Introduction to Permutations | Mathematics (Maths) for JEE Main & Advanced PDF Download

 Page 1


 
 
 
 
206 Permutations and Combinations  
 
 
 
 
 
 
 
 
 
 
P Pe er rm mu ut ta at ti io on ns s  
 5.1 The Factorial. 
 Factorial notation: Let n be a positive integer. Then, the continued product of first n 
natural numbers is called factorial n, to be denoted by n ! or  n . Also, we define 0 ! = 1.  
 when n is negative or a fraction, n ! is not defined. 
 Thus, n ! = n (n – 1) (n – 2) ......3.2.1. 
 Deduction: n ! = n(n – 1) (n – 2) (n – 3) ......3.2.1  
             = ] 1 . 2 . 3 )...... 3 )( 2 )( 1 [( ? ? ? n n n n = ] ! ) 1 [( ? n n 
 Thus, ) ! 2 ( 3 ! 3 ), ! 4 ( 5 ! 5 ? ? ? ? and ) ! 1 ( 2 ! 2 ? ? 
 Also, 1 ! 0 ) ! 0 ( 1 ! 1 ? ? ? ? . 
 5.2 Exponent of Prime p in n ! . 
 Let p be a prime number and n be a positive integer. Then the last integer amongst 1, 2, 3, 
.......(n – 1), n which is divisible by p is p
p
n
?
?
?
?
?
?
, where 
?
?
?
?
?
?
p
n
 denote the greatest integer less than or 
equal to 
p
n
. 
 For example: 3
3
10
?
?
?
?
?
?
?
, 5
3
15
, 2
5
12
?
?
?
?
?
?
?
?
?
?
?
?
?
?
 etc.  
 Let ) (n E
p
 denotes the exponent of the prime p in the positive integer n. Then, 
 ) ) 1 ( .......... 3 . 2 . 1 ( ) ! ( n n E n E
p p
? ? = 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
p
p
n
p p p E
p
....... 3 . 2 . = 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
p
n
E
p
n
p
...... 3 . 2 . 1 
[ ? Remaining integers between 1 and n are not divisible by p] 
 Now the last integer amongst 1, 2, 3,.....
?
?
?
?
?
?
p
n
 which is divisible by p is  
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
p
p
n
p p p E
p
n
p
n
p
p n
p
2 2
.... 3 , 2 ,
/
 because the remaining natural numbers from 1 to 
?
?
?
?
?
?
p
n
 
are not divisible by p = 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2 2
...... 3 . 2 . 1
p
n
E
p
n
p
n
p
 
Page 2


 
 
 
 
206 Permutations and Combinations  
 
 
 
 
 
 
 
 
 
 
P Pe er rm mu ut ta at ti io on ns s  
 5.1 The Factorial. 
 Factorial notation: Let n be a positive integer. Then, the continued product of first n 
natural numbers is called factorial n, to be denoted by n ! or  n . Also, we define 0 ! = 1.  
 when n is negative or a fraction, n ! is not defined. 
 Thus, n ! = n (n – 1) (n – 2) ......3.2.1. 
 Deduction: n ! = n(n – 1) (n – 2) (n – 3) ......3.2.1  
             = ] 1 . 2 . 3 )...... 3 )( 2 )( 1 [( ? ? ? n n n n = ] ! ) 1 [( ? n n 
 Thus, ) ! 2 ( 3 ! 3 ), ! 4 ( 5 ! 5 ? ? ? ? and ) ! 1 ( 2 ! 2 ? ? 
 Also, 1 ! 0 ) ! 0 ( 1 ! 1 ? ? ? ? . 
 5.2 Exponent of Prime p in n ! . 
 Let p be a prime number and n be a positive integer. Then the last integer amongst 1, 2, 3, 
.......(n – 1), n which is divisible by p is p
p
n
?
?
?
?
?
?
, where 
?
?
?
?
?
?
p
n
 denote the greatest integer less than or 
equal to 
p
n
. 
 For example: 3
3
10
?
?
?
?
?
?
?
, 5
3
15
, 2
5
12
?
?
?
?
?
?
?
?
?
?
?
?
?
?
 etc.  
 Let ) (n E
p
 denotes the exponent of the prime p in the positive integer n. Then, 
 ) ) 1 ( .......... 3 . 2 . 1 ( ) ! ( n n E n E
p p
? ? = 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
p
p
n
p p p E
p
....... 3 . 2 . = 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
p
n
E
p
n
p
...... 3 . 2 . 1 
[ ? Remaining integers between 1 and n are not divisible by p] 
 Now the last integer amongst 1, 2, 3,.....
?
?
?
?
?
?
p
n
 which is divisible by p is  
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
p
p
n
p p p E
p
n
p
n
p
p n
p
2 2
.... 3 , 2 ,
/
 because the remaining natural numbers from 1 to 
?
?
?
?
?
?
p
n
 
are not divisible by p = 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2 2
...... 3 . 2 . 1
p
n
E
p
n
p
n
p
 
  
 Similarly we get 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
S
p
p
n
p
n
p
n
p
n
n E ..... ) ! (
3 2
 
 where S is the largest natural number. Such that 
1 ?
? ?
S S
p n p .        
 5.3 Fundamental Principles of Counting . 
 (1) Addition principle : Suppose that A and B are two disjoint events (mutually exclusive); 
that is, they never occur together. Further suppose that A occurs in m ways and B in n ways. 
Then A or B can occur in m + n ways. This rule can also be applied to more than two mutually 
exclusive events. 
 
Example: 1 A college offers 7 courses in the morning and 5 in the evening. The number of ways a student can 
select exactly one course, either in the morning or in the evening 
 (a) 27 (b) 15 (c) 12 (d) 35 
Solution: (c) The student has seven choices from the morning courses out of which he can select one course in 7 
ways.  
 For the evening course, he has 5 choices out of which he can select one course in 5 ways. 
Hence he has total number of 7 + 5 = 12 choices. 
 
 (2) Multiplication principle : Suppose that an event X can be decomposed into two stages A 
and B. Let stage A occur in m ways and suppose that these stages are unrelated, in the sense 
that stage B occurs in n ways regardless of the outcome of stage A. Then event X occur in mn 
ways. This rule is applicable even if event X can be decomposed in more than two stages. 
 Note : ? The above principle can be extended for any finite number of operations and may be 
stated as under : 
If one operation can be performed independently in m different ways and if second 
operation can be performed independently in n different ways and a third 
operation can be performed independently in p different ways and so on, then the 
total number of ways in which all the operations can be performed in the stated 
order is (m × n × p × .....)  
  
Example: 2 In a monthly test, the teacher decides that there will be three questions, one from each of exercise 7, 
8 and 9 of the text book. If there are 12 questions in exercise 7, 18 in exercise 8 and 9 in exercise 9, in 
how many ways can three questions be selected 
(a) 1944 (b) 1499 (c) 4991 (d) None of these 
Solution: (a) There are 12 questions in exercise 7. So, one question from exercise 7 can be selected in 12 ways. 
Exercise 8 contains 18 questions. So, second question can be selected in 18 ways. There are 9 
questions in exercise 9. So, third question can be selected in 9 ways. Hence, three questions can be 
selected in 12 × 18 × 9 = 1944 ways. 
 
 5.4 Definition of Permutation. 
 The ways of arranging or selecting a smaller or an equal number of persons or objects at a 
time from a given group of persons or objects with due regard being paid to the order of 
arrangement or selection are called the (different) permutations. 
 For example : Three different things a, b and c are given, then different arrangements 
which can be made by taking two things from three given things are ab, ac, bc, ba, ca, cb.  
Page 3


 
 
 
 
206 Permutations and Combinations  
 
 
 
 
 
 
 
 
 
 
P Pe er rm mu ut ta at ti io on ns s  
 5.1 The Factorial. 
 Factorial notation: Let n be a positive integer. Then, the continued product of first n 
natural numbers is called factorial n, to be denoted by n ! or  n . Also, we define 0 ! = 1.  
 when n is negative or a fraction, n ! is not defined. 
 Thus, n ! = n (n – 1) (n – 2) ......3.2.1. 
 Deduction: n ! = n(n – 1) (n – 2) (n – 3) ......3.2.1  
             = ] 1 . 2 . 3 )...... 3 )( 2 )( 1 [( ? ? ? n n n n = ] ! ) 1 [( ? n n 
 Thus, ) ! 2 ( 3 ! 3 ), ! 4 ( 5 ! 5 ? ? ? ? and ) ! 1 ( 2 ! 2 ? ? 
 Also, 1 ! 0 ) ! 0 ( 1 ! 1 ? ? ? ? . 
 5.2 Exponent of Prime p in n ! . 
 Let p be a prime number and n be a positive integer. Then the last integer amongst 1, 2, 3, 
.......(n – 1), n which is divisible by p is p
p
n
?
?
?
?
?
?
, where 
?
?
?
?
?
?
p
n
 denote the greatest integer less than or 
equal to 
p
n
. 
 For example: 3
3
10
?
?
?
?
?
?
?
, 5
3
15
, 2
5
12
?
?
?
?
?
?
?
?
?
?
?
?
?
?
 etc.  
 Let ) (n E
p
 denotes the exponent of the prime p in the positive integer n. Then, 
 ) ) 1 ( .......... 3 . 2 . 1 ( ) ! ( n n E n E
p p
? ? = 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
p
p
n
p p p E
p
....... 3 . 2 . = 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
p
n
E
p
n
p
...... 3 . 2 . 1 
[ ? Remaining integers between 1 and n are not divisible by p] 
 Now the last integer amongst 1, 2, 3,.....
?
?
?
?
?
?
p
n
 which is divisible by p is  
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
p
p
n
p p p E
p
n
p
n
p
p n
p
2 2
.... 3 , 2 ,
/
 because the remaining natural numbers from 1 to 
?
?
?
?
?
?
p
n
 
are not divisible by p = 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2 2
...... 3 . 2 . 1
p
n
E
p
n
p
n
p
 
  
 Similarly we get 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
S
p
p
n
p
n
p
n
p
n
n E ..... ) ! (
3 2
 
 where S is the largest natural number. Such that 
1 ?
? ?
S S
p n p .        
 5.3 Fundamental Principles of Counting . 
 (1) Addition principle : Suppose that A and B are two disjoint events (mutually exclusive); 
that is, they never occur together. Further suppose that A occurs in m ways and B in n ways. 
Then A or B can occur in m + n ways. This rule can also be applied to more than two mutually 
exclusive events. 
 
Example: 1 A college offers 7 courses in the morning and 5 in the evening. The number of ways a student can 
select exactly one course, either in the morning or in the evening 
 (a) 27 (b) 15 (c) 12 (d) 35 
Solution: (c) The student has seven choices from the morning courses out of which he can select one course in 7 
ways.  
 For the evening course, he has 5 choices out of which he can select one course in 5 ways. 
Hence he has total number of 7 + 5 = 12 choices. 
 
 (2) Multiplication principle : Suppose that an event X can be decomposed into two stages A 
and B. Let stage A occur in m ways and suppose that these stages are unrelated, in the sense 
that stage B occurs in n ways regardless of the outcome of stage A. Then event X occur in mn 
ways. This rule is applicable even if event X can be decomposed in more than two stages. 
 Note : ? The above principle can be extended for any finite number of operations and may be 
stated as under : 
If one operation can be performed independently in m different ways and if second 
operation can be performed independently in n different ways and a third 
operation can be performed independently in p different ways and so on, then the 
total number of ways in which all the operations can be performed in the stated 
order is (m × n × p × .....)  
  
Example: 2 In a monthly test, the teacher decides that there will be three questions, one from each of exercise 7, 
8 and 9 of the text book. If there are 12 questions in exercise 7, 18 in exercise 8 and 9 in exercise 9, in 
how many ways can three questions be selected 
(a) 1944 (b) 1499 (c) 4991 (d) None of these 
Solution: (a) There are 12 questions in exercise 7. So, one question from exercise 7 can be selected in 12 ways. 
Exercise 8 contains 18 questions. So, second question can be selected in 18 ways. There are 9 
questions in exercise 9. So, third question can be selected in 9 ways. Hence, three questions can be 
selected in 12 × 18 × 9 = 1944 ways. 
 
 5.4 Definition of Permutation. 
 The ways of arranging or selecting a smaller or an equal number of persons or objects at a 
time from a given group of persons or objects with due regard being paid to the order of 
arrangement or selection are called the (different) permutations. 
 For example : Three different things a, b and c are given, then different arrangements 
which can be made by taking two things from three given things are ab, ac, bc, ba, ca, cb.  
 
 
 
 
   
 Therefore the number of permutations will be 6. 
 5.5 Number of Permutations without Repetition. 
 (1) Arranging n objects, taken r at a time equivalent to filling r places from n  things 
 
 r-places :    
 Number of choices : 
 The number of ways of arranging = The number of ways of filling r places. 
 = ) 1 ).......( 2 ( ) 1 ( ? ? ? ? r n n n n = 
r
n
P
r n
n
r n
r n r n n n n
?
?
?
?
? ? ? ? ?
)! (
!
)! (
) )! )(( 1 ).....( 2 ( ) 1 (
 
 (2) The number of arrangements of n different objects taken all at a time = ! n P
n
n
? 
 Note : ?  
1
1
0
. ; 1
!
!
?
?
? ? ?
r
n
r
n n
P n P
n
n
P 
         ? 0
! ) (
1
; 1 ! 0 ?
?
?
r
 or ) ( ! ) ( N r r ? ? ? ? 
 
Example: 3  If 2 : 1 :
5 4
? P P
n n
, then ? n                                   [MP PET 1987; Rajasthan PET  
1996]  
 (a) 4 (b) 5 (c) 6 (d) 7 
Solution: (c) 
2
1
5
4
?
P
P
n
n
 ? 
2
1
!
! ) 5 (
! ) 4 (
!
?
?
?
? n
n
n
n
 ? 2 4 ? ? n ? 6 ? n . 
Example: 4 In a train 5 seats are vacant then how many ways can three passengers sit                 [Rajasthan PET 
1985; MP PET 2003] 
 (a) 20 (b) 30 (c) 60 (d) 10 
Solution: (c) Number of ways are = 60
2
120
! 2
! 5
! ) 3 5 (
! 5
3
5
? ? ?
?
? P . 
Example: 5 How many words comprising of any three letters of the word “UNIVERSAL” can be formed 
 (a) 504 (b) 405 (c) 540 (d) 450 
Solution: (a) Required numbers of words = 504
! 6
! 9
! ) 3 9 (
! 9
3
9
? ?
?
? P . 
Example: 6 How many numbers of five digits can be formed from the numbers 2, 0, 4, 3, 8 when repetition of 
digit is not allowed  
[MP PET 2000] 
 (a) 96 (b) 120 (c) 144 (d) 14 
Solution: (a) Given numbers are 2, 0, 4, 3, 8 
 Numbers can be formed = {Total – Those beginning with 0} 
                                 = {5 ! – 4 !} = 120 – 24 = 96. 
Example: 7 How many numbers can be made with the help of the digits 0, 1, 2, 3, 4, 5 which are greater than 
3000 (repetition is not allowed)       [IIT 1976] 
 (a) 180 (b) 360 (c) 1380 (d) 1500 
Solution: (c) All the 5 digit numbers and 6 digit numbers are greater than 3000. Therefore number of 5 digit 
numbers                           = 600
5
5
5
6
? ? P P . 
{Since the case that 0 will be at ten thousand place should be omit}. Similarly number of 6 digit 
numbers 6 ! – 5 ! = 600. 
n (n–1) (n – 
2 
)(n – 
3) 
1 2 3 4 
n – (r –
1) 
r 
Page 4


 
 
 
 
206 Permutations and Combinations  
 
 
 
 
 
 
 
 
 
 
P Pe er rm mu ut ta at ti io on ns s  
 5.1 The Factorial. 
 Factorial notation: Let n be a positive integer. Then, the continued product of first n 
natural numbers is called factorial n, to be denoted by n ! or  n . Also, we define 0 ! = 1.  
 when n is negative or a fraction, n ! is not defined. 
 Thus, n ! = n (n – 1) (n – 2) ......3.2.1. 
 Deduction: n ! = n(n – 1) (n – 2) (n – 3) ......3.2.1  
             = ] 1 . 2 . 3 )...... 3 )( 2 )( 1 [( ? ? ? n n n n = ] ! ) 1 [( ? n n 
 Thus, ) ! 2 ( 3 ! 3 ), ! 4 ( 5 ! 5 ? ? ? ? and ) ! 1 ( 2 ! 2 ? ? 
 Also, 1 ! 0 ) ! 0 ( 1 ! 1 ? ? ? ? . 
 5.2 Exponent of Prime p in n ! . 
 Let p be a prime number and n be a positive integer. Then the last integer amongst 1, 2, 3, 
.......(n – 1), n which is divisible by p is p
p
n
?
?
?
?
?
?
, where 
?
?
?
?
?
?
p
n
 denote the greatest integer less than or 
equal to 
p
n
. 
 For example: 3
3
10
?
?
?
?
?
?
?
, 5
3
15
, 2
5
12
?
?
?
?
?
?
?
?
?
?
?
?
?
?
 etc.  
 Let ) (n E
p
 denotes the exponent of the prime p in the positive integer n. Then, 
 ) ) 1 ( .......... 3 . 2 . 1 ( ) ! ( n n E n E
p p
? ? = 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
p
p
n
p p p E
p
....... 3 . 2 . = 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
p
n
E
p
n
p
...... 3 . 2 . 1 
[ ? Remaining integers between 1 and n are not divisible by p] 
 Now the last integer amongst 1, 2, 3,.....
?
?
?
?
?
?
p
n
 which is divisible by p is  
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
p
p
n
p p p E
p
n
p
n
p
p n
p
2 2
.... 3 , 2 ,
/
 because the remaining natural numbers from 1 to 
?
?
?
?
?
?
p
n
 
are not divisible by p = 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2 2
...... 3 . 2 . 1
p
n
E
p
n
p
n
p
 
  
 Similarly we get 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
S
p
p
n
p
n
p
n
p
n
n E ..... ) ! (
3 2
 
 where S is the largest natural number. Such that 
1 ?
? ?
S S
p n p .        
 5.3 Fundamental Principles of Counting . 
 (1) Addition principle : Suppose that A and B are two disjoint events (mutually exclusive); 
that is, they never occur together. Further suppose that A occurs in m ways and B in n ways. 
Then A or B can occur in m + n ways. This rule can also be applied to more than two mutually 
exclusive events. 
 
Example: 1 A college offers 7 courses in the morning and 5 in the evening. The number of ways a student can 
select exactly one course, either in the morning or in the evening 
 (a) 27 (b) 15 (c) 12 (d) 35 
Solution: (c) The student has seven choices from the morning courses out of which he can select one course in 7 
ways.  
 For the evening course, he has 5 choices out of which he can select one course in 5 ways. 
Hence he has total number of 7 + 5 = 12 choices. 
 
 (2) Multiplication principle : Suppose that an event X can be decomposed into two stages A 
and B. Let stage A occur in m ways and suppose that these stages are unrelated, in the sense 
that stage B occurs in n ways regardless of the outcome of stage A. Then event X occur in mn 
ways. This rule is applicable even if event X can be decomposed in more than two stages. 
 Note : ? The above principle can be extended for any finite number of operations and may be 
stated as under : 
If one operation can be performed independently in m different ways and if second 
operation can be performed independently in n different ways and a third 
operation can be performed independently in p different ways and so on, then the 
total number of ways in which all the operations can be performed in the stated 
order is (m × n × p × .....)  
  
Example: 2 In a monthly test, the teacher decides that there will be three questions, one from each of exercise 7, 
8 and 9 of the text book. If there are 12 questions in exercise 7, 18 in exercise 8 and 9 in exercise 9, in 
how many ways can three questions be selected 
(a) 1944 (b) 1499 (c) 4991 (d) None of these 
Solution: (a) There are 12 questions in exercise 7. So, one question from exercise 7 can be selected in 12 ways. 
Exercise 8 contains 18 questions. So, second question can be selected in 18 ways. There are 9 
questions in exercise 9. So, third question can be selected in 9 ways. Hence, three questions can be 
selected in 12 × 18 × 9 = 1944 ways. 
 
 5.4 Definition of Permutation. 
 The ways of arranging or selecting a smaller or an equal number of persons or objects at a 
time from a given group of persons or objects with due regard being paid to the order of 
arrangement or selection are called the (different) permutations. 
 For example : Three different things a, b and c are given, then different arrangements 
which can be made by taking two things from three given things are ab, ac, bc, ba, ca, cb.  
 
 
 
 
   
 Therefore the number of permutations will be 6. 
 5.5 Number of Permutations without Repetition. 
 (1) Arranging n objects, taken r at a time equivalent to filling r places from n  things 
 
 r-places :    
 Number of choices : 
 The number of ways of arranging = The number of ways of filling r places. 
 = ) 1 ).......( 2 ( ) 1 ( ? ? ? ? r n n n n = 
r
n
P
r n
n
r n
r n r n n n n
?
?
?
?
? ? ? ? ?
)! (
!
)! (
) )! )(( 1 ).....( 2 ( ) 1 (
 
 (2) The number of arrangements of n different objects taken all at a time = ! n P
n
n
? 
 Note : ?  
1
1
0
. ; 1
!
!
?
?
? ? ?
r
n
r
n n
P n P
n
n
P 
         ? 0
! ) (
1
; 1 ! 0 ?
?
?
r
 or ) ( ! ) ( N r r ? ? ? ? 
 
Example: 3  If 2 : 1 :
5 4
? P P
n n
, then ? n                                   [MP PET 1987; Rajasthan PET  
1996]  
 (a) 4 (b) 5 (c) 6 (d) 7 
Solution: (c) 
2
1
5
4
?
P
P
n
n
 ? 
2
1
!
! ) 5 (
! ) 4 (
!
?
?
?
? n
n
n
n
 ? 2 4 ? ? n ? 6 ? n . 
Example: 4 In a train 5 seats are vacant then how many ways can three passengers sit                 [Rajasthan PET 
1985; MP PET 2003] 
 (a) 20 (b) 30 (c) 60 (d) 10 
Solution: (c) Number of ways are = 60
2
120
! 2
! 5
! ) 3 5 (
! 5
3
5
? ? ?
?
? P . 
Example: 5 How many words comprising of any three letters of the word “UNIVERSAL” can be formed 
 (a) 504 (b) 405 (c) 540 (d) 450 
Solution: (a) Required numbers of words = 504
! 6
! 9
! ) 3 9 (
! 9
3
9
? ?
?
? P . 
Example: 6 How many numbers of five digits can be formed from the numbers 2, 0, 4, 3, 8 when repetition of 
digit is not allowed  
[MP PET 2000] 
 (a) 96 (b) 120 (c) 144 (d) 14 
Solution: (a) Given numbers are 2, 0, 4, 3, 8 
 Numbers can be formed = {Total – Those beginning with 0} 
                                 = {5 ! – 4 !} = 120 – 24 = 96. 
Example: 7 How many numbers can be made with the help of the digits 0, 1, 2, 3, 4, 5 which are greater than 
3000 (repetition is not allowed)       [IIT 1976] 
 (a) 180 (b) 360 (c) 1380 (d) 1500 
Solution: (c) All the 5 digit numbers and 6 digit numbers are greater than 3000. Therefore number of 5 digit 
numbers                           = 600
5
5
5
6
? ? P P . 
{Since the case that 0 will be at ten thousand place should be omit}. Similarly number of 6 digit 
numbers 6 ! – 5 ! = 600. 
n (n–1) (n – 
2 
)(n – 
3) 
1 2 3 4 
n – (r –
1) 
r 
 
 
 
 
  
Now the numbers of 4 digit numbers which are greater than 3000, having 3, 4 or 5 at first place, this 
can be done in 3 ways and remaining 3 digit may be filled from remaining 5 digits i.e., required 
number of 4 digit numbers are 180 3
3
5
? ? P . 
 Hence total required number of numbers = 600 + 600 + 180 = 1380.  
 
 5.6 Number of Permutations with Repetition. 
 (1) The number of permutations (arrangements) of n different objects, taken r at a time, 
when each object may occur once, twice, thrice,........upto r times in any arrangement = The 
number of ways of filling r places where each place can be filled by any one of n objects. 
 r – places :  
 Number of choices :  
 The number of permutations = The number of ways of filling r places = 
r
n) ( 
 (2) The number of arrangements that can be formed using n objects out of which p are 
identical (and of one kind) q are identical (and of another kind), r are identical (and of another 
kind) and the rest are distinct is 
! ! !
!
r q p
n
.  
Example: 8 The number of arrangement of the letters of the word “CALCUTTA”    [MP PET 1984] 
 (a) 2520 (b) 5040 (c) 10080 (d) 40320 
Solution: (b) Required number of ways 5040
! 2 ! 2 ! 2
! 8
? ? .  [since here 2C’s, 2T’s and 2A’s]  
Example: 9 The number of 5 digit telephone numbers having at least one of their digits repeated is    [Pb. CET 2000] 
 (a) 90,000 (b) 100,000 (c) 30,240 (d) 69,760 
Solution: (d) Using the digits 0, 1, 2,.......,9 the number of five digit telephone numbers which can be formed is 
5
10 .  
 (since repetition is allowed) 
 The number of five digit telephone numbers which have none of the digits repeated = 30240
5
10
? P 
 ? The required number of telephone numbers = 69760 30240 10
5
? ? . 
Example: 10 How many words can be made from the letters of the word ‘COMMITTEE’         [MP PET 2002; RPET 
1986]  
 (a) 
2
) ! 2 (
! 9
  (b) 
3
) ! 2 (
! 9
 (c) 
! 2
! 9
 (d) 9 ! 
Solution: (b) Number of words = 
3
) ! 2 (
! 9
! 2 ! 2 ! 2
! 9
? [Since here total number of letters is 9 and 2M’s, 2T’s and 2E’s] 
 
 5.7 Conditional Permutations. 
 (1) Number of permutations of n dissimilar things taken r at a time when p particular 
things always occur           = ! r C
p r
p n
?
?
 
 (2) Number of permutations of n dissimilar things taken r at a time when p particular 
things never occur          = ! r C
r
p n ?
 
n 
n n 
n 
1 2 3 4 
n    
r 
Page 5


 
 
 
 
206 Permutations and Combinations  
 
 
 
 
 
 
 
 
 
 
P Pe er rm mu ut ta at ti io on ns s  
 5.1 The Factorial. 
 Factorial notation: Let n be a positive integer. Then, the continued product of first n 
natural numbers is called factorial n, to be denoted by n ! or  n . Also, we define 0 ! = 1.  
 when n is negative or a fraction, n ! is not defined. 
 Thus, n ! = n (n – 1) (n – 2) ......3.2.1. 
 Deduction: n ! = n(n – 1) (n – 2) (n – 3) ......3.2.1  
             = ] 1 . 2 . 3 )...... 3 )( 2 )( 1 [( ? ? ? n n n n = ] ! ) 1 [( ? n n 
 Thus, ) ! 2 ( 3 ! 3 ), ! 4 ( 5 ! 5 ? ? ? ? and ) ! 1 ( 2 ! 2 ? ? 
 Also, 1 ! 0 ) ! 0 ( 1 ! 1 ? ? ? ? . 
 5.2 Exponent of Prime p in n ! . 
 Let p be a prime number and n be a positive integer. Then the last integer amongst 1, 2, 3, 
.......(n – 1), n which is divisible by p is p
p
n
?
?
?
?
?
?
, where 
?
?
?
?
?
?
p
n
 denote the greatest integer less than or 
equal to 
p
n
. 
 For example: 3
3
10
?
?
?
?
?
?
?
, 5
3
15
, 2
5
12
?
?
?
?
?
?
?
?
?
?
?
?
?
?
 etc.  
 Let ) (n E
p
 denotes the exponent of the prime p in the positive integer n. Then, 
 ) ) 1 ( .......... 3 . 2 . 1 ( ) ! ( n n E n E
p p
? ? = 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
p
p
n
p p p E
p
....... 3 . 2 . = 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
p
n
E
p
n
p
...... 3 . 2 . 1 
[ ? Remaining integers between 1 and n are not divisible by p] 
 Now the last integer amongst 1, 2, 3,.....
?
?
?
?
?
?
p
n
 which is divisible by p is  
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
p
p
n
p p p E
p
n
p
n
p
p n
p
2 2
.... 3 , 2 ,
/
 because the remaining natural numbers from 1 to 
?
?
?
?
?
?
p
n
 
are not divisible by p = 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2 2
...... 3 . 2 . 1
p
n
E
p
n
p
n
p
 
  
 Similarly we get 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
S
p
p
n
p
n
p
n
p
n
n E ..... ) ! (
3 2
 
 where S is the largest natural number. Such that 
1 ?
? ?
S S
p n p .        
 5.3 Fundamental Principles of Counting . 
 (1) Addition principle : Suppose that A and B are two disjoint events (mutually exclusive); 
that is, they never occur together. Further suppose that A occurs in m ways and B in n ways. 
Then A or B can occur in m + n ways. This rule can also be applied to more than two mutually 
exclusive events. 
 
Example: 1 A college offers 7 courses in the morning and 5 in the evening. The number of ways a student can 
select exactly one course, either in the morning or in the evening 
 (a) 27 (b) 15 (c) 12 (d) 35 
Solution: (c) The student has seven choices from the morning courses out of which he can select one course in 7 
ways.  
 For the evening course, he has 5 choices out of which he can select one course in 5 ways. 
Hence he has total number of 7 + 5 = 12 choices. 
 
 (2) Multiplication principle : Suppose that an event X can be decomposed into two stages A 
and B. Let stage A occur in m ways and suppose that these stages are unrelated, in the sense 
that stage B occurs in n ways regardless of the outcome of stage A. Then event X occur in mn 
ways. This rule is applicable even if event X can be decomposed in more than two stages. 
 Note : ? The above principle can be extended for any finite number of operations and may be 
stated as under : 
If one operation can be performed independently in m different ways and if second 
operation can be performed independently in n different ways and a third 
operation can be performed independently in p different ways and so on, then the 
total number of ways in which all the operations can be performed in the stated 
order is (m × n × p × .....)  
  
Example: 2 In a monthly test, the teacher decides that there will be three questions, one from each of exercise 7, 
8 and 9 of the text book. If there are 12 questions in exercise 7, 18 in exercise 8 and 9 in exercise 9, in 
how many ways can three questions be selected 
(a) 1944 (b) 1499 (c) 4991 (d) None of these 
Solution: (a) There are 12 questions in exercise 7. So, one question from exercise 7 can be selected in 12 ways. 
Exercise 8 contains 18 questions. So, second question can be selected in 18 ways. There are 9 
questions in exercise 9. So, third question can be selected in 9 ways. Hence, three questions can be 
selected in 12 × 18 × 9 = 1944 ways. 
 
 5.4 Definition of Permutation. 
 The ways of arranging or selecting a smaller or an equal number of persons or objects at a 
time from a given group of persons or objects with due regard being paid to the order of 
arrangement or selection are called the (different) permutations. 
 For example : Three different things a, b and c are given, then different arrangements 
which can be made by taking two things from three given things are ab, ac, bc, ba, ca, cb.  
 
 
 
 
   
 Therefore the number of permutations will be 6. 
 5.5 Number of Permutations without Repetition. 
 (1) Arranging n objects, taken r at a time equivalent to filling r places from n  things 
 
 r-places :    
 Number of choices : 
 The number of ways of arranging = The number of ways of filling r places. 
 = ) 1 ).......( 2 ( ) 1 ( ? ? ? ? r n n n n = 
r
n
P
r n
n
r n
r n r n n n n
?
?
?
?
? ? ? ? ?
)! (
!
)! (
) )! )(( 1 ).....( 2 ( ) 1 (
 
 (2) The number of arrangements of n different objects taken all at a time = ! n P
n
n
? 
 Note : ?  
1
1
0
. ; 1
!
!
?
?
? ? ?
r
n
r
n n
P n P
n
n
P 
         ? 0
! ) (
1
; 1 ! 0 ?
?
?
r
 or ) ( ! ) ( N r r ? ? ? ? 
 
Example: 3  If 2 : 1 :
5 4
? P P
n n
, then ? n                                   [MP PET 1987; Rajasthan PET  
1996]  
 (a) 4 (b) 5 (c) 6 (d) 7 
Solution: (c) 
2
1
5
4
?
P
P
n
n
 ? 
2
1
!
! ) 5 (
! ) 4 (
!
?
?
?
? n
n
n
n
 ? 2 4 ? ? n ? 6 ? n . 
Example: 4 In a train 5 seats are vacant then how many ways can three passengers sit                 [Rajasthan PET 
1985; MP PET 2003] 
 (a) 20 (b) 30 (c) 60 (d) 10 
Solution: (c) Number of ways are = 60
2
120
! 2
! 5
! ) 3 5 (
! 5
3
5
? ? ?
?
? P . 
Example: 5 How many words comprising of any three letters of the word “UNIVERSAL” can be formed 
 (a) 504 (b) 405 (c) 540 (d) 450 
Solution: (a) Required numbers of words = 504
! 6
! 9
! ) 3 9 (
! 9
3
9
? ?
?
? P . 
Example: 6 How many numbers of five digits can be formed from the numbers 2, 0, 4, 3, 8 when repetition of 
digit is not allowed  
[MP PET 2000] 
 (a) 96 (b) 120 (c) 144 (d) 14 
Solution: (a) Given numbers are 2, 0, 4, 3, 8 
 Numbers can be formed = {Total – Those beginning with 0} 
                                 = {5 ! – 4 !} = 120 – 24 = 96. 
Example: 7 How many numbers can be made with the help of the digits 0, 1, 2, 3, 4, 5 which are greater than 
3000 (repetition is not allowed)       [IIT 1976] 
 (a) 180 (b) 360 (c) 1380 (d) 1500 
Solution: (c) All the 5 digit numbers and 6 digit numbers are greater than 3000. Therefore number of 5 digit 
numbers                           = 600
5
5
5
6
? ? P P . 
{Since the case that 0 will be at ten thousand place should be omit}. Similarly number of 6 digit 
numbers 6 ! – 5 ! = 600. 
n (n–1) (n – 
2 
)(n – 
3) 
1 2 3 4 
n – (r –
1) 
r 
 
 
 
 
  
Now the numbers of 4 digit numbers which are greater than 3000, having 3, 4 or 5 at first place, this 
can be done in 3 ways and remaining 3 digit may be filled from remaining 5 digits i.e., required 
number of 4 digit numbers are 180 3
3
5
? ? P . 
 Hence total required number of numbers = 600 + 600 + 180 = 1380.  
 
 5.6 Number of Permutations with Repetition. 
 (1) The number of permutations (arrangements) of n different objects, taken r at a time, 
when each object may occur once, twice, thrice,........upto r times in any arrangement = The 
number of ways of filling r places where each place can be filled by any one of n objects. 
 r – places :  
 Number of choices :  
 The number of permutations = The number of ways of filling r places = 
r
n) ( 
 (2) The number of arrangements that can be formed using n objects out of which p are 
identical (and of one kind) q are identical (and of another kind), r are identical (and of another 
kind) and the rest are distinct is 
! ! !
!
r q p
n
.  
Example: 8 The number of arrangement of the letters of the word “CALCUTTA”    [MP PET 1984] 
 (a) 2520 (b) 5040 (c) 10080 (d) 40320 
Solution: (b) Required number of ways 5040
! 2 ! 2 ! 2
! 8
? ? .  [since here 2C’s, 2T’s and 2A’s]  
Example: 9 The number of 5 digit telephone numbers having at least one of their digits repeated is    [Pb. CET 2000] 
 (a) 90,000 (b) 100,000 (c) 30,240 (d) 69,760 
Solution: (d) Using the digits 0, 1, 2,.......,9 the number of five digit telephone numbers which can be formed is 
5
10 .  
 (since repetition is allowed) 
 The number of five digit telephone numbers which have none of the digits repeated = 30240
5
10
? P 
 ? The required number of telephone numbers = 69760 30240 10
5
? ? . 
Example: 10 How many words can be made from the letters of the word ‘COMMITTEE’         [MP PET 2002; RPET 
1986]  
 (a) 
2
) ! 2 (
! 9
  (b) 
3
) ! 2 (
! 9
 (c) 
! 2
! 9
 (d) 9 ! 
Solution: (b) Number of words = 
3
) ! 2 (
! 9
! 2 ! 2 ! 2
! 9
? [Since here total number of letters is 9 and 2M’s, 2T’s and 2E’s] 
 
 5.7 Conditional Permutations. 
 (1) Number of permutations of n dissimilar things taken r at a time when p particular 
things always occur           = ! r C
p r
p n
?
?
 
 (2) Number of permutations of n dissimilar things taken r at a time when p particular 
things never occur          = ! r C
r
p n ?
 
n 
n n 
n 
1 2 3 4 
n    
r 
 
 
 
 
   
 (3) The total number of permutations of n different things taken not more than r at a time, 
when each thing may be repeated any number of times, is 
1
) 1 (
?
?
n
n n
r
. 
 (4) Number of permutations of n different things, taken all at a time, when m specified 
things always come together is ! ) 1 ( ! ? ? ? m n m 
 (5) Number of permutations of n different things, taken all at a time, when m specified 
things never come together is ! ) 1 ( ! ! ? ? ? ? m n m n 
 (6) Let there be n objects, of which m objects are alike of one kind, and the remaining 
) ( m n ? objects are alike of another kind. Then, the total number of mutually distinguishable 
permutations that can be formed from these objects is 
! ) ( ) ! (
!
m n m
n
? ?
. 
 Note : ?  The above theorem can be extended further i.e.,  if there are n objects, of which 
1
p are alike of one kind; 
2
p are alike of another kind; 
3
p are alike of 3
rd
 kind;......: 
r
p are alike 
of r
th
 kind such that n p p p
r
? ? ? ? ......
2 1
; then the number of permutations of these n objects is 
) ! ( ...... ) ! ( ) ! (
!
2 1 r
p p p
n
? ? ?
.   
Important Tips 
 
? Gap method : Suppose 5 males A, B, C, D, E are arranged in a row as × A × B × C × D × E  ×. There will be six 
gaps between these five. Four in between and two at either end. Now if three females P, Q,R are to be arranged so 
that no two are together we shall use gap method i.e., arrange them in between these 6 gaps. Hence the answer 
will be  
3
6
P . 
? Together : Suppose we have to arrange 5 persons in a row which can be done in 5 ! = 120 ways. But if two 
particular persons are to be together always, then we tie these two particular persons with a string. 
Thus we have 5 – 2 + 1 (1 corresponding to these two together) = 3 +1 = 4 units, which can be 
arranged in 4! ways. Now we loosen the string and these two particular can be arranged in 2 ! ways. 
Thus total arrangements = 24 × 2 = 48.   
Never together = Total –  Together = 120 – 48 = 72. 
 
Example: 11 All the letters of the word ‘EAMCET’ are arranged in all possible ways. The number of such 
arrangement in which two vowels are not adjacent to each other is                   
[EAMCET 1987; DCE 2000]   
 (a) 360 (b) 114 (c) 72 (d) 54 
Solution: (c) First we arrange 3 consonants in 3 ! ways and then at four places (two places between them and two 
places on two sides) 3 vowels can be placed in 
! 2
1
3
4
? P ways. 
 Hence the required ways = 3 ! × 72
! 2
1
3
4
? ? P . 
Example: 12 The number of words which can be made out of the letters of the word ‘MOBILE’ when consonants 
always occupy odd places is        [Rajasthan PET 1999]    
 (a) 20 (b) 36 (c) 30 (d) 720