Chapter 16 : Permutations - Rd Sharma Solutions for Class 11-science Maths CBSE

Chapter 16 - Permutations Exercise Ex. 16.1

Question 1
Solution 1

Question 2
Solution 2
Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9

Question 10
Solution 10
Question 11
Solution 11

Question 12

Solution 12
Question 13
Solution 13
Question 14
Solution 14
Question 15
Solution 15
Question 16
Solution 16

Question 17
Solution 17
Question 18
Solution 18

Chapter 16 - Permutations Exercise Ex. 16.2

Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5

Solution 5

Question 6
Solution 6

Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14
Solution 14
Question 15
Solution 15
Question 16
Solution 16
Question 17

How many three-digit numbers are there whit no digit repeated?

Solution 17

Question 18

Solution 18
Question 19
Solution 19
Question 20
Solution 20
Question 21
Solution 21
Question 22
Solution 22
Question 23
Solution 23
Question 24
Solution 24
Question 25
Solution 25
Question 26
How many odd numbers less than 1000 can be formed by using the digits 0, 3, 5, 7 when repetition of digits is not allowed?
Solution 26
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.
Question 27
Solution 27
Question 28
Solution 28
Question 29
Solution 29
Question 30
Solution 30
Question 31

Serial numbers for an item produced in a factory are to be made using two letters followed by four digits (0 to 9). If the letters are to be are to be taken from six letters of English alphabet without repetition and the digits are also not repeated in a serial number, how many serial numbers are possible?

Solution 31

Question 32
Solution 32
Question 33
Solution 33
Question 34
Solution 34
Question 35

How many for digit natural numbers not exceeding 4321 can be formed with the digits 1, 2, 3, and 4, if the digits can repeat?

Solution 35

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.

Question 36

How many numbers of six digits can be formed from the digits 0, 1, 3, 5, 7, and 9 when on digit is repeated? How many of them are divisible by 10?

Solution 36

Question 37
Solution 37
Question 38
Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

Solution 47

Question 48

In how many ways can 5 different balls be distributed among three boxes?

Solution 48

  

Question 49

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

There are 10 lamps in a hall. Each one of them can be switched on independently. Find the number of ways in which the hall can be illuminated.

Solution 52

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.

Chapter 16 - Permutations Exercise Ex. 16.3

Question 1
Solution 1
Question 2
Solution 2
Question 3

Solution 3
Question 4
Solution 4
Question 5

Solution 5

Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10
Solution 10
Question 11
Solution 11

Question 12
Solution 12
Question 13
Solution 13


Question 14
Solution 14
Question 15
Solution 15
Question 16
Solution 16
Question 17
Solution 17

Question 18
Solution 18
Question 19
Solution 19

Question 20
Solution 20
Question 21
Solution 21
Question 22
Solution 22
Question 23
Solution 23
Question 24
Solution 24
Question 25
Solution 25
Question 26
Solution 26
Question 27
Solution 27
Question 28
Solution 28

Question 29
Solution 29
Question 30
Solution 30
Question 31
Solution 31
Question 32
Solution 32
Question 33
Solution 33
Question 34
Solution 34
Question 35

All the letters of the word ‘EAMCOT’ are arranged in different possible ways. Find the number of arrangements in which no two vowels are adjacent to each other.

Solution 35

Chapter 16 - Permutations Exercise Ex. 16.4

Question 1
Solution 1
Question 2

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

Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10
Solution 10
Question 11
Solution 11
Question 12
Solution 12

Chapter 16 - Permutations Exercise Ex. 16.5

Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10
Solution 10
Question 11
Solution 11
Question 12
Solution 12
Question 13
Solution 13
Question 14
Solution 14
Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

How many permulations of the letters of the world 'MADHUBANI' do not begain with M but end with I?

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Chapter 16 - Permutations Exercise Ex. 16VSAQ

Question 1

Solution 1

Question 2
Solution 2
Question 3
Solution 3
Question 4

Solution 4
Question 5
Solution 5
Question 6

Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10
Solution 10
Question 11

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.

Question 12
Solution 12
Question 13

Solution 13
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