Permutations and Combinations

Permutations and 

Combinations 

Quantitative Aptitude & Business Statistics

The Fundamental Principle of 

• If there are 

Multiplication 

• n 1 ways of doing one operation, 

• n 2 ways of doing a second 

operation, n 3 ways of doing a 

third operation , and so forth, 

Quantitative Aptitude & Business 

Statistics:Permutations and Combinations 

2

• then the sequence of k 

operations can be performed in 

n 1 n 2 n 3 ….. n k ways. 

• N= n 1 n 2 n 3 … . n k 



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

Full(F) 

Medium 

(M) 

Compact 

(C) 

Solution 



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. 

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) = 

n r 



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 

Since 

We have 

n! 

( n − r)! 

= 

= 

= 

n 

Pr 

P 

n 

r n r 

n 

P r n r 

, P , P ( n , r ) 

n( 

n −1)( 

n − 2)....( 

n − 



, P , P( 

n, 

r) 

n( 

n −1)( 

n − 2)....( 

n − r + 1)( 

n − r)... 

3⋅ 

2 ⋅1 

( n − r)... 

3⋅ 

2 ⋅1 

P n 

r 

= 

( n 

n! 

− 

r 

r)! 

+ 

1) 

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 

= 7 4 = 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 

P 

10 

10 



18

• 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. 



D 

A 

C 

B 

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!. 

Hence 

P 

= 



4! 

4 

= 

( 4 

− 

1)! 

= 

3! 

20

• The number of distinct circular 

arrangements of n objects is 

(n —1)! 

Solution (b) 

• 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


• 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! 


• 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! 

! ! ⋅⋅⋅ 

⋅ 

Quantitative Aptitude & 2Business 


1 r k k k 

! 

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 P 3 30 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 

P 3 30 is 3! times too large. 

• Hence, if we divide P 30 

3 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 C 3 30 and 

C 

= 

30 

3 

= P 

3! 

30! 

27! 

3! 

1 30 

3 

= 



4060 

38

• In general, each unordered relement 

subset of a given nelement 

set (r≤ n) is called a 

combination. 

• The number of combinations of 

n elements taken r at a time is 

denoted by C n r or nCr or C(n, r) . 



39

• A general equation relating 

combinations to permutations 

is 

C 

n 

r 

= 

1 

r! 

P 

n 

r 

= 

n! 

( n − r)! 

r! 



40

• Note: 

• (1) C n n = C n 0 = 1 

• (2) C n 1 = n 

• (3) C n n = C n n-r 



41

Example8 

• If 167 C 90+167 C x =168 C x then x 

is 

• Solution: nC r-1+nC r=n+1 C r 

• Given 167 C 90+167c x =168C x 

• We may write 

• 167C 91-1 + 167 C 91=167+1 C 61 

• =168 C 91 

• X=91 



42

Example9 

• If 20 C 3r= 20C 2r+5 ,find r 

• Using nC r=nC n-r in the right –side 

of the given equation ,we find , 

• 20 C 3 r =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 +999C 97 

• = 999C 901+999C 97 

• X=999 



44

Example11 

• If 13 C 6 + 2 13 C 5 +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 

• n-r+1 =7/3 * r 

r−1 

84 

36 

• 3/2 (r+1)+1 =7/3 * r 

• r=3 

nC 

nC 

126 

84 

r + 1 = = 

r 

nC 

nC 

3 

2 

r 

= 

= 



7 

3 

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 

C 52 5= 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 

• = C 6 3 = 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 

= C6 3 ÷ 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 C 25 4 


• By the fundamental counting principle, the 

total number of different teams that can be 

formed is 

25 × C 24 4=265650. 



55

(B) Conditional 


• 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 = C 5 2 = 10. 



58

• The number of ways of selecting 

3 women out of 8 =C 8 3 = 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


• Boys can be chosen in C 15 2 = 105 

ways 

• Girls can be chosen in C 10 2 = 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 C15 1 × 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. 


62 


• the total number of teams is

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


• An unrestricted choice of 6 

out of 12 gives C 12 6= 924. 



64

• (b) two of the friends is a couple 

and will not attend separately, 



65

B Solution 

• If the couple attend, the 

remaining 4 may then be chosen 

from the other 10 in C10 4 ways. 

• If the couple does not attend, 

then He simply chooses 6 from 

the other 10 in C10 6 ways. 

• total number of ways is C10 4 + 

C10 6 = 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 C 30 10 ways . 



68

• Group B can be selected from the 

remaining 20 students in C 20 10 

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 C30 10 × C20 10 × C10 10 

by 3!, 

• hence the total number of ways 

of forming the three groups is 

1 

3! 

30 

3 

20 

10 

× 

C × C × 



C 

10 

10 

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 P r = n P r+1 and 

n C r = n C r-1 

Solution : Given n P r = 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

• 5. If n 

P : 

n+1 

P =3:4 

then 

13 12 

value of n is 

• A) 15 

• B) 14 

• C) 13 

• D) 12 



82

• 5. If n 

P : 

n+1 

P =3:4 

then 

13 12 

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

18 18 

• 8. If C then the value 

n = Cn+2 

of n is ______ 

A) 0 

B) –2 

C) 8 

D) None of above 



88

18 18 

• 8. If C then the value 

n = Cn+2 

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

Permutations and 


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