Some concepts for Combinatorics

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Permtatins an Cmbina-

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PERMuTATIoNS ANd CoMBINATIoNSCHAPTER

7LEARNINg oBJECTIvES

(i) Fundamental principle of counting : If an event can occur in m different ways, following which another event can occur in

n different ways, then the total number of occurrence of the events in the given order is m × n.

(ii) The number of permutations of n different things taken r at a time, where repetition is not allowed, is denoted by nPr and is

given by nPr = , where .

(iii) n! = 1 × 2 × 3 × ...× n

n! = n × (n – 1) !

(i) The number of permutations of n different things, taken r at a time, where repetition is allowed, is n r .

() The number of permutations of n objects taken all at a time, where p1

objects are of rst kind, p2

objects are of the second

kind, ..., pk

objects are of the k th kind and rest, if any, are all different is .

(i) The number of combinations of n different things taken r at a time, denoted by nCr , is given by

INTRoduCTIoN

In this chapter, we shall discuss the problems of arranging of certain things, taking particular number of things at a time. 12

and 21 are different arrangements of 1 and 2 but represents the same group of 1 and 2. The order of things plays an important

role in arrangements. Using permutations and combinations techniques we can nd the number of arrangements without listing

them.

The concepts of permutations and combinations can be traced back to the advent of Jainism in India and perhaps even earlier.

The credit, however, goes to the Jains who treated its subject matter as a self-contained topic in mathematics, under the name

Vikalpa. Among the Jains, Mahavira, (around 850 A.D.) is perhaps the world’s rst mathematician credited with providing the

general formulae for permutations and combinations.

When making certain decisions, it is often important to analyze the possibilities associated with the situation. For example,

if your school offers two sections of a required math course and three sections of a required english course, then how many

schedules are possible in enrolling for these courses ? We can answer this question by constructing a tree diagram, as shown in

gure.

We start by listing the two math sections, and then we branch out from these points to the three English sections. Reading

each branch from left to right, we determine that there are six schedules, as shown in gure. Note that the number of possible

schedules is the product of the number of math choices and the number of English choices. This example illustrate the basic

principle of counting.

FuNdAMENTAL PRINCIPLE oF CouNTINg

(a) Mltiplicatin principle : If an operation can be performed in ‘m’ different ways; following which a second operation can be

performed in ‘n’ different ways, then the two operations in succession can be performed in m × n ways. This can be extended




2 IIT-JEE Maths Sihiinayaka Ecatinal Acaemy Rajenra Naar Chwk Link Ra Bilaspr Ph-07752- 237799/238799


to any innite number of operations.

(b) Aitin principle : If an operation can be performed in ‘m’ different ways and another operation, which is independent of

the rst operation, can be performed in ‘n’ different ways. Then either of the two operations can be performed in (m + n) ways.

This can be extended to any nite number of mutually exclusive operations.

Example 1 :

A person wants to go from station P to station R via station Q. There are 4 routes from P to Q and 5 routes from Q to R. In howmany ways can he travel from P to R.

Sl. He can go from P to Q in 4 ways and Q to R in 5 ways

So number of ways of travel from P to R is 4 × 5 = 20.

Example 2 :

A Hall has 3 gates. In how many ways can a man enter the hall through one gate and come out through a different gate?

Sl. Suppose the gates are A, B and C. Now there are 3 ways (A, B or C) of entering into the hall. After entering into the hall, the

man come out through a different gate in 2 ways. Hence, by the multiplication principle, total number of ways is

⇒3 × 2 = 6 ways.Example 3 :

A person wants to leave station Q. There are 4 routes from station Q to P and 5 routes from Q to R. In how many ways can he

travel the station Q.

Sl. He can go from Q to P in 4 ways and Q to R in 5 ways

he can leave station Q in 4 + 5 = 9 ways

FACToRIALS

If n is a natural number then the product of all natural number upto n is called factorial n and it is denoted by n !

Thus n ! = n (n – 1) (n – 2) ..... 3.2.1.

It is obvious to note that

n! = n (n – 1)! = n (n – 1) (n – 2)! = n (n – 1) (n – 2) (n – 3)! etc.

1 ! = 1, 2 ! = 2 3 ! = 6, 4 ! = 24, 5 ! = 120, 6 ! = 720

Imprtant Prperties

(i) n! = n (n – 1) ! (ii) (2n)! = 2n. n! [1. 3. 5. 7. . . . (2n – 1)]

(iii) 0! = 1! = 1 (i) factorials of negative integers are not dened

PERMuTATIoNS:

One important application of the fundamental counting principle is in determining the number of ways that n elements can be

arranged (in order). An ordering of n elements is called a permutation of the elements.

A permutation of n different elements is an ordering of the elements such that one element is rst, one is second, one is third,

and so on.

Example 4 :

How many permutations are possible for the letters A, B, C, D, E, and F?

Sl. Consider the following reasoning.

First position: Any of the six letters.Second position: Any of the remaining ve letters

Third position: Any of the remaining four letters.

Fourth position: Any of the remaining three letters.

Fifth position: Any of the remaining two letters.

Sixth position: The one remaining letter.

So, the numbers of choices for the six positions are as follows.

The total number of permutations of the six letters is 6! = 720.

Nmber f Permtatins f n Elements

The number of permutations of n elements is given by

n . (n – 1) . . . 4 . 3 . 2 . 1 = n!

In other words, there are n! different ways that n elements can be ordered.

Occasionally, you are interested in ordering a subset of a collection of elements rather than the entire collection. For example,

you might want to choose (and order) r elements out of a collection of n elements. Such an ordering is called a permutation of

n elements taken r at a time.



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Example 5 :

Eight horses are running in a race. In how many different ways can these horses come in rst, second, and third? (Assume that

there are no ties.)

Sl. Here are the different possibilities.

Win (rst position): Eight choices

Place (second position): Seven choicesShow (third position): Six choices

Using the Fundamental Counting Principle, multiply these three numbers together to obtain the following.

So, there are 8 . 7 . 6 = 336 different orders.

Permtatins f n Elements Taken r at a Time

The number of permutations of n elements taken r at a time is

Using this formula, you can rework previous example to nd that the number of permutations of eight horses taken three at a

time is

which is the same answer as obtained in previous example.

Remember that for permutations, order is important. So, if you are looking at the possible permutations of the letters A, B, C,

and D taken three at a time, the permutations (A, B, D) and (B, A, D) would be different because the order of the elements is

different.

Suppose, however, that you are asked to nd the possible permutations of the letters A, A, B, and C. The total number of per -

mutations of the four letters would be 4P4

= 4!. However, not all of these arrangements would be distinguishable because there

are two A’s in the list. To nd the number of distinguishable permutations, you can use the following formula.

distinishable Permtatins

Suppose a set of n objects has n1

of one kind of object, n2

of a second kind, n3

of a third kind, and so on, with n = n1

+ n2

+ n3

+

. . . + nk , Then the number of distinguishable permutations of the n objects is

Example 6 :

In how many distinguishable ways can the letters in BANANA be written?

Sl. This word has six letters, of which three are A’s, two are N’s, and one is a B. Thus, the number of distinguishable ways the let-

ters can be written is

Circlar permtatins:

(i) Arranements rn a circlar table :

Consider ve persons A, B, C, D and E to be seated on the circumference of a circular table in order (which has no head). Now,

shifting A, B, C, D and E one position in anticlockwise direction we will get arrangements as follows:

we see tha t arrangements in all

gures are same.

∴ The number of circular permutations of n different things taken all at a time is (n – 1) !, if clockwise and anticlockwise

orders are taken as different.






(ii) Arrangementsofbeadsorowers(alldifferent)aroundacircularnecklaceorgarland:

Consider ve beads A, B, C, D and E in a necklace or ve owers A,

B, C and D, E in a garland etc. If the necklace or garland on the left is

turned over we obtain the arrangement on the right, i.e., anticlockwise

and clockwise order of arrangements are not different.

Thus the number of circular permutations of ‘n’ different things taken

all at a time is (n – 1)!, if clockwise and anticlockwise orders are taken

to be some.

Example 7 :

The number of ways in which 10 persons can sit round a circular table so that none of them has the same neighbors in any two

arrangements.

Sl. 10 persons can sit round a circular table in 9! ways. But here clockwise and anticlockwise orders will give the same neighbors.

Hence the required number of ways = 9!

Example 8 :

In how many different ways can ve boys and ve girls form a circle such that

the boys and girls are alternate?

Sl. After xing up one boy on the table the remaining can be arranged

in 4! ways. There will be 5 places, one place each between two

boys which can be lled by 5 girls in 5! ways.

Hence by the principle of multiplication, the required number of ways

= 4! × 5! = 2880.

Example 9 :

In how many ways can 5 boys and 5 girls be seated at a round table no two girls may be together ?

Sl. Leaving one seat vacant between two boys may be seated in 4! ways. Then at remaining 5 seats, 5 girls any sit in 5! ways. Hence

the required number = 4! × 5!

Example 10 :

How many necklace of 12 beads each can be made from 18 beads of various colours?

Sl. In the case of necklace there is no distinction between the clockwise and anticlockwise arrangements, thus the required number

of circular permutations =

TRY IT YouRSELFQ.1 How many four digit numbers can be made by using the digits 1, 2, 3, 7, 8, 9 when

(a) repetition of a digit is allowed (b) repetition of a digit is not allowed

Q.2 How many six digit numbers are there in which sum of the digits is even ?

Q.3 In how many ways can 3 persons stay in 5 hotels ? In how many of these each person stays in a different hotel.

Q.4 If 5Pr = 6P

r – 1, then show that r = 4.

Q.5 In a railway compartment 6 seats are vacant on a berth. In how many ways can 3 passengers sit on them ?

Q.6 How many different words can be formed from the letters of the word “LUCKNOW” when

(a) all the letters are taken(b) words begin with L

(c) the letters L and W respectively occupy the rst and last places

() the vowels are always together.

Q.7 In how many ways can 7 boys be seated at a round table so that two particular boys are

(a) next to each other (b) separated.

Q.8 Find the number of ways in which 5 boys and 4 girls can be arranged on a circular table such that no two girls sit together and

two particular boys are always together.

Q.9 A person invites a party of 10 friends at dinner and place them

(i) 5 at one round table, 5 at the other round table (ii) 4 at one round table and 6 at other round table

Then nd the ratio, of number of circular permutation of case (i) to case (ii).

ANSWERS(1) 1296, 360 (2) 45 × 104 (3) 125, 60 (5) 6P

3(6) (a) 7! (b) 6! (c) 5! () 6!2! (7) (a) 240 (b) 480 (8) 288 (9) 24/25

CoMBINATIoNS :



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When one counts the number of possible permutations of a set of elements, order is important. As a nal topic in this section,

we look at a method of selecting subsets of a larger set in which order is not important. Such subsets are called combinations of

n elements taken r at a time. For instance, the combinations {A, B, C} and {B, A, C} are equivalent because both sets contain

the same three elements, and the order in which the elements are listed is not important. Hence, you would count only one of

the two sets. A common example of how a combination occurs is a card game in which the player is free to reorder the cards

after they have been dealt.

dIFFERENCE BETWEEN PERMuTATIoN ANd CoMBINATIoN

The process of selecting things is called combination and that of arranging things is called permutations.

To understand the difference clearly consider an example.

If we have 4 objects A, B, C and D the possible selection (or combination) and arrangements (or permutations) of 3 objects out

of 4 are given below. This will help you to understand clearly the difference between permutation and combinations.

Selection Arrangement

↓ ↓

Combination Permutation

ABC ABC, ACB, BAC, BCA, CAB, CBA

ABD ABD, ADB, BAD, BDA, DAB, DBA

ACD ACD, ADC, CAD, CDA, DAC, DCABCD BCD, BDC, CBD, CDB, DBC, DCB

Ttal 4 cmbinatins 24 Permtatins

CouNTINg FoRMuLAS FoR CoMBINATIoN

(a) Selectin f bjects witht Repetitin

The number of combinations of n different things taken r at a time is denoted by nCr or C (n, r)

nCr = ; (0 ≤ r ≤ n) = = ; n ∈ N and r ∈ W

If r > n, then nCr = 0

Imprtant reslts :

(i) nCn

= 1 , nC0

= 1, (ii) nCr = (iii) nC

r = nC

n – 1,

(iv) nCx

= nCy ⇒ x + y = n (v) nC

r = nC

r+1= n+1C

r (vi) nC

r = . n–1C

r–1

(vii) nCr = (n – r + 1) nC

r–1(viii) nC

1= nC

n–1= n

(b) Selectin f objects with Repetitin :

The total number of selections of r things from n different things when each thing may be repeated any number of times isn+r+1C

r

(c) Restricte Selectin/Arranement

(i) The number of combinations of n different things taken r at a time ,

When k particular objects occurs is n–k Cr–k

.

If k particular objects never occur is n–k Cr .

(ii) The number of combinations of n objects, of which p are indentical, taken r at a time isn–pC

r + n–pC

r–1+ n–pC

r–2+ ...........+ n–pC

0if r ≤ p.






n–pCr + n–pC

r – 1+ r–pC

r–2+ ..........+ n–pC

r–pif r > p.

() Selectin frm istinct bjects :

The number of ways (or combinations) of n different things selecting at least one of them is nC1

+ nC2

+ nC3

+ .....+ nCn

= 2n – 1.

This can also be stated as the total number of combination of n different things.

(e) Selectin frm ientical bjects

The number of combination of n identical things taking r(r ≤ n) at a time is 1.The number of ways of selecting r things out of n alike things is n + 1 (where r = 0, 1, 2 .....n).

The number of ways to select some or all out (p + q + r) things where p are alike of rst kind, q are alike of second kind and r

are alike of third kind is (p + 1) (q + 1) (r + 1) – 1

(f) Selectin when bth ientical an istinct bjects are present :

If out of (p + q + r + t) things, p are alike one kind, q are alike of second kind, r are alike of third kind and t are different, then

the total number of combinations is (p + 1)(q + 1)(r + 1) 2t – 1

Example 11 :

If nCr – 1

= 36, nCr = 84 and nC

r + 1= 126, then nd r.

Sl. ⇒

⇒ 3n – 3r + 3 = 7r ⇒ 10r – 3n = 3 . . . (i)

and

⇒ ⇒ 2n – 2r = 3r + 3 ⇒ 5r – 2n = – 3 ⇒ 10 r – 4n = – 6

Substracting (ii) from (i) n = 9

10r – 27 = 3 ⇒ 10 r = 30 ⇒ r = 3

Example 12 :

Ramesh has 6 friends. In how many ways can he invite one or more of them at a dinner ?

Sl. He can invite one, two three, four, ve or six friends at the dinner. So total number of ways of his invitation

= 6C1

+ 6C2

+ 6C4

+ 6C5

+ 6C6

= 26 – 1 = 63

Example 13 : How many combination of 4 letters can be made of the letters of the word ‘JAIPUR’ ?

Sl. Here 4 things are to be selected out of 6 different things. So the number of combinations

= 6C4

= = 15

Example 14 :

From 4 gentlemen and 6 ladies a committee of ve is to be selected. Find the number of ways in which the committee can be

formed so that gentlemen are in majority.

Sl. The committee will consist of 4 gentlemen and 1 lady or 3 gentlemen and 2 ladies

∴ the number of committees = 4C4

× 6C1

+ 4C3

× 6C2

= 66

Example 15 :

8 pens are to be selected from pens of 3 colours (pens of each colour being available any number of times), then nd the total

number of selections.

Sl. 3+8–1C8

= 10C8

= = 45

Example 16 :

There are n different books and p copies of each in a library. Find the number of ways in which one or more than one book can

be selected.

Sl. Total cases = p + 1 (if selected or not)

Required number of ways = (p + 1)(p +1)......n terms – 1 = (p + 1) n – 1



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