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

• Last Updated : 11 Feb, 2021

### 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

### (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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