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Permutations and
Combinations

Quantitative Aptitude & Business Statistics

The Fundamental Principle of
Multiplication
• If there are
• n1 ways of doing one operation,
• n2 ways of doing a second
operation, n3 ways of doing a
third operation , and so forth,

Quantitative Aptitude & Business
2
Statistics:Permutations and Combinations

• then the sequence of k
operations can be performed in
n1 n2 n3….. nk ways.
• N= n1 n2 n3….. nk

3

Example 1
• A used car wholesaler has agents
who classify cars by size (full,
medium, and compact) and age (0
- 2 years, 2- 4 years, 4 - 6 years,
and over 6 years).
• Determine the number of possible
automobile classifications.

4

Solution 0-2
2-4
Full(F) 4-6
>6
0-2
2-4
Medium 4-6
>6
(M)
0-2
2-4
Compact 4-6
(C) >6
The tree diagram enumerates all possible
classifications, the total number of which
is 3x4= 12. Quantitative Aptitude & Business
5

Example 2
• Mr. X has 2 pairs of trousers, 3
shirts and 2 ties.
• He chooses a pair of trousers, a
shirt and a tie to wear everyday.
• Find the maximum number of
days he does not need to repeat
his clothing.

6

Solution

• The maximum number of days
he does not need to repeat his
clothing is 2×3×2 = 12
7

1.2 Factorials

• The product of the first n
consecutive integers is denoted
by n! and is read as “factorial n”.
• That is n! = 1×2×3×4×…. ×(n-1)
×n
• For example,
• 4!=1x2x3x4=24,
• 7!=1×2×3×4×5×6×7=5040.
• Note 0! defined to be 1.
8

•The product of any number of
consecutive integers can be
expressed as a quotient of two
factorials, for example,
• 6×7×8×9 = 9!/5! = 9! / (9 – 4)!
• 11×12×13×14×15= 15! / 10!

=15! / (15 – 5)!
In particular,
• n×(n – 1)×(n – 2)×...×(n – r + 1)
9

1.3 Permutations

• (A) Permutations
• A permutation is an arrangement
of objects.
• abc and bca are two different
permutations.

10

• 1. Permutations with repetition

– The number of permutations of r
objects, taken from n unlike objects,
– can be found by considering the
number of ways of filling r blank
spaces in order with the n given
objects.
– If repetition is allowed, each blank
space can be filled by the objects in n
different ways.
11

1 2 3 4 r
n n n n n

• Therefore, the number of
permutations of r objects,
taken from n unlike objects,
• each of which may be
repeated any number of times
= n × n × n ×.... × n(r factors) =
nr
12

2. Permutations without repetition

• If repetition is not allowed,
the number of ways of filling
each blank space is one less
than the preceding one.

1 2 3 4 r
n n-1 n-2 n-3 n-r+1

13

Therefore, the number of
permutations of r objects, taken
from n unlike objects, each of
which can only be used once in
each permutation
=n(n— 1)(n—2) .... (n—r + 1)
Various notations are used to
represent the number of
permutations of a set of n
elements taken r at a time;
14

• some of them are n
P , Pr , P (n, r )
r n

n!
( n − r )!
Since
n( n − 1)(n − 2)....(n − r + 1)(n − r )...3 ⋅ 2 ⋅ 1
=
( n − r )...3 ⋅ 2 ⋅ 1
Prn , n Pr , P (n, r )

= n( n − 1)(n − 2)....(n − r + 1)
=P r
n

n!
We have P =
n

(n − r )!
r
15

Example 3

• How many 4-digit numbers can be
made from the figures 1, 2, 3, 4, 5,
6, 7 when
• (a) repetitions are allowed;
• (b) repetition is not allowed?

16

• Solution
• (a) Number of 4-digit numbers
= 74 = 2401.
• (b) Number of 4 digit numbers
=7 ×6 ×5 ×4 = 840.

17

Example 4
• In how many ways can 10 men
be arranged
• (a) in a row,
• (b) in a circle?
• Solution
• (a) Number of ways is
= 3628800
10
P
10

18

• Suppose we arrange
the 4 letters A, B, C
and D in a circular A
arrangement as
shown.
D B
• Note that the
arrangements ABCD,
BCDA, CDAB and C
DABC are not
distinguishable.

19

• For each circular arrangement
there are 4 distinguishable
arrangements on a line.
• If there are P circular
arrangements, these yield 4P
arrangements on a line, which
we know is 4!.

4!
Hence P = = (4 − 1)!= 3!
4
20

Solution (b)

• The number of distinct circular
arrangements of n objects is
(n —1)!
• Hence 10 men can be arranged
in a circle in 9! = 362 880 ways.

21

(B) Conditional
Permutations
• When arranging elements in
order , certain restrictions may
apply.
• In such cases the restriction
should be dealt with first..

22

Example 5
How many even numerals between 200
and 400 can be formed by using 1, 2, 3, 4,
5 as digits
(a) if any digit may be repeated;
(b) if no digit may be repeated?

23

• Solution (a)
• Number of ways of choosing the
hundreds’ digit = 2.
tens’ digit = 5.
unit digit = 2.
• Number of even numerals
between 200 and 400 is
2 × 5 × 2 = 20.

24

•Solution (b)
•If the hundreds’ digit is 2,
then the number of ways of choosing
an even unit digit = 1,
and the number of ways of choosing a
tens’ digit = 3.
•the number of numerals formed
1×1×3 = 3.

25

If the hundreds’ digit is 3, then the
number of ways of choosing an
even. unit digit = 2, and the
number of ways of choosing a tens’
digit = 3.
• number of numerals formed
= 1×2×3 = 6.
• the number of even numerals
between 200 and 400 = 3 + 6 =
9
26

Example 6
In how many ways can
7 different books be
arranged on a shelf
(a) if two particular
books are together;

27

• Solution (a)
• If two particular books are
together, they can be considered
as one book for arranging.
• The number of arrangement of 6
books
= 6! = 720.
• The two particular books can be
arranged in 2 ways among
themselves.
• The number of arrangement of 7
books with two particular books
28

(b) if two particular books are
separated?
• Solution (b)
• Total number of arrangement of 7
books = 7! = 5040.
• the number of arrangement of 7
books with 2 particular books
separated = 5040 －1440 = 3600.

29

(C) Permutation with
Indistinguishable Elements

• In some sets of elements there
may be certain members that
are indistinguishable from each
other.
• The example below illustrates
how to find the number of
permutations in this kind of
situation.
30

Example 7
In how many ways can the letters of
the word “ISOS CELES” be
arranged to form a new “word” ?

• Solution
• If each of the 9 letters of
“ISOSCELES” were different,
there would be P= 9! different
possible words.
31

• However, the 3 S’s are
indistinguishable from each
other and can be permuted in 3!
different ways.
• As a result, each of the 9!
arrangements of the letters of
“ISOSCELES” that would
otherwise spell a new word will
be repeated 3! times.

32

• To avoid counting repetitions
resulting from the 3 S’s, we must
divide 9! by 3!.
• Similarly, we must divide by 2! to
avoid counting repetitions
resulting from the 2
indistinguishable E’s.
• Hence the total number of words
that can be formed is
9! ÷3! ÷2! = 30240
33

• If a set of n elements has k1
indistinguishable elements of one
kind, k2 of another kind,
and so on for r kinds of elements,
then the number of permutations of
the set of n elements is

n!
k1!k 2 !⋅ ⋅ ⋅ ⋅ k r !
34

1.4 Combinations

• When a selection of objects is
made with no regard being paid to
order, it is referred to as a
combination.
• Thus, ABC, ACB, BAG, BCA, CAB,
CBA are different permutation, but
they are the same combination of
letters.

35

• Suppose we wish to appoint a
committee of 3 from a class of 30
students.
• We know that P330 is the number of
different ordered sets of 3 students
each that may be selected from
among 30 students.
• However, the ordering of the
students on the committee has no
significance,
36

• so our problem is to determine
the number of three-element
unordered subsets that can be
constructed from a set of 30
elements.
• Any three-element set may be
ordered in 3! different ways, so
P330 is 3! times too large.
• Hence, if we divide P330 by 3!,the
result will be the number of
unordered subsets of 30
elements taken 3 at a time.
37

• This number of unordered
subsets is also called the
number of combinations of 30
elements taken 3 at a time,
denoted by C330 and

1 30
C = P3
30
3
3!
30!
= = 4060
27!3!Quantitative Aptitude & Business
38

• In general, each unordered r-
element subset of a given n-
element set (r≤ n) is called a
combination.
• The number of combinations of
n elements taken r at a time is
denoted by Cnr or nCr or C(n, r) .

39

• A general equation relating
combinations to permutations
is
1 n n!
C r = Pr =
n

r! (n − r )!r!

40

• Note:
• (1) Cnn = Cn0 = 1
• (2) Cn1 = n
• (3) Cnn = Cnn-r

41

Example8
• If 167 C 90+167 C x =168 C x then x
is
• Solution: nCr-1+nCr=n+1 Cr
• Given 167 C90+167c x =168C x
• We may write
• 167C91-1 + 167 C91=167+1 C61
• =168 C91
• X=91
42

Example9
• If 20 C 3r= 20C 2r+5 ,find r
• Using nCr=nC n-r in the right –side
of the given equation ,we find ,
• 20 C 3r =20 C 20-(2r+5)
• 3r=15-2r
• r=3

43

Example 10
• If 100 C 98 =999 C 97 +x C 901 find x.
• Solution 100C 98 =999C 98 +999C97
• = 999C901+999C97
• X=999

44

Example11
• If 13 C 6 + 2 13 C5 +13 C 4 =15 C x ,the value of
x is
• Solution :
• 15C x= 13C 6 + 13 C 5 + 13 C 4 =
• =(13c 6+13 C 5 ) +
• (13 C 5 + 13 C 4)
• = 14 C 6 +14 C 5 =15C6
• X=6 or x+6 =15
• X=6 or 8

45

Example12
• If n C r-1=36 ,n Cr =84 and n C r+1 =126 then
find r
• Solution
nCr 84 7
= =
nCr −1 36 3
• n-r+1 =7/3 * r

• 3/2 (r+1)+1 =7/3 * r
• nCr +1 126 3 r=3
= =
nCr 84 2

46

Example 13

• How many different 5-card
hands can be dealt from a deck
of 52 playing cards?

47

Solution

• Since we are not concerned with
the order in which each card is
dealt, our problem concerns the
number of combinations of 52
elements taken 5 at a time.
• The number of different hands is
C525＝ 2118760.

48

Example 14

6 points are given and no three of
them are collinear.
(a) How many triangles can be
formed by using 3 of the given
points as vertices?

49

Solution:

• Solution
• (a) Number of triangles
• = number of ways
• of selecting 3 points out of 6
• = C63 = 20.

50

• b) How many pairs of triangles
can be formed by using the 6
points as vertices ?

51

• Let the points be A, B, C, D, E, F.
• If A, B, C are selected to form a
triangles, then D, E, F must form
the other triangle.
• Similarly, if D, E, F are selected to
form a triangle, then A, B, C must
form the other triangle.

52

• Therefore, the selections A, B,
C and D, E, F give the same pair
of triangles and the same
applies to the other selections.
• Thus the number of ways of
forming a pair of triangles
= C63 ÷ 2 = 10

53

Example 15

• From among 25 boys who play
basketball, in how many different
ways can a team of 5 players be
selected if one of the players is to
be designated as captain?

54

Solution

• A captain may be chosen from any of the 25
players.
• The remaining 4 players can be chosen in C254
different ways.
• By the fundamental counting principle, the
total number of different teams that can be
formed is
25 × C244＝265650.

55

(B) Conditional
Combinations

• If a selection is to be
restricted in some way, this
restriction must be dealt with
first.
• The following examples
illustrate such conditional
combination problems.
56

A committee of 3 men
and 4 women is to be
selected from 6 men and
9 women.
If there is a married
couple among the 15
persons, in how many
ways can the committee
be selected so that it
contains the married
57

• Solution
• If the committee contains the
married couple, then only 2 men
and 3 women are to be selected
from the remaining 5 men and 8
women.
• The number of ways of selecting 2
men out of 5 = C52 = 10.

58

• The number of ways of selecting
3 women out of 8 =C83 = 56.
• the number of ways of selecting
the committee = lO × 56 = 560.

59

Example 17

• Find the number of ways a team
of 4 can be chosen from 15 boys
and 10 girls if
(a) it must contain 2 boys and 2
girls,

60

• Solution (a)
• Boys can be chosen in C152 = 105
ways
• Girls can be chosen in C102 = 45
ways.
• Total number of ways is 105 × 45
= 4725.

61

(b) it must contain at least 1 boy and 1
girl.

• Solution :
• If the team must contain at least 1
boy and 1 girl it can be formed in
the following ways:
• (I) 1 boy and 3 girls, with C151 × C103
= 1800 ways,
• (ii) 2 boys and 2 girls, with 4725
ways,
• (iii) 3 boys and 1 girl, with C153 ×
C101 = 4550 ways.

• the total number of teams is
62

Example 18

• Mr. .X has 12 friends and
wishes to invite 6 of them to a
party. Find the number
of ways he may do this if
(a) there is no restriction on
choice,

63

• Solution (a)
• An unrestricted choice of 6
out of 12 gives C126＝ 924.

64

two of the friends is a couple
• (b)
and will not attend separately,

65

B Solution

• If the couple attend, the
remaining 4 may then be chosen
from the other 10 in C104 ways.
• If the couple does not attend,
then He simply chooses 6 from
the other 10 in C106 ways.
• total number of ways is C104 +
C106 = 420.

66

Example 19
Find the number of ways in which
30 students can be divided into
three groups, each of 10 students,
if the order of the groups and the
arrangement of the students in a
group are immaterial.

67

• Solution
• Let the groups be denoted by A,
B and C. Since the arrangement
of the students in a group is
immaterial,
• group A can be selected from
the 30 students in C3010 ways .

68

• Group B can be selected from the
remaining 20 students in C2010
ways.
• There is only 1 way of forming
group C from the remaining 10
students.

69

• Since the order of the groups is
immaterial, we have to divide
the product C3010 × C2010 × C1010
by 3!,
• hence the total number of ways
of forming the three groups is

1
× C3 × C10 × C10
30 20 10

3!
70

Example20
• If n Pr = 604800 10 C r =120 ,find
the value of r

• We Know that nC r .r P r = nPr .
• We will use this equality to find r
• 10Pr =10Cr .r|
• r |=604800/120=5040=7 |
• r=7

71

Example 21
• Find the value of n and r
• n Pr = n P r+1 and
n C r = n C r-1
Solution : Given n Pr = n P r+1
n –r=1 (i)
n C r = n C r-1 n-r = r-1 (ii)
Solving i and ii
r=2 and n=3

72

Multiple choice Questions

73

1. Eleven students are
participating in a race. In how
many ways the first 5 prizes can
be won?
A) 44550
B) 55440
C) 120
D) 90
74

1. Eleven students are
participating in a race. In how
many ways the first 5 prizes can
be won?
A) 44550
B) 55440
C) 120
D) 90
75

• 2. There are 10 trains plying between
Calcutta and Delhi. The number of ways in
which a person can go from Calcutta to Delhi
and return
• A) 99.
• B) 90
• C) 80
• D) None of these

76

• 2. There are 10 trains plying between
Calcutta and Delhi. The number of ways in
which a person can go from Calcutta to Delhi
and return
• A) 99.
• B) 90
• C) 80

77

• 3. 4P4 is equal to
• A) 1
• B) 24
• C) 0

78

• 3. 4P4 is equal to
• A) 1
• B) 24
• C) 0

79

• 4.In how many ways can 8
persons be seated at a round
table?
• A) 5040
• B) 4050
• C) 450
• D) 540

80

• 4.In how many ways can 8
persons be seated at a round
table?
• A) 5040
• B) 4050
• C) 450
• D) 540

81

n n+1
• 5. If
P13 : P12 =3 : then
4
value of n is

• A) 15
• B) 14
• C) 13
• D) 12

82

n n+1
• 5. If
P13 : P12 =3 : then
4
value of n is

• A) 15
• B) 14
• C) 13
• D) 12

83

• 6.Find r if 5Pr = 60
• A) 4
• B) 3
• C) 6
• D) 7

84

• 6.Find r if 5Pr = 60
• A) 4
• B) 3
• C) 6
• D) 7

85

• 7. In how many different ways can
seven persons stand in a line for a
group photograph?
• A) 5040
• B) 720
• C) 120
• D) 27

86

• 7. In how many different ways can
seven persons stand in a line for a
group photograph?
• A) 5040
• B) 720
• C) 120
• D) 27

87

• 8. If 18 Cn = 18 Cn+ 2 then the value
of n is ______
A) 0
B) –2
C) 8
D) None of above

88

• 8. If 18 Cn = 18 Cn+ 2 then the value
of n is ______
A) 0
B) –2
C) 8
D) None of above

89

• 9. The ways of selecting 4 letters
from the word EXAMINATION is
• A) 136.
• B) 130
• C) 125

90

• 9. The ways of selecting 4 letters
from the word EXAMINATION is
• A) 136.
• B) 130
• C) 125

91

• 10 If 5Pr = 120, then the value of
r is
• A) 4,5
• B) 2
• C) 4

92

• 10 If 5Pr = 120, then the value of
r is
• A) 4,5
• B) 2
• C) 4

93

THE END

Permutations and
Combinations

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