number of permutations of all digits of the number 12134.
5 digits: 1 repeated twice
Total = 5! / 2! = 60
 
Q19. How many numbers between 100 and 1000 can be formed using digits 1 to 9 without repetition?
First digit = 9 options
Second = 8, Third = 7
Total = 9 × 8 × 7 = 504
 
Q20. How many circular permutations of 6 people are possible such that 2 specific people must not sit together?
Total = 5! = 120
Together = (5−1)! × 2! = 24 × 2 = 48
Not together = 120 − 48 = 72
 
Q21. In how many ways can the word "INFINITY" be arranged if I's must come together?
Treat 3 I’s as one → 6 letters left
Letters: III, N(2), F, T, Y
Total = 6! / 2! = 360
 
Q22. Find the number of ways to arrange the word “STATISTICS”.
Letters: S(3), T(3), A(1), I(2), C(1)
Total = 10! / (3! × 3! × 2!) = 3628800 / 72 = 50400
 
Q23. In how many ways can 3 boys and 3 girls be arranged in a row so that boys and girls alternate?
Patterns: BGBGBG or GBGBGB
Each pattern = 3! × 3! = 36
Total = 36 + 36 = 72
 
Q24. How many 5-digit even numbers can be formed from 1, 2, 3, 4, 5, 6 with no repetition?
Last digit (even) = 2, 4, 6 → 3 choices
Remaining = 5! = 120
Total = 3 × 120 = 360
 
Q25. Find number of ways in which 8 different books can be arranged on a shelf so that 2 particular books are never together.
Total = 8! = 40320
Together = 7! × 2! = 10080
Not together = 40320 − 10080 = 30240
 

Why Use These Solutions?

These solutions have been crafted by subject experts to ensure accuracy and clarity. Difficult problems are simplified into manageable steps, allowing students to develop a strong conceptual foundation in permutations. Whether you’re preparing for school exams or entrance tests, these solutions are a reliable guide to mastering the topic.

Students are encouraged to download the PDF version of these solutions for offline practice and revision.

Frequently Asked Questions

Chapter 16 of RD Sharma Class 11 Maths focuses on permutations, which deal with arranging a group of items in a specific order. The order of objects is important in permutations, and the chapter covers different types, such as permutations of distinct objects, identical objects, and circular permutations.

 

Permutations involve arrangements where order matters.

Combinations are selections where order does not matter.

For example:
Arranging ABC and CBA are different permutations, but the same combination.

Factorial notation and simplification

Fundamental principle of counting

Permutations of distinct objects

Permutations with identical items

Circular permutations

Restricted permutations (e.g., items together or not together)

Understand the situation – is order important?

Identify if repetition is allowed or not

Use factorial identities to simplify expressions

Break down constraints like "together", "not together", or fixed positions

Practice word problems like seating, forming numbers/words, etc.

The following topics from the chapter are crucial:

Permutations of n different things taken r at a time

Permutations when some objects are identical

Restricted permutations (e.g., two people must sit together)

Permutations in circular arrangements

Try to solve all examples and miscellaneous exercises from RD Sharma for complete mastery.

Yes.
Permutations (along with combinations) are important in JEE, NDA, CUET, and other entrance exams. Mastery of this chapter builds a strong base for probability and advanced counting principles used later in Class 11 and 12.

Permutations are used in:

Seating arrangements (e.g., weddings, boards)

Forming passwords or PINs

Scheduling tasks or events

Genetics and bioinformatics

Computer algorithms that generate sequences