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RD Sharma Solution for Class 11 Science Mathematics Chapter 16 - Permutations

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Our RD Sharma Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 11 exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 16 - Permutations.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RD Sharma Solutions for CBSE Class 11 Mathematics will strengthen your fundamentals in Mathematics and will help you in your attempts to score more marks in the final examination. CBSE Class 11 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

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RD Sharma Solution for Class 11 Science Mathematics Chapter 16 - Permutations Page/Excercise 16.1

Solution 1


Solution 2

Solution 3(i)

Solution 3(ii)

Solution 3(iii)

Solution 4(i)

Solution 4(ii)

Solution 4(iii)

Solution 4(iv)


Solution 5

Solution 6


Solution 7

Solution 8

Solution 9

Solution 10

Solution 11(i)


Solution 11(ii)

Solution 12

RD Sharma Solution for Class 11 Science Mathematics Chapter 16 - Permutations Page/Excercise 16.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 4

Solution 5


Solution 6

Solution 7

Solution 8

Solution 9


Solution 10


Solution 11


Solution 12


Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19(i)

Solution 19(ii)

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Any number less than 1000 may be any of a number from one-digit number, two-digit number and three-digit number.

One-digit odd number:

3 possible ways are there. These numbers are 3 or 5 or 7.

Two-digit odd number:

Tens place can be filled up by 3 ways (using any of the digit among 3, 5 and 7) and then the ones place can be filled in any of the remaining 2 digits.

So, there are 3 2 = 6 such 2-digit numbers.

Three-digit odd number:

Ignore the presence of zero at ones place for some instance.

Hundreds place can be filled up in 3 ways (using any of any of the digit among 3, 5 and 7), then tens place in 3 ways by using remaining 3 digits (after using a digit, there will be three digits) and then the ones place in 2 ways.

So, there are a total of 3 3 2 = 18 numbers of 3-digit numbers which includes both odd and even numbers (ones place digit are zero). In order to get the odd numbers, it is required to ignore the even numbers i.e. numbers ending with zero.

To obtain the even 3-digit numbers, ones place can be filled up in 1 way (only 0 to be filled), hundreds place in 3 ways (using any of the digit among 3, 5, 7) and then tens place in 2 ways (using remaining 2 digits after filling up hundreds place).

So, there are a total of 1 3 2 = 6 even 3-digit numbers using the digits 0, 3, 5 and 7 (repetition not allowed)

So, number of three-digit odd numbers using the digits 0, 3, 5 and 7 (repetition not allowed) = 18 -  6 = 12.

Therefore, odd numbers less than 1000 can be formed by using the digits 0, 3, 5, 7 when repetition of digits is not allowed are 3 + 6 + 12 = 21.

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

The given digits are 1, 2, 3 and 4. These digits can be repeated while forming the numbers. So, number of required four digit natural numbers can be found as follows.

Consider four digit natural numbers whose digit at thousandths place is 1.

Here, hundredths place can be filled in 4 ways. (Using the digits 1 or 2 or 3 or 4)

Similarly, tens place can be filled in 4 ways. (Using the digits 1 or 2 or 3 or 4)

Ones place can be filled in 4 ways. (Using the digits 1 or 2 or 3 or 4)

Number of four digit natural numbers whose digit at thousandths place is 1 = 4 4 4 = 64

Similarly, number of four digit natural numbers whose digit at thousandths place is 2 = 4 4 4  = 64

Now, consider four digit natural numbers whose digit at thousandths place is 4:

Here, if the digit at hundredths place is 1, then tens place can be filled in 4 ways and ones place can also be filled in 4 ways.

If the digit at hundredths place is 2, then tens place can be filled in 4 ways and ones place can also be filled in 4 ways.

If the digit at hundredths place is 3 and the digit at tens place is 1, then ones place can be filled in 4 ways.

If the digit at hundredths place is 3 and the digit at tens place is 2, then ones place can be filled only in 1 way so that the number formed is not exceeding 4321.

Number of four digit natural numbers not exceeding 4321 and digit at thousandths place is 3 = 4 4 + 4 4 + 4 + 1 = 37

Thus, required number of four digit natural numbers not exceeding 4321 is 64 + 64 + 64 + 37 = 229.

Solution 34

Solution 35

Solution 36


Solution 37

Solution 38

Solution 39

Solution 40

Solution 41

Solution 42

Solution 43

Solution 44

Solution 45

Solution 46

  

Solution 47(i)

Solution 47(ii)

Solution 47(iii)

Solution 48

Each lamps has two possibilities either it can be switched on or off.

There are 10 lamps in the hall.

So the total numbers of possibilities are 210.

To illuminate the hall we require at least one lamp is to be switched on.

There is one possibility when all the lamps are switched off. If all the bulbs are switched off then hall will not be illuminated.

So the number of ways in which the hall can be illuminated is 210-1.

RD Sharma Solution for Class 11 Science Mathematics Chapter 16 - Permutations Page/Excercise 16.3

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 1(i)

Solution 2


Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8


Solution 9

Solution 10



Solution 11

Solution 12

Solution 13

Solution 14


Solution 15

Solution 16


Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25


Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

RD Sharma Solution for Class 11 Science Mathematics Chapter 16 - Permutations Page/Excercise 16.4

Solution 1

Solution 2

T h e r e space a r e space 7 space l e t t e r s space i n space t h e space w o r d space apostrophe space S T R A N G E apostrophe comma space i n c l u d i n g space 2 space v o w e l s space left parenthesis A comma E right parenthesis space a n d space 5 space c o n s o n a n t s space left parenthesis S comma T comma R comma N comma G right parenthesis.
left parenthesis i right parenthesis space C o n s i d e r i n g space 2 space v o w e l s space a s space o n e space l e t t e r comma space w e space h a v e space 6 space l e t t e r s space w h i c h space c a n space b e space a r r a n g e d space i n space to the power of 6 p subscript 6 equals 6 factorial space w a y s
A comma E space c a n space b e space p u t space t o g e t h e r space i n space 2 factorial space w a y s.

H e n c e comma space r e q u i r e d space space n u m b e r space o f space w o r d s space
equals space 6 factorial cross times 2 factorial
equals space 6 cross times 5 cross times 4 cross times 3 cross times 2 cross times 1 cross times 2
equals 720 cross times 2 space
equals 1440.

left parenthesis i i right parenthesis space T h e space t o t a l space n u m b e r space o f space w o r d s space f o r m e d space b y space u sin g space a l l space t h e space l e t t e r s space o f space t h e space w o r d s space apostrophe S T R A N G E apostrophe
i s space to the power of 7 p subscript 7 equals 7 factorial
equals 7 cross times 6 cross times 5 cross times 4 cross times 3 cross times 2 cross times 1
equals 5040.
S o comma space t h e space t o t a l space n u m b e r space o f space w o r d s space i n space w h i c h space v o w e l s space a r e space n e v e r space t o g e t h e r
equals T o t a l space n u m b e r space o f space w o r d s space minus space n u m b e r space o f space w o r d s space i n space w h i c h space v o w e l s space a r e space a l w a y s space space t o g e t h e r
equals 5040 minus 1440
equals 3600

left parenthesis i i i right parenthesis space T h e r e space a r e space 7 space l e t t e r s space i n space t h e space w o r d space apostrophe S T R A N G E apostrophe. space o u t space o f space t h e s e space l e t t e r s space apostrophe A apostrophe space a n d space apostrophe E apostrophe space a r e space t h e space v o w e l s.
T h e r e space a r e space 4 space o d d space p l a c e s space i n space t h e space w o r d space apostrophe S T R A N G E apostrophe. space T h e space t w o space v o w e l s space c a n space b e space a r r a n g e d space i n space to the power of 4 p subscript 2 space w a y s.
T h e space r e m a i n i n g space 5 space c o n s o n a n t s space c a n space b e space a r r a n g e d space a m o n g space t h e m s e l v e s space i n space to the power of 5 p subscript 5 space w a y s.
T h e space t o t a l space n u m b e r space o f space a r r a n g e m e n t s
equals to the power of 4 p subscript 2 cross times to the power of 5 p subscript 5
equals fraction numerator 4 factorial over denominator 2 factorial end fraction cross times 5 factorial
equals 1440

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

RD Sharma Solution for Class 11 Science Mathematics Chapter 16 - Permutations Page/Excercise 16.5

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 1(v)

Solution 1(vi)

Solution 1(vii)

Solution 1(viii)

Solution 1(ix)

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

RD Sharma Solution for Class 11 Science Mathematics Chapter 16 - Permutations Page/Excercise 16VSAQ

Solution 1


Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

First we will find the least factorial term divisible by 14.

As 7!=7x6x5! is divisible by 14 leaving remainder zero.

 

Hence terms 7! onwards can be written as multiple of 7!.

8!=8x7!, 9!=9x8x7!... like ways 200! can also be written as multiple of 7!.

So all the terms 7! onwards are divisible by 14 leaving remainder zero.

 

1! + 2! + 3! + 4! + 5! + 6!

=1+2+6+24+120+720

=873

 

Hence remainder obtained when 1! + 2! + 3! + ….+ 200! is divided by 14 is 5.

Solution 12

Solution 13

TopperLearning provides step-by-step solutions for each question in each chapter in the RD Sharma textbook for class 11. Access the CBSE Class 11 Mathematics Chapter 16 - Permutations for free. The textbook questions have been solved by our subject matter experts to help you understand how to answer them. Our RD Sharma Textbook Solutions will help you to study and revise, and you can easily clear your fundamentals of Chapter 16 - Permutations.

Text Book Solutions

CBSE XI Science - Mathematics

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