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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,
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then the sequence of k operations
can be performed in n1 n2 n3.. nkways.
N= n1 n2 n3.. nk
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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.
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Solution
Full(F)
Compact
(C)
Medium
(M)
0-2
2-4
4-6
>6
0-2
2-4
4-6
>6
0-2
2-4
4-6
>6
The tree diagram enumerates all possible
classifications, the total number of which
is 3x4= 12.
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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.
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Solution
The maximum number of days he
does not need to repeat his clothing
is 232 = 12
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1.2 Factorials
The product of the first nconsecutive 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!=1234567=5040.
Note 0! defined to be 1.
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The product of any number of
consecutive integers can be
expressed as a quotient of twofactorials, for example,
6789 = 9!/5! = 9! / (9 4)!
1112131415= 15! / 10!
=15! / (15 5)!
In particular,
n(n 1)(n 2)...(n r + 1)
=
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1.3 Permutations
(A) Permutations A permutation is an arrangement of
objects.
abc and bca are two differentpermutations.
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1. Permutations with repetition
The number of permutations of r objects,
taken from n unlike objects,
can be found by considering the number ofways 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 differentways.
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Therefore, the number of
permutations of r objects, takenfrom n unlike objects,
each of which may be repeatedany number of times
= n n n .... n(r factors)= nr
n n n n n
1 2 3 4 r
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2. Permutations without repetition
If repetition is not allowed, thenumber of ways of filling eachblank space is one less than thepreceding one.
n n-1 n-2 n-3 n-r+1
1 2 3 4 r
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Therefore, the number of permutations ofr objects, taken from n unlike objects,
each of which can only be used once in
each permutation
=n(n 1)(n2) .... (nr + 1)
Various notations are used to represent
the number of permutations of a set ofn elements taken r at a time;
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some of them are
),(,, rnPPPrn
n
r ),(,, rnPPP
rn
n
r ),(,, rnPPP
rn
n
r ),(,, rnPPP
rn
n
r ),(,, rnPPP
rn
n
r ),(,, rnPPP
rn
n
r
),(,, rnPPPrn
n
r
n
rP
rnnnn
rn
rnrnnnn
rn
n
=
+=
+=
)1)....(2)(1(123)...(
123)...)(1)....(2)(1(
)!(!
Since
We have
)!(
!
rn
nP
n
r
=
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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?
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Solution
(a) Number of 4-digit numbers= 74= 2401.
(b) Number of 4 digit numbers
=7 6 5 4= 840.
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Example 4
In how many ways can 10 men bearranged
(a) in a row,
(b) in a circle?
Solution
(a) Number of ways is
= 3628800
10
10P
S
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Suppose we arrange
the 4 letters A, B, C
and D in a circular
arrangement as shown. Note that the
arrangements ABCD,
BCDA, CDAB and DABC
are not distinguishable.
A
D
C
B
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For each circular arrangement thereare 4 distinguishable arrangements
on a line. If there are P circular arrangements,
these yield 4P arrangements on aline, which we know is 4!.
!3)!14(4
!4===PHence
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The number of distinct circular
arrangements of n objects is(n1)!
Hence 10 men can be arranged
in a circle in 9! = 362 880 ways.
Solution (b)
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(B) Conditional Permutations When arranging elements in order ,
certain restrictions may apply.
In such cases the restriction shouldbe dealt with first..
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Example 5How many even numerals between200 and 400 can be formed by using1, 2, 3, 4, 5 as digits
(a) if any digit may be repeated;(b) if no digit may be repeated?
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Solution (a)
Number of ways of choosing the
hundreds digit = 2. Number of ways of choosing the tens
digit = 5.
Number of ways of choosing the unit
digit = 2.
Number of even numerals between
200 and 400 is
2 5 2 = 20.
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Solution (b)
If the hundreds digit is 2,
then the number of ways of choosingan even unit digit = 1,
and the number of ways of choosing a
tens digit = 3.
the number of numerals formed
113 = 3.
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If the hundreds digit is 3, then the
number of ways of choosing aneven. 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 numeralsbetween 200 and 400 = 3 + 6 =
9
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Example 6In how many ways can 7different books be
arranged on a shelf(a) if two particular booksare together;
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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 bearranged in 2 ways among themselves.
The number of arrangement of 7 books
with two particular books together= 6! x 2 = 1440.
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(b) if two particular books areseparated?
Solution (b)
Total number of arrangement of 7books = 7! = 5040.
the number of arrangement of 7
books with 2 particular books
separated = 5040 1440 = 3600.
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(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.
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However, the 3 Ss 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 otherwisespell a new word will be repeated 3!
times.
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To avoid counting repetitions resulting
from the 3 Ss, we must divide 9! by 3!.
Similarly, we must divide by 2! to avoid
counting repetitions resulting from the
2 indistinguishable Es.
Hence the total number of words thatcan be formed is
9! 3! 2! = 30240
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If a set of n elements has k1
indistinguishable elements of one kind,
k2of another kind,
and so on for r kinds of elements, then
the number of permutations of the set of
n elements is
!!!
!
21 rkkk
n
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1.4 Combinations
When a selection of objects is madewith 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 arethe same combination of letters.
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Suppose we wish to appoint a
committee of 3 from a class of 30students.
We know that P330 is the number of
different ordered sets of 3 studentseach that may be selected from among
30 students.
However, the ordering of the students
on the committee has no significance,
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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 P330is 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.
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This number of unordered subsets is
also called the number of
combinations of 30 elements taken 3at a time, denoted by C330 and
4060!3!27
!30!3
1 303
30
3
==
= PC
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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 Cnror nCr or C(n, r) .
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A general equation relating
combinations to permutationsis
!)!(
!
!
1
rrn
nP
rC
n
r
n
r
==
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Example8
If 167 C 90+167 C x =168 C xthen x is
Solution: nCr-1+nCr=n+1 Cr Given 167 C90+167cx=168C x
We may write
167C91-1
+ 167 C91
=167+1 C61
=168 C91
X=91
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Example9
If 20 C 3r= 20C 2r+5,find r
Using nCr=nC n-r in the right side ofthe given equation ,we find ,
20 C 3r =20 C 20-(2r+5)
3r=15-2r
r=3
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Example 10
If 100 C 98=999 C 97 +x C 901 find x.
Solution 100C 98=999C 98+999C97 = 999C901+999C97
X=999
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Example12
If n C r-1=36 ,n Cr =84 and n C r+1 =126 thenfind r
Solution
n-r+1 =7/3 * r
3/2 (r+1)+1 =7/3 * r
r=3
3
7
36
84
1
==
r
r
nC
nC
2
3
84
1261==
+
r
r
nC
nC
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Example 13 How many different 5-card hands
can be dealt from a deck of 52
playing cards?
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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 isC525 2118760.
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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?
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Solution:
Solution
(a) Number of triangles = number of ways
of selecting 3 points out of 6
= C63
= 20.
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b) How many pairs of triangles can beformed by using the 6 points as
vertices ?
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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 theother triangle.
Similarly, if D, E, F are selected toform a triangle, then A, B, C must
form the other triangle.
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Example 15 From among 25 boys who play
basketball, in how many different ways
can a team of 5 players be selected ifone of the players is to be designated
as captain?
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Solution
A captain may be chosen from any of the
25 players.
The remaining 4 players can be chosen inC254
different ways.
By the fundamental counting principle, the
total number of different teams that can
be formed is25 C244265650.
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(B) Conditional Combinations
If a selection is to be restricted in
some way, this restriction must bedealt with first.
The following examples illustrate
such conditional combination
problems.
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Example 16
A committee of 3 men and 4women is to be selectedfrom 6 men and 9 women.
If there is a married coupleamong the 15 persons, inhow many ways can thecommittee be selected sothat it contains the marriedcouple?
Solution
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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.
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The number of ways of selecting 3
women out of 8 =C83= 56.
the number of ways of selecting the
committee = lO 56 = 560.
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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,
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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.
(b) it must contain at least 1 boy and
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(b) t ust co ta at east boy a d
1 girl.
Solution : If the team must contain at least 1 boy
and 1 girl it can be formed in thefollowing ways:
(I) 1 boy and 3 girls, with C151 C10
3=
1800 ways,
(ii) 2 boys and 2 girls, with 4725 ways,
(iii) 3 boys and 1 girl, with C15
3 C10
1
=4550 ways.
the total number of teams is
1800 + 4725 + 4550 = 11075.
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Example 18 Mr. .X has 12 friends and wishes to
invite 6 of them to a party. Find thenumber
of ways he may do this if
(a) there is no restriction on choice,
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Solution (a)
An unrestricted choice of 6out of 12 gives C126 924.
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(b) two of the friends is a couple andwill not attend separately,
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B Solution
If the couple attend, the remaining 4
may then be chosen from the other
10 in C104ways.
If the couple does not attend, then
He simply chooses 6 from the other
10 in C10
6
ways. total number of ways is C104 + C
106
= 420.
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Example 19
Find the number of ways inwhich 30 students can bedivided into three groups, eachof 10 students, if the order ofthe groups and thearrangement of the students in
a group are immaterial.
Solution
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Let the groups be denoted by A, Band C. Since the arrangement of thestudents in a group is immaterial,
group A can be selected from the 30students in C3010ways .
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Group B can be selected from the
remaining 20 students in C2010ways.
There is only 1 way of forming group
C from the remaining 10 students.
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Since the order of the groups is
immaterial, we have to divide theproduct C3010 C
2010 C
1010by 3!,
hence the total number of ways of
forming the three groups is
10
10
20
10
30
3!31 CCC
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Example20
If n Pr = 604800 10 C r =120 ,findthe 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
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Example 21
Find the value of n and r
n Pr= n P r+1and
n Cr= n C r-1
Solution : Givenn Pr= n P r+1
n r=1 (i)
n Cr= n C
r-1n-r = r-1 (ii)
Solving i and ii
r=2 and n=3
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Multiple choice Questions
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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
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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
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2. There are 10 trains plying between
Calcutta and Delhi. The number of ways in
which a person can go from Calcutta toDelhi and return
A)99.
B) 90
C) 80 D) None of these
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2. There are 10 trains plying between
Calcutta and Delhi. The number of ways in
which a person can go from Calcutta toDelhi and return
A)99.
B) 90
C) 80 D) None of these
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3. 4P4 is equal to
A) 1
B) 24
C) 0
D) None of these
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3. 4P4 is equal to
A) 1
B) 24
C) 0
D) None of these
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4.In how many ways can 8 persons
be seated at a round table?
A) 5040
B) 4050
C) 450
D) 540
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4.In how many ways can 8 persons
be seated at a round table?
A) 5040
B) 4050
C) 450
D) 540
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5. If then
value of n is
A) 15
B) 14
C) 13
D) 12
n n+113 12P : P =3: 4
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5. If then
value of n is
A) 15
B) 14
C) 13
D) 12
n n+113 12
P : P =3: 4
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6.Find r if 5Pr = 60
A) 4
B) 3
C) 6
D) 7
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6.Find r if 5Pr = 60
A) 4
B) 3
C) 6
D) 7
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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
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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
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8. If then the value of
n is ______
A)0
B) 2
C) 8
D) None of above
18 18
n n+2C = C
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8. If then the value of
n is ______
A)0
B) 2
C) 8
D) None of above
18 18
n n+2C = C
7/25/2019 Permutation 2
90/94
Quantitative Aptitude & Business
Statistics:Permutations and
Combinations90
9. The ways of selecting 4 letters
from the word EXAMINATION is
A) 136.
B) 130
C) 125
D) None of these
7/25/2019 Permutation 2
91/94
Quantitative Aptitude & Business
Statistics:Permutations and
Combinations91
9. The ways of selecting 4 letters
from the word EXAMINATION is
A) 136.
B) 130
C) 125
D) None of these
7/25/2019 Permutation 2
92/94
Quantitative Aptitude & Business
Statistics:Permutations and
Combinations92
10 If 5Pr = 120, then the value of r
is
A) 4,5
B) 2
C) 4
D) None of these
7/25/2019 Permutation 2
93/94
Quantitative Aptitude & Business
Statistics:Permutations and
Combinations93
10 If 5Pr = 120, then the value of r
is
A) 4,5
B) 2
C) 4
D) None of these
7/25/2019 Permutation 2
94/94
THE END
Permutations and
Combinations