MC0063(A)-Unit3-fi

DiscreteMathematics Unit3

SikkimManipalUniversity PageNo:75

Unit3 CombinatoricsStructure

3.1 Introduction

Objectives

3.2 PrinciplesofCounting

3.3 GeneratingFunction

3.4 PartitionsandCompositions

3.5 Orderings

3.6 PrincipleofInclusionandExclusion

SelfAssessmentQuestions(SAQs)

3.7 Summary

3.8 TerminalQuestions

3.9 Answers

3.1Introduction

Combinatorics is the branch of discrete mathematics concerned with

countingproblems. Techniquesforcountingare important inMathematics

andComputer Scienceespecially in theanalysis of algorithmsand in the

theory of probability. In this unit we present the counting techniques,

particularly those for permutations and combinations. We also look into

theseaspectsusing thegenerating functionwhich is theadvanced tool to

analyzethecomputerprograms.Furtherwediscusstheapplicationsofthe

principleofinclusionandexclusion.

Objectives

Attheendoftheunitthestudentmustbeableto:

Learntheprinciplesofcountingwithcertainnaturalobjects.

Applythetechniquesofgeneratingfunctiontopartitionsandcompositions.

Write thepermutationsinlexicographicandreverselexicographicorder.

Applytheprinciplesofinclusionandexclusiontovariousmodels.



3.2PrincipleofCounting

3.2.1(i). RuleofSum:IftheobjectAmaybechooseninmways,andBin

nways,theneitherAorB(exactlyone)maybechooseninm+nways.

Thiscanbegeneralisedforanypobjects.

(ii).RuleofProduct: If theobjectAmaybe choosen inmwaysand the

objectBinnways,thenbothAandBmaybechooseninthisorderinmn

ways.Thiscanbegeneralisedforanypobjects.

3.2.2Example

If thereare42ways toselecta representativeforclassAand50ways to

selectarepresentativefortheclassB,then

i) By the rule of product, there are 42 50 ways to select the

representativeforboththeclassAandclassB

ii) Bytheruleofsum,therewillbe42+50waystoselectarepresentative

foreitherclassAorclassB.

3.2.3Example

Supposea licenseplatecontains2 lettersfollowedbyfourdigits,with the

firstdigitisnotzero.Howmanydifferentlicenseplatescanbeprinted?.

Solution:Eachlettercanbeprintedin26differentways.

Sincethefirstdigitisotherthanzero,thiscanbeselectedin9ways.

Second,thirdandfourthdigitscanbeselectedin10ways.

Thereforebytheruleofproduct,thereare 26 26 9 10 10 10ways.

Specialcase:Allaredistinct

Firstlettercanbeprintedin26ways.

Secondlettercanbeprintedin25ways.

Firstdigitcanbeprintedin9ways(otherthan0).

Seconddigitcanbeprintedin9ways(anyonefrom0to9exceptchoosen

firstdigit)



Thirddigitcanbeprintedin8ways

Fourthdigitcanbeprintedin7ways.

Thereforebytheruleofproduct,thereare26 25 9 9 8 7ways.

3.2.4Permutationofdistinctthings

Letusrecollectthatthefirstofthemembersofan rpermutationofndistinctthingsmaybechooseninnways.Thesecondischoosenin(n1)ways,,therth ischooseninn(n1)ways.

So by the repeated application of product rule, the number required is

n(n1).(n(r 1))ways,n r,itisdenotedbyp(n,r).

Ifr=n,thenp(n,n)=n(n1)(nn+1)=n (n 1)2.1=n!.

Thereforep(n,r)=1.2)...rn(

1.2)...1r(n)...(2n)(1n(n -

- - -

=)!rn(

!n -

=)rn,rn(p

)n,n(p - -

or p(n,n)=p(n,r).p(n r,nr).

3.2.5Problem

Provethatp(n,r)=p(n1,r)+r.p(n1,r1)

Solution:Writep(n,r)=n(n 1)(n(r1))

=(n1)(n 2)..n(r1)[(nr)+r],whichis

equaltop(n1,r)+r.p(n1,r 1),onmultiplication.

3.2.6Note:

i) The number of permutations of n objects taken r at a time with

unlimited repetition, which is same as the number of ways of filling

rblankspaceswithnobjects.Afterchoosingtheobjectinnways,the

nextobjectcanalsobechoosenin nwaysandsoon. Therefore, in

thiscasetherearen n . n=nr=U(n,r)ways.

rtimes



ii) Consider nobjectsofwhichm1 are first kind,m2 areof secondkind,

.,mkareofkthkind,then nmk

1ii =

= .

iii) Leta1,a2,,akberealnumbers.Considerthepower

(a1+a2++ak)n=(a1+a2++ak)(a1+a2++ak)(a1+a2++ak).

Afterperformingthisproductbutbeforecollecting like terms,a typical

term in this product has the form 1 2 kn n n1 2 ka a ...a . The coefficient of

1 2 kn n n1 2 ka a ...a aftercollectingliketermsisequaltothenumberofways

ofpickingn1factorsequaltoa1andn2 factorsequaltoa2andsoon,as

we multiply n copies of a1+ a2 + + ak. This is precisely the

multinomialcoefficient

1 2 k

n

n n ...n

=!!....mm!m

n!

k21.

3.2.7Theorem

Thenumberofdistinguishable permutationsof nobjects inwhich the first

object appears in m1 times, second object in m2 ways,. and so on, is

!!....mm!mn!

k21,wheremk isthekthobjectappearsinmk times.

Proof:Letxbethenumberrequired.Inpermutationamongx,makem1all

distinct. Since m1 objects can be permuted among themselves, one

permutation will give rise to m1!. Therefore x permutations gives x.m1!

permutations.Nowmakem2 identicalobjectsalldistinct.Thenwegetxm1!

m2! Permutations of n objects in which m3 are alike, mk are alike.

Continuing this process we get x.m1! m2! mk! as the number of

permutationsofnobjectsofwhicharealldistinctandhenceequalton!.



Thereforex=!!....mm!m

n!

k21.

3.2.8Example

(i) Find the number of different letter arrangements can be formed using

BHANUSHASHANK.

Solution:Totalnumberoflettersn=13(withrepetitions)

NumberofAs=3

NumberofHs=3

NumberofNs=2.

NumberofSs=2

AndthelettersB,U,K,eachis1.

Thereforetherequirednumberofpermutationsis13!

3!3!2!2!1!1!1!.

(ii):Writethecoefficientofx3y2z2 in(x+y+z)9.

Solution: This issameashowmanywaysonecanchoosexfrom three

brackets,ayfromtwobracketsandazfromtwobracketsintheexpansion

(x+y+z)(x+y+z)(x+y+z)(9times).

Thiscanbedonein9 9!

15120.322 3!2!2!

= =

3.2.9Note

The r objects of each rcombination can be permuted among r! different

rpermutations, eachofwhich corresponds to a single combination. If the

number or rcombinations of n objects without repetition (denoted by

C(n,r)). Then

r!C(n,r)=p(n,r)=r)!(n

n!



C(n,r)=r!r)!(n

n! =

r

n.

Itcanbeeasilyverifiedthat C(n1,r1)+C(n1,r)=C(n,r).

3.2.10Example

Howmanywaysmayonerightandoneleftshoebeselectedfromsixpairs

ofshoeswithoutobtainapair.

Solution: Anyoneoftheleftshoecanbeselectedinsixways.Wehave

fivechoiceforselectingarightshoewithoutobtainingapair.Thereforethe

totalnumberofwaysselectingoneleftandonerightshoeis=6 5=30

ways.

3.2.11Example

Anewnationalflagistobedesignedwithsixverticalstripsinyellow,green,

blue,andred.Inhowmanywayscanthisbedonesothatnotwoadjacent

stripshavethesamecolour.

Solution:Thefirststripcanbeselectedinfourdifferentways.Sincenotwo

adjacent stripshave the samecolour, the secondstripcanbe selected in

threedifferentways.Inasimilarway,3rd ,4rth,5thand6thstripsareselected

in three different ways.Therefore the total number of ways selecting the

differentcoloursinthestripsare4 3 3 3 3 3=4 35=972ways.

3.2.12Problem

i) Howmanypositiveintegerslessthanonemillioncanbeformedusing

7s8sand9sonly?

ii) Howmanyusing0s,8sand9sonly?.

Solution:

i) Wefindthenumberofintegersusedfrom1to9,99,999.

Numberofsingledigits(lessthan10)are7,8,9.

Numberofintegersformedusingtwodigitsare3 3=32.



Similarly, numberof integerswith 3digits is3 3 3=33,, the

numberofintegerswith6digitsis36.

Therefore thetotalnumberofpositive integers less than1millioncan

beformedusing7,8,9only=3+32+33+34+35+36=1092.

ii) Number of positive integers containing one digit is 2 (zero is not

considered)numberofpositiveintegerscontainingtwodigits=2 31.,

andsoon,numberofpositiveintegerscontainingsixdigitsis2 35.

Therefore the total number of integers containing 0, 8, 9 is = 2 + 2

(3+32++35)=728.

3.2.13Problem

Findthesumofallthefourdigitnumberthatcanbeobtainedbyusingthe

digits1,2,3,4onceineach.

Solution:Thenumberofpermutations(arrangements)canbemadeusing

4numbers (1,2,3,4) taking4atatimeisp(4,4)=0!4! =24.

Eachnumberoccur6 times inunitplace,6 times in10th place,6 times in

100thplace,6timesin1000place.

Thereforesumofthenumbersintheunitplaceis = 6.1+6.2+6.3+6.4=60

Totalsumofthedigitsinthe10thplace =60 10



Thereforetotalsumofall24numbers =66,660.

3.2.14Note

Supposethatrselectionsaretobemadefromnitemswithoutregardtothe

orderandthatunlimitedrepetitionsareallowed,assumingat leastrcopies

ofnitems.Thenumberofwaysoftheseselectioncanbemadeis



C(n+r1,r)=1)!(nr!

1)!r(n+ .

3.2.15Example

Inhowmanywayscanaladywearfiveringsonthefingers(notthethumb)

ofherrighthand?

Solution:Therearefiveringsandfourfingers.Fiveringscanbepermuted

in p (5, 5) ways. The number of unrestricted combinations of 4 objects

taken5atatimeis

- +

5

154=

5

8.Thereforethetotalnumberofways

=5!

5

8=6720.

3.3GeneratingFunctions

Considerthethreedistinctobjectsa,bandcandformthepolynomial

(1+ax)(1+bx)(1+cx)=1+(a+b+c)x+(ab+bc+ca)x2+abcx3.

If we consider the C(3,1) ways of selecting one object (a or b or c) and

represent it a + b + c which is the coefficient of first power of x = x1.

Similarly the 3 ways of selecting 2 objects (ab or bc or ca) may be

representedab+bc+ca,whichisthecoefficientofx2.

Next, there isonlyonewayofselectingall threeobjects,namelyabc, the

coefficientofx3. Inthelefthandside,thefactor1+axcanbeconsidered

asrepresentingsymbolicallythetwowaysofselectingaornot,the1(orx0)

representingthenonselectionofaandaxrepresentingtheselectionof

a.Thefactors(1+bx)(1+cx)canberepresentedinsimilarway.

The product of the three factors (1 + ax)(1 + bx)(1 + cx) indicates the

selectionornonselectionofalltheobjectsa,bandc.



Considerthegeneralcaseofnobjectsa1,a2,,anbyexpandinginpowers

ofx,thepolynomial(1+a1x)(1+a2x)(1+a3x)(1+anx)=1+(a1+a2+

+an)x+(a1a2+a1a3+)x2+(a1a2a3+)x3++(a1a2an)xn.

The coefficient of x2 on the right hand side represents the rcombination

choosenfromnobjectsa1,a2,,an.

Takinga1=a2=.=an=1weget

(1+x)n=1+nx+2

1)n(n - x2+.+xn.

=C(n,0)+C(n,1)x+C(n,2)x2+.+C(n,n)xn..(i).

The number of rcombinations is equal to the coefficient of xr in the

expansionof(1+x)r.Eachexpressionisalsotermedasenumeration.

If theobject isallowedunlimitedrepetitionthenthecorrespondingfactor in

theenumerationmusthaveeverypowerofxpresentandsois(1+x+x2+

+xi+)=(1x)1. Thustheenumerationofrcombinationsofnobjects

withunlimitedrepetitionis(1x)1(1x)1.(1x)1=(1x)n

=1+nx+2!

1)n(n + x2+

3!2)1)(nn(n + +

x3++r!

1)r1).....(nn(n + + x2+..

=

=

- +

0r

rxr

1rn.

The coefficient of xr in (1x)n is C(n + r 1, r) which is the number of

rcombinationswithunlimitedrepetition.

Intheexpression(i),replaceC(n,r)byr!

r)p(n,.

(1+x)n=1+1!

1)p(n,x+

2!2)p(n,

x2++r!

r)p(n,xr++

n!n)p(n,

xn,here

p(n,r)isthecoefficientofr!xr .



Theexpressiona0+a1 1!x +a2 2!

x2 +.+ai i!xi +.

=

=0i

i

i i!xa ,iscalledtheexponentialgeneratingfunction,herear=p(n,r).

Intheenumerationforrpermutationsofnobjectswithunlimitedrepetition,

the factor for eachobjectmust represent the fact that theobjectmaynot

appearorappearonce,twice,,etc.

Thus each factor is (1 +1!x +

2!x2 +

3!x3 +. +

i!xi + .) and the

exponentialenumeratoris(1+1!x +

2!x2 +)n=enx=

=0r

ir

i!xn .

3.3.1Example

Howmanycombinationsofthreeobjectscanbeformedifoneobjectcanbe

selectedatmostonce,thesecondobjectatmosttwiceandthethirdatmost

threetimes.

Solution:Thefirstobjectcanappeareitheronceornotatall,sothefactor

in theenumeratormustcontainx0 andx1 butnootherterms. Thesecond

objectwillhavex2,xandx0.Similarlythefactorforthethirdobjectwillhave

x0,x1,x2,x3.

Thereforetheenumerationis

(1+x)(1+x+x2)(1+x+x2+x3)=1+3x+5x2+6x3+5x4+3x5+x6.

Thecoefficientofxarea,b,c

Thecoefficientofx2areab,ac,bb,bc,cc

Thecoefficientofx3areabc,abb,acc,bbc,bcc,ccc

Thecoefficientofx4areabbc,accc,abcc,bbcc,bccc

Thecoefficientofx5areabbcc,abccc,bbccc

Thecoefficientofx6areabbccc.



3.4Partitions,compositions

Representationofanypositiveintegernasasumofoneormorepositive

integers(ai)thatis.,n=a1 +a2 +.+am. Notethatthisrepresentationis

unordereddivisions(senseofcombination)iscalledpartition,whereasthe

ordereddivision(senseofpermutation)iscalledcomposition.

3.4.1Example

Thepartitionsandcompositionsoftheinteger5.

i) Thereare7unrestrictedpartitions5,4+1,3+2,3+1+1,2+2+1,

2 + 1 + 1 +1, 1 + 1+ 1 + 1 + 1 and two of these namely,

4+1and3+2haveexactlytwoparts.

ii) Thereare 16 unrestricted compositionsof n=5are 5,4+1, 1+4,

3+2,2+3,3+1+1,1+3+1,1+1+3,2+2+1,2+1+2,

1+2+2,2+1+1+1,1+2+1+1,1+1+2+1,1+1+1+2,

1+1+1+1+1andfourofthesehaveexactlytwoparts.

Some times we omit the sign + in representation of partition or

composition.Thatis2+1+1+1willbe2111or213.

3.4.2GeneratingFunctionforpartition

Letpn be thenumberof unrestrictedpartitionsof nso that thegenerating

functionisp(x)=p0+p1x+p2x2+.+pnxn+

Consider the polynomial 1 + x + x2 + x3 + . + xk + . + xn. The

appearance of xk can be interpreted as the existence of just k ones in a

partitionoftheintegern.

Similarlythepolynomial1+x2+x4+.+x2k+.isinterpretedastwosin

the partition and in particular, the coefficient of x2k = (x2)k represents the

casejustktwosinthepartition.

Ingeneral,1+xr+x2r+.+xkr+.representsrsinthepartition.



Thereforegeneratingfunctionforpartitionsofnisobtainedby

p(x)=(1+x+x2+.+xk+.)(1+x2+x4+.+x2k+.)

(1+xr+x2r+.+xkr.).

=(1x)1(1x2)1.(1xr)1.

Theenumerationforpartitioninwhichnotermofapartisgreaterthankis

(1x)1(1x2)1.(1xk)1(i)

Theenumerationforpartitionwherenotermofapartisgreaterthank1is

(1x)1(1x2)1.(1xk1)1(ii).

Subtracting(ii)from(i)givesthegeneratingfunctionforpartitionofnintok

parts,which isxk(1x)1(1x2)1 .(1xk)1. Thecoefficientofxn gives

numberofpartitionsofnintoexactlykparts.

Thepartitionwitheverypartoddareenumeratedbythegeneratingfunction

containingonlyfactorscorrespondingtooddnumber,1,3,5,is

(1 x)1(1 x3)1(1 x5)1 Thepartitionwithunequal terms in apart is

enumeratingbythegeneratingfunction:

(1+x)(1+x2)(1+x3)

Each factor can only contain two terms, one which indicates that the

correspondingnumberdoesnotappearinthepartitionandtheotherthatthe

numberappearsonce.

3.4.3 Prove that the number of partitions of n inwhich no integer occurs

morethantwiceasapartisequaltothenumberofpartitionsofnintoparts

notdivisibleby3.

Solution: The generating function corresponding to no integer occurs

morethantwiceis(1+x+x2)(1+x2+x4)(1+x3+x6) (1)

Thegeneratingfunctioncorrespondingtonopartisdivisibleby3is

(1x)1(1x2)1(1x4)1(1x5)1(1x7)1(1x8)1. (2)

1x3=(1x)(1+x+x2)



Therefore(1)becomes

(1x3)(1x)1(1x6)(1x2)1(1x9)(1x3)1(1x12)(1x4)1(1x15) (1x5)1(1x18)(1x6)1

=(1x)1(1x2)1(1x4)4(1x5)1(1x7)1,whichisequation(2).

For instance, taken=6. Thepartitions inwhichno integeroccursmore

thantwiceare6,51,42,411,33,321,221,1(theseare8innumber).

Thepartitionsinwhichnopartisdivisibleby3are51,42,411,222,2211,

214,16.

3.4.4Generatingfunctionforcompositions

Considertheunrestrictedcompositionsofninwhichnonesinarow.Since

there is no restriction on the number of parts, wemay or may not put a

marker in any of the (n1) spaces between the ones in order to form

groups.Thiscanbedonein2n1ways.

Ifwerestrictthecompositionstohaveexactlymparts,then(m1)markers

are needed to form groups and the number of ways of placing (m1)

markersinthe(n1)spacesbetweentheonesis

-

-

1m

1n.

Let us use the generating function to obtain this. Let Cm(x) be the

enumeration for competitions of n with exactly m parts, where Cm(x) =

n

nmnxC and Cmn is the coefficient of x

n in this series is the number of

compositionsofnintoexactlymparts.Eachpartofanycompositioncanbe

one,two,threeoranygreaternumbersothatthefactorintheenumeration

mustcontaineachofthesepowersofx,andsois x+x2+x3+x4+.+xk

+. =x(1+x+x2+.+xk1+xk+.)=x(1x)1.

Sincethereareexactlymparts,thegeneratingfunctionistheproductofm

suchfactorsCm(x)=(x+x2+x3+x4+.+xk+.)m

=xm(1x)m



=xm

=

- +

0i

ixi

1im.

Replacingm+ibyr inthesummationweget

Cm(x)=

=

- -

mr

rxmr

1r=

=

- -

mr

rx1m

1r.

Thereforethecoefficientofxn inthisenumerationis

-

-

1m

1n.

The generating the compositions with no restriction, C(x) =

=1mm(x)C =

=

- 1m

mm x)(1x .Taket=x1

x -

intheseries,weobtain

C(x)=t+t2+t3+=t1

t -

=2x1

x -

=

=

-

1n

n1n x2 .

Thecoefficientofxn intheenumerationis2n1.

3.4.5FerrersGraph

The representationof permutationsbyanarrayof dots knownas Ferrers

graph.

Forinstance,theferrersgraphforthepartition:5322(whichisinstandard

formwiththelargestpartfirst)is

o

o

oo

o oooo

oo

oo



A Ferrersgraphhasthefollowingproperties:

i) Thereisonerowforeachpart

ii) Thenumberofdotsinarowisthesizeofthatpart

iii) Anupperrowalwayscontainatleastasmanydotsasalowerrow

iv) Therowsarealignedontheleft.

ThepartitionobtainedbyreadingtheFerrersgraphbycolumnsiscalledthe

Ferrersconjugatepartition.

In the above illustration, the same by rows and by columns is called self

conjugate.Thepartition4321isselfconjugate.

3.5Orderings

3.5.1Lexicographicorder

Let {1, 2, 3,..., n} be n objects to be permuted. For two permutations

a1,a2,,anandb1,b2,...,bn,wesaythata1,a2,...,ancomesbeforeb1,b2,

.,bn intheLexicographicorderifforsome1 m



3.5.4Problem

ObtaintheproceduretofindnextpermutationinLexicographicorder.

Solution:Considerapermutationk1,k2,...,knofndigits.Toobtainthenext

permutationinLexicographicorder,weproceedasfollows:

Step(i):Findthelargestisuchthatki1



3.5.7Example

Findthenextpermutationto4123inthereverseLexicographicorder.

Solution:Considertheorderk1 k2 k3 k4

4 1 2 3

i) i=2,sothat1=k2



3.6.3Example

Thirtycarswereassembledinafactory.Theoptionsavailablewerearadio,

anairconditioner,andwhitewalltires. It isknownthat15ofthecarshave

radios,8ofthemhaveairconditioners,and6ofthemhavewhitewalltires.

Moreover,3ofthemhaveallthreeoptions.Findoutatleasthowmanycars

donothaveanyoptionsatall.

Sollution:LetA1,A2andA3denotethesetsofcarswiththegivenoptions

respectively.

|A1|=15,|A2|=8,|A3|=6,|A1 A2 A3|=3.

Nowbytheprincipleofinclusionandexclusion,|A1 A2 A3|=15+8+6

|A1 A2| |A1 A3||A2 A3|+3=32|A1 A2||A1 A3||A2 A3|

32333=23.

(hereweusedthefact:|Ai Aj Ak| |Ai Aj|foranyi,j,k)

Thereforethereareatmost23carshaveoneormoreoptions.Thismeans

thereareatleast7carsthatdonothaveanyoptions.

3.6.4Example

Determinethenumberofintegersbetween1to250thataredivisiblebyany

oftheintegers2,3,5and7.

Solution:WriteA1={x Z+ /x 250andxisdivisibleby2}

SimilarlyA2,A3,A4 aresetof integers 250 thataredivisibleby3,5and

7respectively.

|A1|=

2250

=125where x denotestheintegersmallerthanorequaltox.

|A2|=

3250

=83,|A3|=

5250

=50,|A4|=

7250

=35,

|A1 A2| =

32250

=41,|A1 A3|=

52250

=25,|A1 A4|=17,|A2 A3|

=16,|A2 A4|=11,|A3 A4|=7,|A1 A2 A3|=

532250

=8,



|A1 A2 A4|=5,|A1 A3 A4|=3,|A2 A3 A4|=2,|A1 A2 A3 A4|=1.

Therefore|A1 A2 A3 A4|=125+83+50+3541251716

117+8+5+3+21=193.

3.6.5Example:

Howmanyarrangementsof thedigits0,1,2,3,4,5,6,7,8,9containat

leastoneofthepatterns289,234or487?

Solution: Let A289 be theevent of havingpattern 289. Similarly A234 and

A487.Wehavetofind|A289orA234orA487|.

Now|A289|=8!,as289consideredasagroupwhichisasingleobjectand

theremainingsevensingledigits.Similarly|A234|=|A487|=8!

Alsosince2cannotbefollowedbyboth3and8,wehave|A289 A234|=0.

Similarly |A289 A487|=0. But |A234 orA487|=6! ,since23487asasingle

objectandremaining5singleobjects. |A289 A234 A487|=0.

Thereforebytheprincipleofinclusionandexclusion

|A289 A234 A487|=8! +8! +8! 006! +0=3 8! 6! .

3.6.6Note

Among the permutations of {1, 2, 3, ., n}, there are some (called

derangements),inwhichnoneofthenintegersappearsinitsnaturalplace.

Inotherwords:(i1,i2,,in)isaderangementifi11,i22,.,and inn.

IfDndenotethenumberofderangementsof{1,2,3,.,n},thenforn=1,

2,3,wehave, D1 =0,D2 =1,D3 =2 (that is, theonlyderangementsof

(1,2,3)are(2,3,1)and(3,1,2)).

3.6.7 TheformulaforDn foranypositiveintegern,

LetUbethesetofn!Permutationsof{1,2,3,,n}.

Foreachi,letAi bethepermutation(b1,b2,.,bn)of{1,2,3,.,n}such

thatbi=i.



ThenA1={(1,b2,,bn)/(b2,b3,.bn)isapermutationof{2,3,,n}}.

Therefore|A1|=(n1)!.Inasimilarway|Ai|=(n1)!foreachi.

Also A1 A2 = {(1, 2, b3, , bn) / (b3, b4, , bn) is a permutation of

{3,4,,n}}.Therefore|A1 A2|=(n2)!.

Inasimilarway, |Ai Aj|=(n2)!forall i, j. Thereforeforanyintegerk,

1 k n.

|A1 A2 . Ak|=(nk)!.

Thisistrueforanykcombinationof{1,2,.,n}.

ThereforeDn= n21 A....AA = ( ) n21 A....AA =|U||A1 A2 . An|

= n!C(n,1)(n1)!+C(n,2)(n2)!+.+(1)nC(n,n)

=n!!1!n +

!2!n

!3!n +.+(1)n

!n!n

=n!

- + + - + - !n)1(....

!31

!21

!111

n=n!e1,ifn isverylarge.

3.6.8Example

Letnbooksbedistributedtonstudents.Supposethatthebooksarereturned

anddistributedtothestudentsagainlateron.Inhowmanywayscanthebooks

bedistributedsothatnostudentwillgetthesamebooktwice?

Solution:Firsttimethebooksaredistributedinn!wayssincenostudents

getsthesamebookthathegotfirsttime,thesecondtimeDnways.

Thereforethetotalnumberofways:

n!Dn=n!n!

- + + - + -

!n1)1(....

!31

!21

!111 n =(n!)2

- + + - + -

!n1)1(....

!31

!21

!111 n .



3.6.9Example:

Find the number of derangements of the integers from 1 to 10 inclusive,

satisfyingtheconditionthatthesetofelementsinthefirst5placesis

(i) 1,2,3,4,5insomeorder

(ii) 6,7,8,9,10insomeorder.

Solution:

(i).D5.D5ways

(ii)(5!)2=14,400derangements.

3.6.10Problem

Obtain the Euler f function, f(n), the number of integers x such that

1 x



SelfAssessmentQuestions1. InhowmanywayscanthelettersofthewordSUNDAYbearranged?

HowmanyofthembeginwithSandendwithY?HowmanyofthemdonotbeginwithSbutendwith?

2. Howmanysolutionsdoes theequation: x+ y + z=17have,wherex,y,zarenonnegativeintegers?

3. Astudentwishestotakeacombinationofthreecourses,onefromeachof three Science subjects. There are 4 Physics, 3 Chemistry and 2Biologycoursesonoffer.Howmanypossiblecombinationsarethere?

4. Howmanydifferent twodigitpositive integerscanbeformedfromthedigits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. (i)When repetition is not allowed,

(ii)Whenrepetitionisallowed.5. Howmany4digitnumberscanbeformedbyusing2,4,6,8when

repetitionofdigitsisallowed?6. Inhowmanywayscan 4 prizesbedistributedamong 5 personswhen

i) Nopersongetsmorethan1prizeii) Apersonmaygetanynumberofprizes.iii) Apersongetsalltheprizes.

7. Out of 15 boys and 9 girls, how many different committees can beformedeachconsistingof6boysand4girls?

8. Howmanycardsmustyoupickupfromastandard52carddecktobesureofgettingatleastoneredcard.

9. Show that thenumberof n isequal to thenumberof partitionsof 2nwhichhaveexactlynparts.

10. Howmanywayscananexaminerassign30marks to8questionssothatnoquestionreceiveslessthan2marks.

3.7Summary

The unit provides the close connection between combinatorics and

computer science for the advantage of both the areas. Computer science

gainsfromcombinatoricsthe toolsnecessaryforanalyzingalgorithmsand

data structures. Computer science also benefits from the computational

problemssuggestedbytheclassicalstructuresofcombinatorics.



3.8TerminalQuestions

1. Find thenumberofwayscanweobtainchangeforRs.10 in termsofRs.5,Rs.2 andRs.1.

2. How many positive integers less than or equal to 70, are relativelyprimeto70?

3. Howmany ndigit ternary (0, 1, 2) sequences are therewith at leastone0,atleastone1,andatleastone2?

4. Suppose there are 100 students in a school and there are 40students taking each language, French, 40 taking Latin, 40 takingGerman,20studentsaretakinganygivenpairoflanguages,and10studentsaretakingallthreelanguages,thenhowmanystudentsaretakingnolanguage?

5. Howmanyarrangementsofsix 0s,five 1s andfour 2s arethereinwhich:i)Thefirst0precedesthefirst1?ii)Thefirst0precedesthefirst1,whichprecedesthefirst2?

3.9Answers

SelfAssessmentQuestions

1. ThewordSUNDAYconsistsof6 letters,whichcanbearranged inP

(6,6)=6!=720ways. If SoccupiesfirstplaceandYoccupies last

place, thenother four lettersU, N, D, A can be arranged in 4! =24

ways. IfSdoesnotoccupythefirstplacebutYoccupies lastplace,

thefirstplacecanbeoccupiedin4waysbyanyoneofU,N,D,A.For

thesecondplace, again4 lettersareavailable, includingS. The3rd,

4th and5th places canbe filledby3, 2, 1ways. Hence the required

numberofarrangements=4 4 3 2 1=96.

2. To find the number of elements in {(x, y, z) / x + y + z = 17}= C

(17+31,17)=C(19,2)=171.

3. 24.

4. (i)90,(ii)100.

5. Thenumberof4digitsnumbersformedis:256.

6. (i)10,(ii)625.,(iii)620.

7. 630630.



8. 27cards.9. TaketheFerrersgraphforapartitionofnandacolumnofndots Con

theleftwegetaFerrersgraphcorrespondingto2nhavingnparts.

Forinstance,n=4andapartitionforitas:211

Ferrersgraphis

Addacolumnofn=4dotstotheleftofthegraphweget

Thecorrespondingpartitionis3221whichhasn=4partsof2n=8.10. Hereeachpartisgreaterthanorequalto2.Alsogivenn=30,m=8.

Apartcanberepresentedasx2 +x3 +=x2 [1+x+x2 +]=x2

(1x)1.Sincethereare8partswehavex16 (1x)8=x16

=

- +

0i

ixi

1im= +

- + 16ixi

1im

Sincen=30,thecoefficientofx30 is(herei=14)

- +

14

1148=

14

21.

TerminalQuestions1. 10ways.2. 24.3. The number of n digit ternary (0, 1, 2) sequences are there with

at least one 0, at least one 1, and at least one 2 is =

3n3 2n+3.4. HereN=100,N(F)=N(L)=N(G)=40,N(FL)=N(LG)=N(FG)=

20 andN(FLG)=10.Therefore thenumberof students takingno

languageis N ( )GLF =100(40+40+40)+(20+20+20) 10=30.5. i)ThenumberofarrangementsisC(15,4) C(10,5).

ii) C(14,5) C(8,4)ways.

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MC0063(A)-Unit3-fi