Summary This is a class notes of permutation and

Vidyamandir Classes Permutation & Combination


Permutation & Combination
INTRODUCTION Section - 1

1.1 Definition of Factorial
Factorial : The continued product of first n natural numbers is called the “n factorial” and is denoted by n!
or n . i.e. n!  n  n  1  n  2  . . . . . 3  2  1
Thus, 4!  4  3  2  1  24
5!  5  4  3  2  1  120

1.2 Properties of factorial
(i) n! is defined for positive integers only.
(ii) Factorials of proper fractions or integers are not defined. Factorial n is defined only for whole
numbers.
(iii) n! = 1  2  3   . . . . . .   n  1  n  1  2  3   . . . . . .   n  1 n = (n – 1) ! n
Thus, n! = n (n – 1) !
(iv) 0! = 1 (by definition)
(v) If two factorials, i.e., x! and y! are equal, then x = y or x = 0, y = 1 or x = 1, y = 0.

Illustrating the Concepts :
n! n!
If 2!  n  2  ! and 4!  n  4  ! are in the ratio 2 : 1, then find the value of n ?


n! n  n  1  n  2  . . . . . 3  2  1 n! n!
= Given : 2!  n  2 ! : 4! n  4 !  2 : 1
2!  n  2  ! 2!  n  2  !
n  n  1  n  2  n  3 . . . . . 3  2  1 n  n  1
= 2 2
2!  n  2  ! 
 n  n  1  n  2   n  3 1
n  n  1  n  2  ! n  n  1 4  3  2 1
= 
2!  n  2 ! 2!
12 2
Similarly,  
 n  2   n  3 1
n! n  n  1  n  2   n  3 n  4  !
=  (n – 2) (n – 3) = 6
4!  n  4 ! 4! n  4  !
 n2 – 5n = 0 n = 0, 5
n (n  1) ( n  2) ( n  3) But, for n = 0, (n – 2) ! and (n – 4) ! are not
=
4!
meaningful.
So, n = 5.

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Illustration - 1  2n  !
The value of n ! is equal to :
(A) {1. 3. 5. . . . . .(2n – 1)}2n (B) {1. 3. 5. . . . . .(2n – 1)} 2n n!
(C) {1. 3. 5. . . . . .(2n + 1)}2n (D) None of these
SOLUTION : (A)
Using the definition of factorials,
 2n ! 1 . 2 . 3 . . . . . . .  2n  1  2n 

n! n!
Separating odd and even terms in numerator, we get :
1 . 3 . 5 . . . . . .  2n  1 2 . 4 . 6 . . . . 2n  {1· 3· 5 . . .. (2n  1)} 2n n!
=
n! n!
= {1 . 3 . 5 . . . . . . . (2n – 1) }2n

1.3 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 n
n which is divisible by p is   p where   denotes the greatest integer less than or equal to n/p.
 p  p
Ep (n!) denote the exponent of p in the positive integer n. Then,
E p  n ! = E p 1 . 2 . 3 . . . . . .  n  1 n 

 n 
= Ep  p . 2p . 3p . . . . . .  p  p 
   
[As Remaining integers between 1 and n are not divisible by p]
n   n 
=  p   E p 1 . 2 . 3 . . . . .  p  
    
n
Now, the last integer amongst, 1, 2 , 3, . . . . .,   which is divisible by p is :
 p
n / p  n 
 p  =  p2  p
p
   
n   n  
 E p  n ! =    E 
p  p . 2 p . . . . . .  2  p 
 p  p  
[As Remaining integers between 1 and [n/p] are not divisible by p.]



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n  n    n 
=       E p  1 . 2 . 3 . . . . .  2 
 p   p 2   
  p  
Continuing in the same manner, we get :
n  n  n 
E p  n ! =     2   . . . . .   t 
 p   p   p 
[where t is the largest positive integer such that pt  n  pt  1 ]
Note : This result is not valid for composite numbers.
For Example :
This result cannot be used to find the exponent of 6 in n ! as 6 is composite of 2 and 3.

Illustrating the Concepts :
Find the exponent the 2 in 50 ! ?
 50   50   50   50   50 
E2 (50!)      2    3    4    5   25  12  6  3  1  47
 2  2  2  2  2 


Illustration - 2 The number of zeroes in 100! are :
(A) 20 (B) 24 (C) 97 (D) 28
SOLUTION : (B)
We know, 10 = 5 × 2
So, to form one 10, we need one 2 and one 5.
Number of 10’s will be same as min {E2 (100!), E5 (100!) }

100  100  100  100  100  100 
E2 100! =       
 2   22   23   24   25   26 

= 50 + 25 + 12 + 6 + 3 + 1 = 97

100  100 
E5 100! =    = 20 + 4 = 24
 5   52 

E10(100!) = min {97, 24} = 24




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FUNDAMENTAL PRINCIPLE OF COUNTING Section - 2

2.1. Addition Principle
If a work can be done in m different ways and another work which is independent of first can
be done in n different ways. then either of the two operations can be performed in (m + n) ways.

Illustrating the Concepts :
There are 15 gates to enter a city from north and 10 gates to enter the city from east. In how many
ways a person can enter the city ?
Number of ways to enter the city from north = 15
Number of ways to enter the city from east = 10
A person can enter the city from north or from east.
Hence, the number of ways to enter the city = 15 + 10 = 25

Illustration - 3 There are 15 students is a class in which 10 are boys and 5 are girls. The class teacher
selects either a boy or a girl for monitor of the class. In how many ways the class teacher can make this
selection ?
(A) 5 (B) 10 (C) 15 (D) 150
SOLUTION : (C)
A boy can be selected for the post of monitor in 10 ways.
A girl can be selected for the post of monitor in 5 ways.
Number of ways in which either a boy or a girl can be selected = 10 + 5 = 15


2.2. Multiplication Principle
If one Operation (I) can be done in m different ways and another Operation (II) can be performed in n
different ways, then total number of ways in which both of these can be performed together is m × n. If there
are more than two operations to be done, then the total number of different ways to do all of them together will
be m × n × p × .........


For example :
Operation A can be performed in “a, b, c” different ways and Operation B can be performed in p, q different
ways, then possible ways to perform Operation A and Operation B together are a - p, a - q, b - p, b - q,
c - p, c - q i.e. each way of Operation A is combined by 2 ways of Operation B.
i.e. Total ways = 2 + 2 + 2 = 2 added 3 times = 2 × 3 = 6.
In general, total ways = m × n where m and n are individual ways of performing the operations.


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