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1. Permutations and
Combinations
Quantitative Aptitude & Business Statistics
2. 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
3. • then the sequence of k
operations can be performed in
n1 n2 n3….. nk ways.
• N= n1 n2 n3….. nk
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
4. 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.
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
5. 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
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Statistics:Permutations and Combinations
6. 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.
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
7. Solution
• The maximum number of days
he does not need to repeat his
clothing is 2×3×2 = 12
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
8. 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.
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
9. •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)
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
10. 1.3 Permutations
• (A) Permutations
• A permutation is an arrangement
of objects.
• abc and bca are two different
permutations.
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
11. • 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.
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
12. 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
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
13. 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
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
14. 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;
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
15. • 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
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
16. 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?
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
17. • Solution
• (a) Number of 4-digit numbers
= 74 = 2401.
• (b) Number of 4 digit numbers
=7 ×6 ×5 ×4 = 840.
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
18. 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
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
19. • 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.
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
20. • 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
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
21. 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.
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
22. (B) Conditional
Permutations
• When arranging elements in
order , certain restrictions may
apply.
• In such cases the restriction
should be dealt with first..
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
23. 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?
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Statistics:Permutations and Combinations
24. • 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.
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
25. •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.
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
26. 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
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
27. Example 6
In how many ways can
7 different books be
arranged on a shelf
(a) if two particular
books are together;
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Statistics:Permutations and Combinations
28. • 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
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Statistics:Permutations and Combinations
29. (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.
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
30. (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.
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
31. 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.
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Statistics:Permutations and Combinations
32. • 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.
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Statistics:Permutations and Combinations
33. • 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
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
34. • 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 !
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
35. 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.
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
36. • 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,
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Statistics:Permutations and Combinations
37. • 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.
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
38. • 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
Statistics:Permutations and Combinations
38
39. • 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) .
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
40. • A general equation relating
combinations to permutations
is
1 n n!
C r = Pr =
n
r! (n − r )!r!
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Statistics:Permutations and Combinations
41. • Note:
• (1) Cnn = Cn0 = 1
• (2) Cn1 = n
• (3) Cnn = Cnn-r
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
42. 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
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Statistics:Permutations and Combinations
43. 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
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
44. Example 10
• If 100 C 98 =999 C 97 +x C 901 find x.
• Solution 100C 98 =999C 98 +999C97
• = 999C901+999C97
• X=999
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
45. 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
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
46. 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
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
47. Example 13
• How many different 5-card
hands can be dealt from a deck
of 52 playing cards?
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
48. 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.
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Statistics:Permutations and Combinations
49. 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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Statistics:Permutations and Combinations
50. Solution:
• Solution
• (a) Number of triangles
• = number of ways
• of selecting 3 points out of 6
• = C63 = 20.
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Statistics:Permutations and Combinations
51. • b) How many pairs of triangles
can be formed by using the 6
points as vertices ?
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Statistics:Permutations and Combinations
52. • 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.
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
53. • 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
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
54. 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?
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Statistics:Permutations and Combinations
55. 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.
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
56. (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.
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Statistics:Permutations and Combinations
57. 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
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
58. • 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.
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
59. • The number of ways of selecting
3 women out of 8 =C83 = 56.
• the number of ways of selecting
the committee = lO × 56 = 560.
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
60. 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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Statistics:Permutations and Combinations
61. • 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.
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
62. (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.
Quantitative Aptitude & Business
• the total number of teams is
62
Statistics:Permutations and Combinations
63. 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,
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Statistics:Permutations and Combinations
64. • Solution (a)
• An unrestricted choice of 6
out of 12 gives C126= 924.
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Statistics:Permutations and Combinations
65. two of the friends is a couple
• (b)
and will not attend separately,
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Statistics:Permutations and Combinations
66. 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.
Quantitative Aptitude & Business
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Statistics:Permutations and Combinations
67. 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.
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Statistics:Permutations and Combinations
68. • 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 .
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Statistics:Permutations and Combinations
69. • 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.
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Statistics:Permutations and Combinations
70. • 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!
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Statistics:Permutations and Combinations
71. 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
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Statistics:Permutations and Combinations
72. 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
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Statistics:Permutations and Combinations
73. Multiple choice Questions
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Statistics:Permutations and Combinations
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
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Statistics:Permutations and Combinations
75. 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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Statistics:Permutations and Combinations
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
• D) None of these
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Statistics:Permutations and Combinations
77. • 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
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Statistics:Permutations and Combinations
78. • 3. 4P4 is equal to
• A) 1
• B) 24
• C) 0
• D) None of these
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Statistics:Permutations and Combinations
79. • 3. 4P4 is equal to
• A) 1
• B) 24
• C) 0
• D) None of these
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Statistics:Permutations and Combinations
80. • 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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Statistics:Permutations and Combinations
81. • 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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Statistics:Permutations and Combinations
82. n n+1
• 5. If
P13 : P12 =3 : then
4
value of n is
• A) 15
• B) 14
• C) 13
• D) 12
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Statistics:Permutations and Combinations
83. n n+1
• 5. If
P13 : P12 =3 : then
4
value of n is
• A) 15
• B) 14
• C) 13
• D) 12
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Statistics:Permutations and Combinations
84. • 6.Find r if 5Pr = 60
• A) 4
• B) 3
• C) 6
• D) 7
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Statistics:Permutations and Combinations
85. • 6.Find r if 5Pr = 60
• A) 4
• B) 3
• C) 6
• D) 7
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Statistics:Permutations and Combinations
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
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Statistics:Permutations and Combinations
87. • 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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Statistics:Permutations and Combinations
88. • 8. If 18 Cn = 18 Cn+ 2 then the value
of n is ______
A) 0
B) –2
C) 8
D) None of above
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Statistics:Permutations and Combinations
89. • 8. If 18 Cn = 18 Cn+ 2 then the value
of n is ______
A) 0
B) –2
C) 8
D) None of above
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Statistics:Permutations and Combinations
90. • 9. The ways of selecting 4 letters
from the word EXAMINATION is
• A) 136.
• B) 130
• C) 125
• D) None of these
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Statistics:Permutations and Combinations
91. • 9. The ways of selecting 4 letters
from the word EXAMINATION is
• A) 136.
• B) 130
• C) 125
• D) None of these
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Statistics:Permutations and Combinations
92. • 10 If 5Pr = 120, then the value of
r is
• A) 4,5
• B) 2
• C) 4
• D) None of these
Quantitative Aptitude & Business
92
Statistics:Permutations and Combinations
93. • 10 If 5Pr = 120, then the value of
r is
• A) 4,5
• B) 2
• C) 4
• D) None of these
Quantitative Aptitude & Business
93
Statistics:Permutations and Combinations