RD Sharma Class 11 Maths Chapter 16 Solutions – Permutations Step-by-Step

RD Sharma Solutions for Class 11 Mathematics Chapter “Permutations” are available here, prepared by expert teachers according to the latest CBSE Class 11th Maths syllabus and NCERT guidelines. This chapter covers the fundamental concept of permutations, which is essential for understanding how to arrange objects in different sequences. On this page, students will find a variety of permutation problems with detailed, step-by-step solutions to help master this important topic.

RD Sharma Solution for Class 11 will assist students in learning how to calculate permutations of distinct and repeated objects, understand factorial notations, and apply permutation formulas in different scenarios. Mastering permutations is crucial for excelling in board exams, competitive exams, and further studies in mathematics and related fields.

RD Sharma Solutions for Class 11 Chapter 16 Permutations

In RD Sharma Solutions for Class 11 Chapter Permutations, students will gain a clear understanding of the fundamental concept of permutations and factorial notation. The chapter covers how to calculate permutations of distinct objects as well as solving problems involving permutations with repetition. Students also learn to apply permutation formulas to real-life scenarios, such as arranging letters, digits, and various objects.

Important topics include handling permutations of n objects taken r at a time, along with detailed, step-by-step solutions to all textbook exercises for thorough practice. This chapter builds a strong foundation in combinatorial mathematics, enabling students to confidently analyze and solve complex permutation problems essential for both board exams and competitive tests.

Download RD Sharma Solutions for Permutations PDF Free

Download comprehensive RD Sharma Class 11 Maths solutions for the chapter on permutations, including solved examples, extra practice questions, and detailed explanations. Access the free PDF to enhance your understanding of permutations, improve accuracy, and boost your exam preparation with trusted, exam oriented solutions.

RD Sharma Class 11 Chapter 16: Permutations - Step-by-Step Solutions

Question 1: Compute:

(i) 30!/28!

Solution:

30!/28! = (30 × 29 × 28!)/28!

= 30 × 29

= 870

(ii) 11! - 10!/9!

Solution:

First, calculate 10!/9! = (10 × 9!)/9! = 10

So, 11! - 10!/9! = 11! - 10

= 39,916,800 - 10

= 39,916,790

(iii) L.C.M. (6!, 7!, 8!)

Solution:

Since 8! = 8 × 7! and 7! = 7 × 6!

We have 8! contains both 7! and 6! as factors

Therefore, L.C.M.(6!, 7!, 8!) = 8! = 40,320

Question 2: Express the product 5 × 6 × 7 × 8 × 9 × 10 as a factorial

Solution:

5 × 6 × 7 × 8 × 9 × 10 = (10!)/(4!)

Since 10! = 10 × 9 × 8 × 7 × 6 × 5 × 4!

So, 5 × 6 × 7 × 8 × 9 × 10 = 10!/4!

Question 3: Convert the product (n+1)(n+2)(n+3)...(2n) into factorial notation

Solution:

The product starts from (n+1) and ends at 2n

This can be written as: (2n)!/n!

Since (2n)! = 2n × (2n-1) × ... × (n+1) × n!

Therefore, (n+1)(n+2)(n+3)...(2n) = (2n)!/n!

Question 4: If P(n, 4) = 12 × P(n, 2), find n

Solution:

P(n, 4) = n!/(n-4)! and P(n, 2) = n!/(n-2)!

Given: n!/(n-4)! = 12 × n!/(n-2)!

Simplifying: (n-2)!/(n-4)! = 12

(n-2)(n-3) = 12

n² - 5n + 6 = 12

n² - 5n - 6 = 0

(n-6)(n+1) = 0

Since n must be positive, n = 6

Question 5: Find the value of r if P(5, r) = P(6, r-1)

Solution:

P(5, r) = 5!/(5-r)! and P(6, r-1) = 6!/(6-(r-1))! = 6!/(7-r)!

Given: 5!/(5-r)! = 6!/(7-r)!

5! × (7-r)! = 6! × (5-r)!

120 × (7-r)! = 720 × (5-r)!

(7-r)!/!(5-r)! = 6

(7-r)(6-r) = 6

42 - 7r - 6r + r² = 6

r² - 13r + 36 = 0

(r-4)(r-9) = 0

Since r ≤ 5, r = 4

Question 6: If nP₄ = 360, find the value of n

Solution:

nP₄ = n!/(n-4)! = n(n-1)(n-2)(n-3)

Given: n(n-1)(n-2)(n-3) = 360

Testing values: When n = 6

6 × 5 × 4 × 3 = 360 ✓

Therefore, n = 6

Question 7: In how many ways can 5 students be arranged in a row?

Solution:

Number of ways to arrange 5 students = 5!

= 5 × 4 × 3 × 2 × 1

= 120 ways

Question 8: How many 3-digit numbers can be formed using digits 1, 2, 3, 4, 5 if no digit is repeated?

Solution:

We need to select 3 digits from 5 digits and arrange them

Number of ways = P(5,3) = 5!/(5-3)! = 5!/2!

= 5 × 4 × 3 = 60

Question 9: Find the number of ways to arrange the letters of the word "INDIA"

Solution:

The word "INDIA" has 5 letters with I appearing twice

Number of arrangements = 5!/2!

= 120/2 = 60

Question 10: How many different signals can be given using 6 flags of different colors, taking one or more at a time?

Solution:

Taking 1 flag: P(6,1) = 6

Taking 2 flags: P(6,2) = 6 × 5 = 30

Taking 3 flags: P(6,3) = 6 × 5 × 4 = 120

Taking 4 flags: P(6,4) = 6 × 5 × 4 × 3 = 360

Taking 5 flags: P(6,5) = 6 × 5 × 4 × 3 × 2 = 720

Taking 6 flags: P(6,6) = 6! = 720

Total = 6 + 30 + 120 + 360 + 720 + 720 = 1956

Question 11: In how many ways can the letters of "SERIES" be arranged so that vowels always come together?

Solution:

"SERIES" has letters S, E, R, I, E, S with E appearing twice and S appearing twice

Vowels: E, I, E (treat as one unit)

Consonants: S, R, S

Total units to arrange: 4 (vowel unit + 3 consonants)

Arrangements of 4 units with S repeated = 4!/2! = 12

Arrangements within vowel unit (E,I,E) = 3!/2! = 3

Total arrangements = 12 × 3 = 36

Question 12: How many numbers greater than 100000 can be formed from digits 2, 3, 0, 3, 4, 2, 3?

Solution:

We have digits: 2, 3, 0, 3, 4, 2, 3 (where 2 appears twice, 3 appears thrice)

For numbers greater than 100000, we need 6-digit numbers

Total 6-digit arrangements = 6!/2!3! = 720/12 = 60

Arrangements starting with 0 = 5!/2!3! = 120/12 = 10

Numbers greater than 100000 = 60 - 10 = 50

Question 13: If P(n, 5) = 20 × P(n, 3), find n

Solution:

P(n, 5) = n!/(n-5)! and P(n, 3) = n!/(n-3)!

Given: n!/(n-5)! = 20 × n!/(n-3)!

(n-3)!/(n-5)! = 20

(n-3)(n-4) = 20

n² - 7n + 12 = 20

n² - 7n - 8 = 0

(n-8)(n+1) = 0

Since n must be positive, n = 8

Question 14: In a class of 27 boys and 14 girls, how many ways can a teacher select 1 boy and 1 girl?

Solution:

Ways to select 1 boy from 27 boys = 27

Ways to select 1 girl from 14 girls = 14

Total ways = 27 × 14 = 378

Question 15: A coin is tossed five times. How many possible outcomes are there?

Solution:

Each toss has 2 possible outcomes (Head or Tail)

For 5 tosses: 2⁵ = 32 possible outcomes

Question 16: How many ways can an examinee answer a set of ten true/false questions?

Solution:

Each question has 2 possible answers (True or False)

For 10 questions: 2¹⁰ = 1024 ways

Question 17: Find the number of words formed by permuting all letters of "INDEPENDENCE"

Solution:

"INDEPENDENCE" has 12 letters:

I-1, N-4, D-2, E-4, P-1, C-1

Number of arrangements = 12!/(1!×4!×2!×4!×1!×1!)

= 479,001,600/(1×24×2×24×1×1)

= 479,001,600/1152

= 415,800

Question 18: In how many ways can 4 parcels be sent to 5 post offices?

Solution:

Each parcel can be sent to any of the 5 post offices

Total ways = 5⁴ = 625

Question 19: If P(n-1, 3) : P(n, 4) = 1 : 9, find n

Solution:

P(n-1, 3) = (n-1)!/(n-4)! and P(n, 4) = n!/(n-4)!

Given ratio: P(n-1, 3)/P(n, 4) = 1/9

[(n-1)!/(n-4)!]/[n!/(n-4)!] = 1/9

(n-1)!/n! = 1/9

1/n = 1/9

Therefore, n = 9

Question 20: How many five-digit telephone numbers can be formed using digits 0-9 if at least one digit is repeated?

Solution:

Total 5-digit numbers using digits 0-9 = 10⁵ = 100,000

5-digit numbers with no repeated digits:

First digit: 9 choices (1-9, can't start with 0)

Second digit: 9 choices (0 and 8 remaining digits)

Third digit: 8 choices

Fourth digit: 7 choices

Fifth digit: 6 choices

Numbers with no repetition = 9 × 9 × 8 × 7 × 6 = 27,216

Numbers with at least one repetition = 100,000 - 27,216 = 72,784