Table of Contents
Chapter 16 Permutations Exercise Ex. 16.1
Question 1
Solution 1
Question 2
Solution 2
Question 3(i)
Solution 3(i)
Question 3(ii)
Solution 3(ii)
Question 3(iii)
Solution 3(iii)
Question 4(i)
Solution 4(i)
Question 4(ii)
Solution 4(ii)
Question 4(iii)
Solution 4(iii)
Question 4(iv)
Solution 4(iv)
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(i)
Solution 11(i)
Question 11(ii)
Solution 11(ii)
Question 12
Solution 12
Chapter 16 Permutations Exercise Ex. 16.2
Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
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
How many three-digit numbers are there whit no digit repeated?Solution 16
Question 17
Solution 17
Question 18
Solution 18
Question 19(i)
Solution 19(i)
Question 19(ii)
Solution 19(ii)
Question 20
Solution 20
Question 21
Solution 21
Question 22
Solution 22
Question 23
Solution 23
Question 24How 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 24Any 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 25
Solution 25
Question 26
Solution 26
Question 27
Solution 27
Question 28
Solution 28
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 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 29
Question 30
Solution 30
Question 31
Solution 31
Question 32
Solution 32
Question 33
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 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.Question 34
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 34
Question 35
Solution 35
Question 36
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
In how many ways can 5 different balls be distributed among three boxes?Solution 46
Question 47(i)
Solution 47(i)
Question 47(ii)
Solution 47(ii)
Question 47(iii)
Solution 47(iii)
Question 48
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 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.
Chapter 16 Permutations Exercise Ex. 16.3
Question 1(ii)
Solution 1(ii)
Question 1(iii)
Solution 1(iii)
Question 1(iv)
Solution 1(iv)
Question 1(i)
Solution 1(i)
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
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 32
Chapter 16 Permutations Exercise Ex. 16.4
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
Chapter 16 Permutations Exercise Ex. 16.5
Question 1(i)
Solution 1(i)
Question 1(ii)
Solution 1(ii)
Question 1(iii)
Solution 1(iii)
Question 1(iv)
Solution 1(iv)
Question 1(v)
Solution 1(v)
Question 1(vi)
Solution 1(vi)
Question 1(vii)
Solution 1(vii)
Question 1(viii)
Solution 1(viii)
Question 1(ix)
Solution 1(ix)
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
How many permulations of the letters of the world ‘MADHUBANI’ do not begain with M but end with I?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
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