16806 per com

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 Statistics:Permutations and Combinations 2

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

• N= n1 n2 n3….. nk


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.


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.

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.


Solution

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


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

Statistics:Permutations and Combinations 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)

• = n! / (n – r)!

1.3 Permutations

• (A) Permutations • A permutation is an arrangement

of objects. • abc and bca are two different

permutations.


• 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.


• 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


n n n n n 1 2 3 4 r

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.


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

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;


• some of them are


),(,, rnPPP rnn

r ),(,, rnPPP rnn

r ),(,, rnPPP rnn

r ),(,, rnPPP rnn

r ),(,, rnPPP rnn

r ),(,, rnPPP rnn

r

),(,, rnPPP rnn

r

nrP

rnnnnrn

rnrnnnnrn

n

=

+−−−=⋅⋅−

⋅⋅−+−−−=

−

)1)....(2)(1(123)...(

123)...)(1)....(2)(1()!(

!

Since

We have )!(

!rn

nP nr −=

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?


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


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


1010P

• 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

• 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!.


!3)!14(4!4

=−==PHence

• 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.


Solution (b)

(B) Conditional Permutations

• When arranging elements in order , certain restrictions may apply.

• In such cases the restriction should be dealt with first..


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?


• 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.


•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.


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


Example 6 In how many ways can

7 different books be arranged on a shelf (a) if two particular books are together;


• 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 together


(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.


(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.


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.


• 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.


• 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


• 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


!!!!

21 rkkkn

⋅⋅⋅⋅

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.


• 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,


• 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 P3

30 is 3! times too large. • Hence, if we divide P3

30 by 3!,the result will be the number of unordered subsets of 30 elements taken 3 at a time. Quantitative Aptitude & Business


• This number of unordered subsets is also called the number of combinations of 30 elements taken 3 at a time, denoted by C3

30 and


4060!3!27

!30!3

1 303

303

==

= PC

• 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 Cn

r or nCr or C(n, r) .


• A general equation relating combinations to permutations is !)!(

!!

1rrn

nPr

C nr

nr −

==


• Note: • (1) Cn

n = Cn0 = 1

• (2) Cn1 = n

• (3) Cnn = Cn

n-r


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


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


Example 10

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

• X=999


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


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

• 3/2 (r+1)+1 =7/3 * r • r=3


37

3684

1

==−r

r

nCnC

23

841261 ==+

r

r

nCnC

Example 13

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


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 C52

5＝ 2118760.


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?


Solution:

• Solution • (a) Number of triangles • = number of ways • of selecting 3 points out of 6 • = C6

3 = 20.


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


• 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.


• 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


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?


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.


(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.


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

couple?


• 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 = C5

2 = 10.


• The number of ways of selecting 3 women out of 8 =C8

3 = 56. • the number of ways of selecting

the committee = lO × 56 = 560.


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,


• Solution (a) • Boys can be chosen in C15

2 = 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 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 × 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


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,


• Solution (a) • An unrestricted choice of 6

out of 12 gives C126＝

924.


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


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 C104 +

C106 = 420.


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.


• 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 C30

10 ways .


• Group B can be selected from the remaining 20 students in C20

10 ways.

• There is only 1 way of forming group C from the remaining 10 students.


• Since the order of the groups is immaterial, we have to divide the product C30

10 × C2010 × C10

10 by 3!,

• hence the total number of ways of forming the three groups is


1010

2010

303!3

1 CCC ×××

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


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


Multiple choice Questions


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 Quantitative Aptitude & Business


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 Quantitative Aptitude & Business


• 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


• 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



• 3. 4P4 is equal to • A) 1 • B) 24 • C) 0 • D) None of these


• 3. 4P4 is equal to • A) 1 • B) 24 • C) 0 • D) None of these


• 4.In how many ways can 8 persons be seated at a round table?

• A) 5040 • B) 4050 • C) 450 • D) 540



• 4.In how many ways can 8 persons be seated at a round table?

• A) 5040 • B) 4050 • C) 450 • D) 540



• 5. If then

value of n is

• A) 15 • B) 14 • C) 13 • D) 12


n n+113 12P : P =3 : 4

• 5. If then

value of n is

• A) 15 • B) 14 • C) 13 • D) 12


n n+113 12P : P =3 : 4

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


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


• 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


• 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


• 8. If then the value of n is ______

A) 0 B) –2 C) 8 D) None of above


18 18n n+2C = C

• 8. If then the value of n is ______

A) 0 B) –2 C) 8 D) None of above


18 18n n+2C = C

• 9. The ways of selecting 4 letters from the word EXAMINATION is



• 9. The ways of selecting 4 letters from the word EXAMINATION is



• 10 If 5Pr = 120, then the value of r is

• A) 4,5 • B) 2 • C) 4 • D) None of these


• 10 If 5Pr = 120, then the value of r is

• A) 4,5 • B) 2 • C) 4 • D) None of these


THE END

Permutations and Combinations

16806 per com

Education

quantitative

10 trains

round table

word examination

hundreds digit

element set

group photograph

persons stand