Class 11 RD Sharma Solutions- Chapter 16 Permutations - Exercise 16.2 | Set 2

Chapter 16 of RD Sharma's Class 11 Mathematics textbook "Permutations" explores the fundamental concept of the permutations and their applications. The Permutations are arrangements of the objects in the specific order and are crucial for the solving problems related to the counting and arrangement in the combinatorics. This chapter covers the various techniques for the calculating permutations including the factorial notation and permutation formulas.

Permutations

The Permutations refer to the different ways in which a set of the objects can be arranged in the specific sequence or order. The number of permutations of n distinct objects is given by n!. In problems involving subsets or choosing a subset from the larger set permutations are used to the calculate the total number of the possible arrangements. Understanding permutations helps in solving the problems related to the probability, scheduling and optimization.

Question 17. How many three-digit numbers are there?

Solution:

Ways of selecting 100th place = ⁹P₁ = 9 (Selecting from all digits except 0)

Ways of selecting 10th place = ¹⁰P₁ = 10 (Selecting from all digits)

Ways of selecting unit pace = ¹⁰P₁ = 10 (Selecting from all digits)

Total 3-digit numbers possible = 9 x 10 x 10 = 900          

Question 18. How many three-digit odd numbers are there?

Solution:

Ways of selecting 100th place = ⁹P₁ = 9 (Selecting from all digits except 0)

Ways of selecting 10th place = ¹⁰P₁ = 10 (Selecting from all digits)

Ways of selecting unit pace = ⁵P₁ = 5 (Selecting from all odd digits)

Total 3-digit odd numbers = 9 x 10 x 5 = 450          

Question 19. How many different five-digit number license plates can be made if

(i) The first digit cannot be zero, and the repetition of digits is not allowed,

(ii) The first-digit cannot be zero, but the repetition of digits is allowed?

Solution:

(i) First digit = ⁹P₁ = 9 ways (all digits except zero)

Second digit = ⁹P₁ = 9 ways (all digits except the first digit of the license plate) 

Third digit = ⁸P₁ = 8 ways (all digits except the first and second digits of the license plate) 

Fourth digit = ⁷P₁ = 7 ways (all digits except the first, second and third digits of the license plate) 

Fifth digit = 6 ways (all digits except the first, second, third and fourth digits of the license plate)  

Total ways = 9 x 9 x 8 x 7 x 6 

= 27216        

(ii) First digit = ⁹P₁ = 9 ways (all digits except zero)

Second digit = ¹⁰P₁ = 10 ways (all digits)

Third digit = ¹⁰P₁ = 10 ways (all digits)

Fourth digit = ¹⁰P₁ = 10 ways (all digits)

Fifth digit = ¹⁰P₁ = 10 ways (all digits)  

Total ways = 9 x 10 x 10 x 10 x 10

= 90,000

Question 20. How many four-digit numbers can be formed with the digits 3, 5, 7, 8, 9 which are greater than 7000, if repetition of digits is not allowed?

Solution:

First digit = ³P₁ = 3 ways (only 7, 8, 9 are possible digits to make it greater than 7000)

Second digit = ⁴P₁ = 4 ways (all 5 digits except the first digit of the selected number)

Third digit = ³P₁ = 3 ways (all 5 digits except the first and second digits of the number)

Fourth digit = ²P₁ = 2 ways (all 5 digits except the first, second and third digits of the number)

Total numbers = 3 x 4 x 3 x 2

= 72

Question 21. How many four-digit numbers can be formed with the digits 3, 5, 7, 8, 9 which are greater than 8000, if repetition of digits is not allowed?

Solution:

First digit = ²P₁ = 2 ways (only 8, 9 are possible digits to make it greater than 8000)

Second digit = ⁴P₁ = 4 ways (all 5 digits except the first digit of the selected number)

Third digit = ³P₁ = 3 ways (all 5 digits except the first and second digits of the number)

Fourth digit = ²P₁ = 2 ways (all 5 digits except the first, second and third digits of the number)

Total numbers = 2 x 4 x 3 x 2

= 48

Question 22. In how many ways can six persons be seated in a row?

Solution:

Let's consider 6 seats in the given row

First seat = ⁶P₁ = 6 options of people

Second seat= ⁵P₁ = 5 options (the person already sitting on first seat can not sit here)  

Third seat = ⁴P₁ = 4 options (people sitting on first and second can't sit on third)

Fourth seat = ³P₁ = 3 options

Fifth seat = ²P₁ = 2 options

Sixth seat = ¹P₁ = 1 option

Total ways = 6 x 5 x 4 x 3 x 2 x 1 = 6! = 720

Question 23. How many 9-digit numbers of different digits can be formed?

Solution:

1st digit = ⁹P₁ = 9 possibilities (zero is not possible)

2nd digit = ⁹P₁ = 9 possibilities (except first digit)

3rd digit = ⁸P₁ = 8 possibilities (except first & second digits selected)

4th digit = ⁷P₁ = 7 possibilities (except first, second, third digits selected)

for all subsequent i-th position possibilities keep on decreasing by 1 so,

eventually for 9th digit, 2 possibilities

Total numbers = 9 x 9 x 8 x ..... 2

= 9 x 9!

= 3265920

Question 24. 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:

Case 1: 1 digit number

³P₁ = 3 ways (since 0 is even so it can not be selected)

Case 2: 2 digit numbers

1st digit = ³P₁ = 3 ways (except 0)

2nd digit = ²P₁ = 2 ways (except first digit and 0)

= 3 x 2 = 6

Case 3: 3 digit numbers  

1st digit = ³P₁ = 3 ways (except 0)

2nd digit = ³P₁ = 3 ways (except the first digit)

3rd digit = ²P₁ = 2 ways (except the first and second digits)

4th digit = ¹P₁ = 1 ways (except the first, second, third digits)

Total numbers including odd and even = 3 x 3 x 2 x 1 = 18

Even such numbers included = No.s ending with zero

= Number of options for 3rd position is 1

= Number of options for 2nd position is 3

= Number of options for 1st position - 2

Even numbers = 1 x 3 x 2 = 6  

= 18 - 6 = 12 (odd numbers)

Case 4: 4 digit numbers

No possibility (since they'll be greater than 1000)  

Total numbers = 3 + 6 + 12 + 0

= 21

Question 25. How many 3-digit numbers are there, with distinct digits, with each digit odd?

Solution:

Odd digits are 1, 3, 5, 7, 9

So, the total number of odd digits = 5

Now, select 3 digits = ⁵P₃ = 10 ways

Arrange these selected digits = 3! ways

Total numbers = 10 x 3! = 60    

Question 26. How many different numbers of six digits each can be formed from the digits 4, 5, 6, 7, 8, 9 when repetition of digits is not allowed?

Solution:

First digit of six-digit number can be selected = 6 ways

Second digit of six-digit number can be selected = 5 ways

Third digit of six-digit number can be selected = 4 ways

Fourth digit of six-digit number can be selected = 3 ways

Fifth digit of six-digit number can be selected = 2 ways

And the last digit of six-digit number can be selected = 1 ways

So, the final arrangements of 6 digits = 6 x 5 x 4 x 3 x 2 x 1 = 720 ways

Question 27. How many different numbers of six digits can be formed from the digits 3,1,7,0,9,5 when repetition of digits is not allowed?

Solution:

 Possibilities for 1st digit = 5 (except 0)

Possibilities for 2nd digit = 5 

Possibilities for 3rd digit = 4

Possibilities for 4th digit = 3

Possibilities for 5th digit = 2

Possibilities for last digit = 1

So, the total numbers = 5 x 5 x 4 x 3 x 2 x 1 = 600    

Question 28. How many four-digit different numbers, greater than 5000 can be formed with the digits 1,2,5,9,0 when repetition of digits is not allowed?

Solution:

For 1st digit, ways = 2 ways (only 5 or 9)

For remaining places = ⁴P₃ ways 

To choose the digits = ⁴P₃ x 3! = 24 arrangements

Total ways = 2 x 24 = 48  

Question 29. 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 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:

Arrangements for 2 letters = 2! x ⁶P₂ = 30

Arrangements for 4 digits = 4! x ¹⁰P₄ = 5040

Required numbers = 30 x 5040 = 151200

Question 30.  A number lock on a suitcase has 3 wheels each labelled with ten digits 0 to 9. If opening of the lock is a particular sequence of three digits with no repeats, how many such sequences will be possible? Also, find the number of unsuccessful attempts to open the lock.

Solution:

Ways of first number = 10

Ways of second number = 9 (except first)

Ways of third number = 8 (except first and second)

Total numbers = 10 x 9 x 8 = 720

Number of unsuccessful attempts = 719 (since only 1 will be correct possibility)  

Question 31. A customer forgets a four-digit code for an Automatic Teller Machine (ATM) in a bank. However, he remembers that this code consists of digits 3, 5, 6, and 9. Find the largest possible number of trials necessary to obtain the correct code.

Solution:

Total number of digits = 4

So, the largest possible number of trials necessary to obtain the correct code = 4! = 24 attempts

Question 32. In how many ways can three jobs I, II, and III be assigned to three persons A, B, and C if one person is assigned only one job and all are capable of doing each job?

Solution:

Total number of jobs = 3

Total number of persons = 3

It is exactly same as arranging 3 objects, in 3 different positions = 3! ways = 6    

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