# Class 11-science RD SHARMA Solutions Maths Chapter 16 - Permutations

## Permutations Exercise Ex. 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

## Permutations Exercise Ex. 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

**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 2^{10}.

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 2^{10}-1.

## Permutations Exercise Ex. 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

## Permutations Exercise Ex. 16.4

### Solution 1

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

### Solution 11

### Solution 12

## Permutations Exercise Ex. 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

## Permutations Exercise Ex. 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.