ICAI Notes- Basic Concepts of Permutations and Combinations | Quantitative Aptitude for CA Foundation PDF Download

 Page 1


BASIC CONCEPTS OF PERMUTATIONS
AND COMBINATIONS
5
CHAPTER
After reading this Chapter a student will be able to understand —
? difference between permutation and combination for the purpose of arranging different
objects;
? number of permutations and combinations when r objects are chosen out of n different
objects.
? meaning and computational techniques of circular permutation and permutation with
restrictions.
F ac t o r ial
Fundamental Principle Permutations
Multiplication
Rule
Properties of
Permutations
Permutations with
restrictions
Circular
Permuations
Combinations
Addition
Rule
Properties of
Combinations
CHAPTER OVERVIEW
© The Institute of Chartered Accountants of India
Page 2


BASIC CONCEPTS OF PERMUTATIONS
AND COMBINATIONS
5
CHAPTER
After reading this Chapter a student will be able to understand —
? difference between permutation and combination for the purpose of arranging different
objects;
? number of permutations and combinations when r objects are chosen out of n different
objects.
? meaning and computational techniques of circular permutation and permutation with
restrictions.
F ac t o r ial
Fundamental Principle Permutations
Multiplication
Rule
Properties of
Permutations
Permutations with
restrictions
Circular
Permuations
Combinations
Addition
Rule
Properties of
Combinations
CHAPTER OVERVIEW
© The Institute of Chartered Accountants of India
BUSINESS MATHEMATICS
5. 2
In this chapter we will learn problem of arranging and grouping of certain things, taking particular
number of things at a time. It should be noted that (a, b) and (b, a) are two different arrangements,
but they represent the same group. In case of arrangements, the sequence or order of things is
also taken into account.
The manager of a large bank has a difficult task of filling two important positions from a group of
five equally qualified employees. Since none of them has had actual experience, he decides to
allow each of them to work for one month in each of the positions before he makes the decision.
How long can the bank operate before the positions are filled by permanent appointments?
Solution to above - cited situation requires an efficient counting of the possible ways in which the
desired outcomes can be obtained. A listing of all possible outcomes may be desirable, but is
likely to be very tedious and subject to errors of duplication or omission. We need to devise
certain techniques which will help us to cope with such problems. The techniques of permutation
and combination will help in tackling problems such as above.
FUNDAMENTAL PRINCIPLES OF COUNTING
(a) Multiplication Rule: If certain thing may be done in ‘m’ different ways and when it has
been done, a second thing can be done in ‘n ‘ different ways then total number of ways of
doing both things simultaneously = m × n.
Eg. if one can going to school by 5 different buses and then come back by 4 different buses then
total number of ways of going to and coming back from school = 5 × 4 = 20.
(b) Addition Rule : It there are two alternative jobs which can be done in ‘m’ ways and in ‘n’
ways respectively then either of two jobs can be done in (m + n) ways.
Eg. if one wants to go school by bus where there are 5 buses or to by auto where there are 4
autos, then total number of ways of going school = 5 + 4 = 9.
Note :- 1)
AND ? Multiply
OR ? Add
2) The above fundamental principles may be generalised, wherever necessary.
Definition: The factorial n, written as n! or n , represents the product of all integers from 1 to n
both inclusive. To make the notation meaningful, when n = o, we define o! or 0 = 1.
Thus, n! = n (n – 1) (n – 2) ….. …3.2.1
Example 1: Find 5!, 4! and 6!
Solution: 5! = 5 × 4 × 3 × 2 × 1 = 120; 4! = 4 × 3 × 2 × 1 = 24; 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720.
© The Institute of Chartered Accountants of India
Page 3


BASIC CONCEPTS OF PERMUTATIONS
AND COMBINATIONS
5
CHAPTER
After reading this Chapter a student will be able to understand —
? difference between permutation and combination for the purpose of arranging different
objects;
? number of permutations and combinations when r objects are chosen out of n different
objects.
? meaning and computational techniques of circular permutation and permutation with
restrictions.
F ac t o r ial
Fundamental Principle Permutations
Multiplication
Rule
Properties of
Permutations
Permutations with
restrictions
Circular
Permuations
Combinations
Addition
Rule
Properties of
Combinations
CHAPTER OVERVIEW
© The Institute of Chartered Accountants of India
BUSINESS MATHEMATICS
5. 2
In this chapter we will learn problem of arranging and grouping of certain things, taking particular
number of things at a time. It should be noted that (a, b) and (b, a) are two different arrangements,
but they represent the same group. In case of arrangements, the sequence or order of things is
also taken into account.
The manager of a large bank has a difficult task of filling two important positions from a group of
five equally qualified employees. Since none of them has had actual experience, he decides to
allow each of them to work for one month in each of the positions before he makes the decision.
How long can the bank operate before the positions are filled by permanent appointments?
Solution to above - cited situation requires an efficient counting of the possible ways in which the
desired outcomes can be obtained. A listing of all possible outcomes may be desirable, but is
likely to be very tedious and subject to errors of duplication or omission. We need to devise
certain techniques which will help us to cope with such problems. The techniques of permutation
and combination will help in tackling problems such as above.
FUNDAMENTAL PRINCIPLES OF COUNTING
(a) Multiplication Rule: If certain thing may be done in ‘m’ different ways and when it has
been done, a second thing can be done in ‘n ‘ different ways then total number of ways of
doing both things simultaneously = m × n.
Eg. if one can going to school by 5 different buses and then come back by 4 different buses then
total number of ways of going to and coming back from school = 5 × 4 = 20.
(b) Addition Rule : It there are two alternative jobs which can be done in ‘m’ ways and in ‘n’
ways respectively then either of two jobs can be done in (m + n) ways.
Eg. if one wants to go school by bus where there are 5 buses or to by auto where there are 4
autos, then total number of ways of going school = 5 + 4 = 9.
Note :- 1)
AND ? Multiply
OR ? Add
2) The above fundamental principles may be generalised, wherever necessary.
Definition: The factorial n, written as n! or n , represents the product of all integers from 1 to n
both inclusive. To make the notation meaningful, when n = o, we define o! or 0 = 1.
Thus, n! = n (n – 1) (n – 2) ….. …3.2.1
Example 1: Find 5!, 4! and 6!
Solution: 5! = 5 × 4 × 3 × 2 × 1 = 120; 4! = 4 × 3 × 2 × 1 = 24; 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720.
© The Institute of Chartered Accountants of India
5. 3 BASIC CONCEPTS OF PERMUTATIONS AND COMBINATIONS
Example 2: Find 9 ! / 6 ! ; 10 ! / 7 !.
Solution: 
? ? ? ? ? ?
? ?
9 ! 9  8  7  6 ! 10 ! 10  9  8  7 !
 =  = 9  8  7 = 504 ;  = 
6 ! 6 ! 7 ! 7 !
 = 10 × 9 × 8 =720
Example 3: Find x if 1/9 ! + 1/10 ! = x/11 !
Solution: 1/9! (1 + 1/10) = x/11 × 10 × 9! or, 11/10 = x/11 × 10 i.e., x = 121
Example 4: Find n if ? n +1=30 n 1
Solution: 30 n +1=30 n 1      (n +1).n n 1 n 1 ? ? ? ? ?
or, n
2
 + n = 30 or, n
2
 + n – 30 or, n
2
 + 6n – 5n – 30 = 0      or,   (n + 6) (n – 5) = 0
either n = 5 or n = –6. (Not possible)
?
 n = 5.
A group of persons want themselves to be photographed. They approach the photographer and
request him to take as many different photographs as possible with persons standing in different
positions amongst themselves. The photographer wants to calculate how many films does he
need to exhaust all possibilities? How can he calculate the number?
In the situations such as above, we can use permutations to find out the exact number of films.
Definition: The ways of arranging or selecting smaller or equal number of persons or objects
from a group of persons or collection of objects with due regard being paid to the order of
arrangement or selection, are called permutations.
Let us explain, how the idea of permutation will help the photographer. Suppose the group
consists of Mr. Suresh, Mr. Ramesh and Mr. Mahesh. Then how many films does the photographer
need? He has to arrange three persons amongst three places with due regard to order. Then the
various possibilities are (Suresh, Mahesh, Ramesh), (Suresh, Ramesh, Mahesh), (Ramesh, Suresh,
Mahesh), (Ramesh, Mahesh, Suresh), (Mahesh, Ramesh, Suresh) and (Mahesh, Suresh, Ramesh ).
Thus there are six possibilities. Therefore he needs six films. Each one of these possibilities is
called a permutation of three persons taken at a time.
This may also be exhibited as follows :
Alternative Place 1 Place2 Place 3
 1 Suresh Mahesh Ramesh
2 Suresh Ramesh Mahesh
3 Ramesh Suresh Mahesh
4 Ramesh Mahesh Suresh
5 Mahesh Ramesh Suresh
6 Mahesh Suresh Ramesh
with this example as a base, we can introduce a general formula to find the number of
permutations.
© The Institute of Chartered Accountants of India
Page 4


BASIC CONCEPTS OF PERMUTATIONS
AND COMBINATIONS
5
CHAPTER
After reading this Chapter a student will be able to understand —
? difference between permutation and combination for the purpose of arranging different
objects;
? number of permutations and combinations when r objects are chosen out of n different
objects.
? meaning and computational techniques of circular permutation and permutation with
restrictions.
F ac t o r ial
Fundamental Principle Permutations
Multiplication
Rule
Properties of
Permutations
Permutations with
restrictions
Circular
Permuations
Combinations
Addition
Rule
Properties of
Combinations
CHAPTER OVERVIEW
© The Institute of Chartered Accountants of India
BUSINESS MATHEMATICS
5. 2
In this chapter we will learn problem of arranging and grouping of certain things, taking particular
number of things at a time. It should be noted that (a, b) and (b, a) are two different arrangements,
but they represent the same group. In case of arrangements, the sequence or order of things is
also taken into account.
The manager of a large bank has a difficult task of filling two important positions from a group of
five equally qualified employees. Since none of them has had actual experience, he decides to
allow each of them to work for one month in each of the positions before he makes the decision.
How long can the bank operate before the positions are filled by permanent appointments?
Solution to above - cited situation requires an efficient counting of the possible ways in which the
desired outcomes can be obtained. A listing of all possible outcomes may be desirable, but is
likely to be very tedious and subject to errors of duplication or omission. We need to devise
certain techniques which will help us to cope with such problems. The techniques of permutation
and combination will help in tackling problems such as above.
FUNDAMENTAL PRINCIPLES OF COUNTING
(a) Multiplication Rule: If certain thing may be done in ‘m’ different ways and when it has
been done, a second thing can be done in ‘n ‘ different ways then total number of ways of
doing both things simultaneously = m × n.
Eg. if one can going to school by 5 different buses and then come back by 4 different buses then
total number of ways of going to and coming back from school = 5 × 4 = 20.
(b) Addition Rule : It there are two alternative jobs which can be done in ‘m’ ways and in ‘n’
ways respectively then either of two jobs can be done in (m + n) ways.
Eg. if one wants to go school by bus where there are 5 buses or to by auto where there are 4
autos, then total number of ways of going school = 5 + 4 = 9.
Note :- 1)
AND ? Multiply
OR ? Add
2) The above fundamental principles may be generalised, wherever necessary.
Definition: The factorial n, written as n! or n , represents the product of all integers from 1 to n
both inclusive. To make the notation meaningful, when n = o, we define o! or 0 = 1.
Thus, n! = n (n – 1) (n – 2) ….. …3.2.1
Example 1: Find 5!, 4! and 6!
Solution: 5! = 5 × 4 × 3 × 2 × 1 = 120; 4! = 4 × 3 × 2 × 1 = 24; 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720.
© The Institute of Chartered Accountants of India
5. 3 BASIC CONCEPTS OF PERMUTATIONS AND COMBINATIONS
Example 2: Find 9 ! / 6 ! ; 10 ! / 7 !.
Solution: 
? ? ? ? ? ?
? ?
9 ! 9  8  7  6 ! 10 ! 10  9  8  7 !
 =  = 9  8  7 = 504 ;  = 
6 ! 6 ! 7 ! 7 !
 = 10 × 9 × 8 =720
Example 3: Find x if 1/9 ! + 1/10 ! = x/11 !
Solution: 1/9! (1 + 1/10) = x/11 × 10 × 9! or, 11/10 = x/11 × 10 i.e., x = 121
Example 4: Find n if ? n +1=30 n 1
Solution: 30 n +1=30 n 1      (n +1).n n 1 n 1 ? ? ? ? ?
or, n
2
 + n = 30 or, n
2
 + n – 30 or, n
2
 + 6n – 5n – 30 = 0      or,   (n + 6) (n – 5) = 0
either n = 5 or n = –6. (Not possible)
?
 n = 5.
A group of persons want themselves to be photographed. They approach the photographer and
request him to take as many different photographs as possible with persons standing in different
positions amongst themselves. The photographer wants to calculate how many films does he
need to exhaust all possibilities? How can he calculate the number?
In the situations such as above, we can use permutations to find out the exact number of films.
Definition: The ways of arranging or selecting smaller or equal number of persons or objects
from a group of persons or collection of objects with due regard being paid to the order of
arrangement or selection, are called permutations.
Let us explain, how the idea of permutation will help the photographer. Suppose the group
consists of Mr. Suresh, Mr. Ramesh and Mr. Mahesh. Then how many films does the photographer
need? He has to arrange three persons amongst three places with due regard to order. Then the
various possibilities are (Suresh, Mahesh, Ramesh), (Suresh, Ramesh, Mahesh), (Ramesh, Suresh,
Mahesh), (Ramesh, Mahesh, Suresh), (Mahesh, Ramesh, Suresh) and (Mahesh, Suresh, Ramesh ).
Thus there are six possibilities. Therefore he needs six films. Each one of these possibilities is
called a permutation of three persons taken at a time.
This may also be exhibited as follows :
Alternative Place 1 Place2 Place 3
 1 Suresh Mahesh Ramesh
2 Suresh Ramesh Mahesh
3 Ramesh Suresh Mahesh
4 Ramesh Mahesh Suresh
5 Mahesh Ramesh Suresh
6 Mahesh Suresh Ramesh
with this example as a base, we can introduce a general formula to find the number of
permutations.
© The Institute of Chartered Accountants of India
BUSINESS MATHEMATICS
5. 4
Number of Permutations when r objects are chosen out of n different objects. (Denoted by 
n
P
r
or 
n
P
r
 or P
(n, r)
 ) :
Let us consider the problem of finding the number of ways in which the first r rankings are
secured by n students in a class. As any one of the n students can secure the first rank, the number
of ways in which the first rank is secured is n.
Now consider the second rank. There are (n – 1) students left and the second rank can be secured
by any one of them. Thus the different possibilities are (n – 1) ways. Now, applying fundamental
principle, we can see that the first two ranks can be secured in n (n – 1) ways by these n students.
After calculating for two ranks, we find that the third rank can be secured by any one of the
remaining (n – 2) students. Thus, by applying the generalized fundamental principle, the first
three ranks can be secured in n (n – 1) (n – 2) ways .
Continuing in this way we can visualise that the number of ways are reduced by one as the rank
is increased by one. Therefore, again, by applying the generalised fundamental principle for r
different rankings, we calculate the number of ways in which the first r ranks are secured by n
students as
 n
P
r
= n {(n – 1)… ? ?
1 nr ? ? }
= n (n – 1) … (n – r + 1)
Theorem : The number of permutations of n things when r are chosen at a time
n
P
r
 =n  ( n – 1 ) ( n – 2 ) … ( n – r + 1 )
where the product has exactly r factors.
1 Number of permutations of n different things taken all n things at a time is given by
n
P
n 
= n (n – 1) (n – 2) …. (n – n + 1)
=n (n – 1) (n – 2) ….. 2.1 = n! ...(1)
2.
n
P
r
 using factorial notation.
n
P
r
 = n. (n – 1) (n – 2) ….. (n – r + 1)
= n (n – 1) (n – 2) ….. (n – r + 1) × 
? ? ?
? ? ?
(n r) (n r 1) 2.1
1.2 ...(n r 1) (n r)
= n!/( n – r )! ...(2)
Thus
?
n!
n
P = 
r
( n  r )!
3. Justification for 0! = 1. Now applying r = n in the formula for 
n
P
r
, we get
© The Institute of Chartered Accountants of India
Page 5


BASIC CONCEPTS OF PERMUTATIONS
AND COMBINATIONS
5
CHAPTER
After reading this Chapter a student will be able to understand —
? difference between permutation and combination for the purpose of arranging different
objects;
? number of permutations and combinations when r objects are chosen out of n different
objects.
? meaning and computational techniques of circular permutation and permutation with
restrictions.
F ac t o r ial
Fundamental Principle Permutations
Multiplication
Rule
Properties of
Permutations
Permutations with
restrictions
Circular
Permuations
Combinations
Addition
Rule
Properties of
Combinations
CHAPTER OVERVIEW
© The Institute of Chartered Accountants of India
BUSINESS MATHEMATICS
5. 2
In this chapter we will learn problem of arranging and grouping of certain things, taking particular
number of things at a time. It should be noted that (a, b) and (b, a) are two different arrangements,
but they represent the same group. In case of arrangements, the sequence or order of things is
also taken into account.
The manager of a large bank has a difficult task of filling two important positions from a group of
five equally qualified employees. Since none of them has had actual experience, he decides to
allow each of them to work for one month in each of the positions before he makes the decision.
How long can the bank operate before the positions are filled by permanent appointments?
Solution to above - cited situation requires an efficient counting of the possible ways in which the
desired outcomes can be obtained. A listing of all possible outcomes may be desirable, but is
likely to be very tedious and subject to errors of duplication or omission. We need to devise
certain techniques which will help us to cope with such problems. The techniques of permutation
and combination will help in tackling problems such as above.
FUNDAMENTAL PRINCIPLES OF COUNTING
(a) Multiplication Rule: If certain thing may be done in ‘m’ different ways and when it has
been done, a second thing can be done in ‘n ‘ different ways then total number of ways of
doing both things simultaneously = m × n.
Eg. if one can going to school by 5 different buses and then come back by 4 different buses then
total number of ways of going to and coming back from school = 5 × 4 = 20.
(b) Addition Rule : It there are two alternative jobs which can be done in ‘m’ ways and in ‘n’
ways respectively then either of two jobs can be done in (m + n) ways.
Eg. if one wants to go school by bus where there are 5 buses or to by auto where there are 4
autos, then total number of ways of going school = 5 + 4 = 9.
Note :- 1)
AND ? Multiply
OR ? Add
2) The above fundamental principles may be generalised, wherever necessary.
Definition: The factorial n, written as n! or n , represents the product of all integers from 1 to n
both inclusive. To make the notation meaningful, when n = o, we define o! or 0 = 1.
Thus, n! = n (n – 1) (n – 2) ….. …3.2.1
Example 1: Find 5!, 4! and 6!
Solution: 5! = 5 × 4 × 3 × 2 × 1 = 120; 4! = 4 × 3 × 2 × 1 = 24; 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720.
© The Institute of Chartered Accountants of India
5. 3 BASIC CONCEPTS OF PERMUTATIONS AND COMBINATIONS
Example 2: Find 9 ! / 6 ! ; 10 ! / 7 !.
Solution: 
? ? ? ? ? ?
? ?
9 ! 9  8  7  6 ! 10 ! 10  9  8  7 !
 =  = 9  8  7 = 504 ;  = 
6 ! 6 ! 7 ! 7 !
 = 10 × 9 × 8 =720
Example 3: Find x if 1/9 ! + 1/10 ! = x/11 !
Solution: 1/9! (1 + 1/10) = x/11 × 10 × 9! or, 11/10 = x/11 × 10 i.e., x = 121
Example 4: Find n if ? n +1=30 n 1
Solution: 30 n +1=30 n 1      (n +1).n n 1 n 1 ? ? ? ? ?
or, n
2
 + n = 30 or, n
2
 + n – 30 or, n
2
 + 6n – 5n – 30 = 0      or,   (n + 6) (n – 5) = 0
either n = 5 or n = –6. (Not possible)
?
 n = 5.
A group of persons want themselves to be photographed. They approach the photographer and
request him to take as many different photographs as possible with persons standing in different
positions amongst themselves. The photographer wants to calculate how many films does he
need to exhaust all possibilities? How can he calculate the number?
In the situations such as above, we can use permutations to find out the exact number of films.
Definition: The ways of arranging or selecting smaller or equal number of persons or objects
from a group of persons or collection of objects with due regard being paid to the order of
arrangement or selection, are called permutations.
Let us explain, how the idea of permutation will help the photographer. Suppose the group
consists of Mr. Suresh, Mr. Ramesh and Mr. Mahesh. Then how many films does the photographer
need? He has to arrange three persons amongst three places with due regard to order. Then the
various possibilities are (Suresh, Mahesh, Ramesh), (Suresh, Ramesh, Mahesh), (Ramesh, Suresh,
Mahesh), (Ramesh, Mahesh, Suresh), (Mahesh, Ramesh, Suresh) and (Mahesh, Suresh, Ramesh ).
Thus there are six possibilities. Therefore he needs six films. Each one of these possibilities is
called a permutation of three persons taken at a time.
This may also be exhibited as follows :
Alternative Place 1 Place2 Place 3
 1 Suresh Mahesh Ramesh
2 Suresh Ramesh Mahesh
3 Ramesh Suresh Mahesh
4 Ramesh Mahesh Suresh
5 Mahesh Ramesh Suresh
6 Mahesh Suresh Ramesh
with this example as a base, we can introduce a general formula to find the number of
permutations.
© The Institute of Chartered Accountants of India
BUSINESS MATHEMATICS
5. 4
Number of Permutations when r objects are chosen out of n different objects. (Denoted by 
n
P
r
or 
n
P
r
 or P
(n, r)
 ) :
Let us consider the problem of finding the number of ways in which the first r rankings are
secured by n students in a class. As any one of the n students can secure the first rank, the number
of ways in which the first rank is secured is n.
Now consider the second rank. There are (n – 1) students left and the second rank can be secured
by any one of them. Thus the different possibilities are (n – 1) ways. Now, applying fundamental
principle, we can see that the first two ranks can be secured in n (n – 1) ways by these n students.
After calculating for two ranks, we find that the third rank can be secured by any one of the
remaining (n – 2) students. Thus, by applying the generalized fundamental principle, the first
three ranks can be secured in n (n – 1) (n – 2) ways .
Continuing in this way we can visualise that the number of ways are reduced by one as the rank
is increased by one. Therefore, again, by applying the generalised fundamental principle for r
different rankings, we calculate the number of ways in which the first r ranks are secured by n
students as
 n
P
r
= n {(n – 1)… ? ?
1 nr ? ? }
= n (n – 1) … (n – r + 1)
Theorem : The number of permutations of n things when r are chosen at a time
n
P
r
 =n  ( n – 1 ) ( n – 2 ) … ( n – r + 1 )
where the product has exactly r factors.
1 Number of permutations of n different things taken all n things at a time is given by
n
P
n 
= n (n – 1) (n – 2) …. (n – n + 1)
=n (n – 1) (n – 2) ….. 2.1 = n! ...(1)
2.
n
P
r
 using factorial notation.
n
P
r
 = n. (n – 1) (n – 2) ….. (n – r + 1)
= n (n – 1) (n – 2) ….. (n – r + 1) × 
? ? ?
? ? ?
(n r) (n r 1) 2.1
1.2 ...(n r 1) (n r)
= n!/( n – r )! ...(2)
Thus
?
n!
n
P = 
r
( n  r )!
3. Justification for 0! = 1. Now applying r = n in the formula for 
n
P
r
, we get
© The Institute of Chartered Accountants of India
5 . 5 BASIC CONCEPTS OF PERMUTATIONS AND COMBINATIONS
n
P
n 
= n!/ (n – n)! = n!/0!
But from Result 1 we find that 
n
P
n
 = n!. Therefore, by applying this
we derive, 0! = n! / n! = 1
Example 1: Evaluate each of 
5
P
3
, 
10
P
2
, 
11
P
5
.
Solution:
5
P
3 
= 5×4× (5–3+1) = 5 × 4 × 3 = 60,
10
P
2
 = 10 × …. × (10–2+1) = 10 × 9 = 90,
11
P
5
 = 11! / (11 – 5)! = 11 × 10 × 9 × 8 × 7 × 6! / 6! = 11 × 10 × 9 × 8 × 7 = 55440.
Example 2: How many three letters words can be formed using the letters of the words
(a) SQUARE and (b) HEXAGON?
(Any arrangement of letters is called a word even though it may or may not have any meaning or pronunciation).
Solution:
(a) Since the word ‘SQUARE’ consists of 6 different letters, the number of permutations of
choosing 3 letters out of six equals 
6
P
3
 = 6 × 5 × 4 = 120.
(b) Since the word ‘HEXAGON’ contains 7 different letters, the number of  permutations is
7
P
3
 = 7 × 6 × 5 = 210.
Example 3: In how many different ways can five persons stand in a line for a group
photograph?
Solution: Here we know that the order is important. Hence, this is the number of permutations
of five things taken all at a time. Therefore, this equals
5
P
5
 = 5! = 5 × 4 × 3 × 2 × 1 = 120 ways.
Example 4: First, second and third prizes are to be awarded at an engineering fair in which 13
exhibits have been entered. In how many different ways can the prizes be awarded?
Solution: Here again, order of selection is important and repetitions are not meaningful as no
exhibit can receive more than one prize. Hence , the answer is the number of permutations of 13
things taken three at a time. Therefore, we find 
13
P
3
 = 13!/10! = 13×12×11 = 1,716 ways.
Example 5: In how many different ways can 3 students be associated with 4 chartered accountants,
assuming that each chartered accountant can take at most one student?
Solution: This equals the number of permutations of choosing 3 persons out of 4. Hence , the
answer is 
4
P
3
 = 4×3×2 = 24.
Example 6: If six times the number permutations of n things taken 3 at a time is equal to seven
times the number of permutations of (n – 1) things taken 3 at a time, find n.
Solution: We are given that 6 × 
n
P
3
 = 7 × 
n-1
P
3 
and we have to solve this equality to find the value
of n. Therefore,
n n 1
6 7
n-3 n-4
?
?
or, 6 n (n – 1) (n – 2) = 7 (n – 1) (n – 2) (n – 3)
or, 6 n = 7 (n – 3)
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