Discrete Mathematics. Silvia Marcaida Bengoechea

Transcription

1 Discrete Mathematics Silvia Marcaida Begoechea

2

3 Cotets Basic combiatorics 7 Combiatorics 7 Lists 8 3 Floor ad ceilig fuctios 9 4 Tree diagrams 4 Rule of product 4 Samples 3 43 Combiatios 6 44 Permutatios 8 45 Combiatios with repetitios 9 5 Factorial powers 6 Classificatios 7 The priciple of iclusio ad exclusio 5 7 Euler s totiet or phi (φ fuctio 8 8 Traslatios 9 9 The Dirichlet pigeohole priciple ad the salutes lemma 3 Combiatorial idetities 33 3

4 4 CONTENTS Combiatorial idetities 33 Combiatorial idetities ad proofs 33 Basic idetities of combiatorial umbers 35 3 Pascal s triagle 35 4 Vadermode s formula ad other idetities 36 Biomial formula 38 3 Multiomial coefficiets 4 3 Multiomial coefficiets 4 3 Combiatorial meaig Multiomial formula Some applicatios 44 4 Geeralized biomial formula 45 3 Geeratig fuctios ad recurrece relatios 49 3 Geeratig fuctios 49 3 Power series 49 3 Geeratig fuctio of a sequece of umbers 5 33 Examples (direct problems 5 34 Examples (iverse problems Operatios with geeratig fuctios 53 3 Geeratig fuctios ad combiatorial problems 54 3 Number of solutios of a equatio 54 3 Applicatios ad examples Recurrece relatios 57

5 CONTENTS 5 33 Combiatorial problems ad recurrece relatios Examples Geeratig fuctios ad recurrece relatios Other methods (algebraic method 6 4 Mai families of umbers 63 4 Fiboacci umbers 63 4 Catala umbers Partitios of atural umbers Defiitio Ordered partitios Partitios ad restricted partitios Partitios of differet summads Partitios with odd umber of summads Ferrers diagrams 7 44 Bell umbers 7 45 Stirlig umbers of the first id Stirlig umbers of the secod id 76 5 Graphs 79 5 Basic cocepts 79 5 Paths Trees Plaarity Colorigs 87

6 6 CONTENTS

7 Chapter Basic combiatorics Combiatorics Combiatorial theory is the study of methods of coutig how may objects there are of a give descriptio, or of coutig i how may ways somethig ca be doe, or of coutig how may ways a certai evet ca occur For example, the followig are some questios for combiatorial theory to aswer: How may objects are there of a give descriptio? (a How may pairs of atural umbers (x, y are there such that x + y 0 or x + y? (b How may teis matches will be played i a touramet i which there are 37 teis players? I how may differet ways ca somethig be doe? (a I how may differet ways ca four people be seated at a circular table? 3 I how may differet ways ca a certai evet occur? (a I how may ways ca 5 poits be obtaied after throwig a die four times?

8 8 Basic combiatorics I geeral, let A be the fiite set of elemets that satisfy a property calculate the umber of elemets of A, that is, the cardiality of A Our aim is to A {x : x satisfies a property P } Calculate A Lists I order to cout the umber of the elemets of a set a first idea is to mae a list of the elemets For example, Solutios of a equatio: (a A {(x, y N N : x + y 5} List: {(0, 5, (, 4, (, 3, (3,, (4,, (5, 0} Therefore, A 6 (b (Geeralizatio A {(x, y N N : x + y, N } List: {(0,, (,, (,, (3, 3,, (,, (, 0} Therefore, A + (c (Variatio A {(x, y N N : x + y, N, x, y }, that is, A is the set of solutios of x + y with x, y itegers If 3, A 0 If 4, list: {(,, (3, 3,, (, } A 3 (d (Geeralizatio A {(x, y, z N N N : x + y + z, N }, that is, A is the set of solutios of x + y + z, x, y, z o-egative itegers List: x y z } {{ } } {{ } x y z 0 } {{ 0 } + Hece, A ( A ( } {{ } 0 0 }{{}

9 3 Floor ad ceilig fuctios 9 A ( + ( + A ( + ( + (e (Geeralizatio A {(x, x,, x m N N N : x + x + + x m, N }, that is, A is the set of solutios of x +x + +x m, x, x,, x m o-egative itegers Remars: Would you try to build the list i this case? (a Buildig the list may be easy, less easy or impracticable (b A list caot have repetitios or abseces (all the elemets must be i the list ad without repetitios (c Before tryig to solve a problem it may be coveiet to aalyze a particular case Multiples of a iteger: (a A set of multiples of 4 amog the 87 first atural umbers A {x : x 87, x 0 ( (mod 4} {4, 8,, 6,, 868} A {4, 4, 4 3, 4 4,, 4 467} A 467 (b A set of multiples of 4 amog the first atural umbers A {4, 4, 4 3, 4 4,, 4 c} where c is the quotiet of dividig by 4 ( 4c + r, 0 r < 4 Thus, A c, that is, c is the iteger part of 4 (c [ 4 ] (c a N, A,a set of multiples of a amog the first atural umbers A,a {a, a, a 3, a 4,, a c} ad A,a c [ a ] 3 Floor ad ceilig fuctios Defiitio 3 Let x R x floor fuctio of x iteger part of x the largest iteger ot greater tha x Defiitio 3 Let x R x ceilig fuctio of x the smallest iteger ot less tha x

10 0 Basic combiatorics Remar 33 These ames ad otatio were itroduced by Keeth E Iverso i 96 Figura: Properties: i x x x Z ii x x + x / Z iii x x, x R iv x x, x R v x x x, x R Defiitio 34 Let x R {x} fractioal part of x x x Defiitio 35 Let x R < x > pseudofractioal part of x x x Remar 36 {x}, < x > [0, Example 37 Recall that A,a c [ a i {,,, } is A,a a { } a a (Notice that { } [0, a ] Notice that the ratio of multiples of a a a { a } a Example 38 Iteger coordiates i a circle Let A be the set of poits whose atural coordiates are i a circle of radius Therefore, A {(x, y N N : x + y } How much is A? Figura First colum: (,, (,,, (, where is the largest iteger such that + Therefore,

11 3 Floor ad ceilig fuctios Secod colum: (,, (,,, (, l where l is the largest iteger such that + l Therefore, j colum: (j,, (j,,, (j, r where r is the largest iteger such that j + r r j r j Therefore, r j Thus, A j j Example 39 Let A be the set of squares of area u i a fourth of a circle How much is A? Figura Area covered by the squares A u area of the fourth of the circle π Hece, 4 A π 4 O the other had, if we move the squares oe up ad oe to the right the circle is covered but the part paited i orage Therefore, π A 4 + Thus, π 4 A π 4 π 4 A π 4 Sice π 4 0 whe, by the sadwich rule, A π 4 whe Thus, A π 4 Example 30 Number of digits i the decimal system Give N calculate the umber of digits of i the decimal system For example, 34 d( 4; 300 d( abcd 9999 < abcd < d( < 0 d( log 0 (0 d( log 0 ( < log 0 (0 d( d( log 0 ( < d( d( log 0 ( d( log 0 ( + Example 3 Number of itegers i (α, β, α, β R Let A α,β set of itegers i the iterval (α, β, α, β R How much is A α,β? For example: α 5, β 47 A α,β α 5, β 8 A α,β 0 Figura

12 Basic combiatorics Remar: Why is the first iteger α + istead of α? If α were iteger the α α But the iterval is ope i α! Therefore, A α,β { α +, α +,, β } { α +, α +,, α + β α } Thus, A α,β β α For example, α 5, β 47 A α,β 5 α 5, β 8 A α,β Tree diagrams 4 Rule of product Rule of product: If a evet X ca happe i x ways ad a distict evet Y i y ways the X ad Y ca happe i xy ways Example 4 If we ca go from Bosto to Chicago i 3 ways (x, y, z ad from Chicago to Dallas i 5 ways (,,3,4,5 the the umber of ways i which we ca go from Bosto to Chicago to Dallas is 5 Figura Figura There are r braches startig from 0 (first geeratio There are r braches startig from each brach of the first geeratio (secod geeratio Ad so o Fially, there are r braches startig from each brach of the geeratio ( geeratio Let R be the total umber of braches i the geeratio, that is, R is the umber of ways from 0 to the top of the tree R r ; R R r r r ; ; R R r i r i Aother versio: Pigeohole

13 4 Tree diagrams 3 If there are c ways of fillig hole, c ways of fillig hole,, c ways of fillig hole the there are c c c ways of fillig the pigeohole Example 4 I order to mae up a meu we ca select amog two differet first dishes: A or B, three differet secod dishes: a,b or c ad two differet desserts: α or β How may differet meus ca we mae up? 3 meus 4 Samples Defiitio 43 A set is a uordered collectio of distict objects The objects are called elemets of the set We use braces to deote a set For example, the set with elemets, ad 3 is deoted {,, 3} Sice the elemets are ot ordered, {,, 3} ad {, 3, } are the same set Defiitio 44 A sequece is a ordered collectio of ot ecessarily distict objects We use bracets to deote a sequece For example, (,, Sice the etries are ordered, (,, ad (,, are differet sequeces Defiitio 45 Let Ω {a,, a } be a set of elemets ad A -sample without repetitio of Ω is a arragemet of the elemets of Ω tae at a time, where two arragemets are regarded as differet if they differ i compositio or i the order of their elemets I other words, a -sample without repetitio of Ω is a sequece of legth that ca be formed with the elemets i Ω without repeatig them Let V, deote the umber of -samples without repetitio give distict objects How much is V,? We ote that we ca choose the first elemet i ways, the secod elemet i ways,, the th elemet i + ways By the rule of product, V, ( ( +! (! Example 46 A certai society has 5 members The members of the society are to elect a presidet, a vice presidet, a secretary, ad a treasurer I how may ways is it possible to select the 4 officers if o member of the society ca hold more tha oe office at a time? We are to fid the umber of samples (without repetitios of 5 members tae 4 at a time This umber is equal to V 5, , 600

14 4 Basic combiatorics Example 47 I a draw there are 3 prizes: a car, a trip ad a baset full of food 00 ticets have bee sold, umbered from to 00 For each ticet a ball has bee itroduced ito a drum The first umber draw wis the baset, the secod wis the trip ad the third, the car How may differet possibilities are there? Baset: 00 possible umbers, trip: 99 possible umbers, car: 98 possible umbers V 00, , 00 If I bought 5 ticets, how may possibilities would I have of wiig the car? Baset: 99 possible umbers, trip: 98 possible umbers, car: 5 possible umbers , 50 How may possibilities do I have of wiig the three prizes? Baset: 5 possible umbers, trip: 4 possible umbers, car: 3 possible umbers How may possibilities do I have of wiig at least oe prize? We study it later Defiitio 48 Let Ω {a,, a } be a set of elemets ad A -sample with repetitio of Ω is a sequece of legth that ca be formed with the elemets i Ω beig able to repeat them Let V R, deote the umber of -samples with repetitio give distict objects How much is V R,? We ote that we ca choose the first elemet i ways, the secod elemet i ways,, the th elemet i ways By the rule of product, Notice that it is ot ecessary V R, Example How may umbers of 3 digits ca be formed i the decimal system? V R 0,3 0 3, 000 How may of them are palidromic? 0 0 V R 0, 0 00

15 4 Tree diagrams 5 How may are eve? V R 0, Example 40 How may football pools coupos must we fill i order to be absolutely sure we wi? Notice that there are 5 matches ad 3 possible results V R 3, , 348, 907 Example 4 Let Ω {a, e, o, u, b, c, d} A set of words of 4 letters that ca be formed with the alphabet i Ω i such a way that the first is a cosoat ad the last a vowel A A set of words of 5 letters that ca be formed with the alphabet i Ω i such a way that there are ot equal cosecutive letters A , 07 Example 4 A is the set of umbers of 5 differet digits startig ad fiishig i a eve digit A , 70 Example 43 Calculate the umber of ways to place 4 distiguishable balls i 3 umbered boxes i such a way that oe of the boxes are empty First idea: Place balls,, 3 oe i each box ad the ball 4 This is ot a good idea because, 3 4 would t be couted Secod idea: Select three balls, place them oe i each box ad place the fourth ball This is ot a good idea because we are coutig, 4 3 ad 4, 3 but they are the same Good idea: Choose the box with two elemets (3, select the balls i that box (?, place the remaiig balls oe i each box ( We study it later

16 6 Basic combiatorics 43 Combiatios Defiitio 44 Let Ω {a,, a } be a set of elemets ad A - combiatio without repetitio of Ω is a arragemet of the elemets of Ω tae at a time, where two arragemets are regarded as differet oly if they differ i compositio I other words, a -combiatio without repetitio of Ω is a subset of elemets of Ω The umber of -combiatios without repetitio of elemets is deoted by the symbol C, There is a simple relatio betwee the umber C, of -combiatios of elemets ad the umber V, of -samples Let s see it with a example Example 45 Let Ω {a, b, c, d, e} Cout the 3-samples without repetitio: Oe way: V 5, Aother way: Choose the three elemets ad the order the elemets: V 5,3 C 5,3 3! C 5,3 V 5,3 3! !! 5! 3! 3! ( 5!3! 3 I geeral, C, V,! ( Example 46 Calculate the umber of ways to place 4 distiguishable balls i 3 umbered boxes i such a way that oe of the boxes are empty As we said i example 43 we ca choose the box with two elemets (3, select the balls i that box (C 4,, place the remaiig balls oe i each box ( Therefore, 3 (4 3 4! 36!! Example 47 (Geeralizatio Calculate the umber of ways to place + distiguishable balls i umbered boxes i such a way that oe of the boxes are empty First we choose the box with two elemets (, select the balls i that box (C +,, place the remaiig balls oe i each box (V, Therefore, C +, V,

17 4 Tree diagrams 7 Example 48 (Geeralizatio Calculate the umber of ways to place + distiguishable balls i umbered boxes i such a way that oe of the boxes are empty Two ways: oe box has three balls ad the remaiig boxes have oe ball each two boxes have balls ad the remaiig boxes have oe ball each Choose the box with tree elemets (, select the balls i that box (C +,3, place the remaiig balls oe i each box (V, Therefore, C +,3 V, Choose the double boxes (C,, select the balls i the first double box (C +,, select the balls i the secod double box (C,, place the remaiig balls oe i each box (V, Therefore, C, C +, C, V, Thus, the solutio is C +,3 V, + C, C +, C, V, Propositio 49 Basic properties of the biomial coefficiets C, ( V,! (! (!!! (!!!, 0,,!! ( (!!!(! ( (!(!, 0,, Choosig elemets amog is the same as rejectig amog 3 ( 0! 4 ( 5 (! ( 0!0!!!! (!!! ( + (,, 0,, ( 6 If, ( 7 If, ( ( Proof- 5 ( ( + (! [ (! + (!! [ + (! (! + ( (!(! (!! (!(! (!! ] (!! ( (!! (!! (!! ] 6 (! ( ( + (!! ( ( ( + ( (! ( ( (!(! 7 ( ( ( (

18 8 Basic combiatorics 44 Permutatios Defiitio 40 A permutatio of elemets or -permutatio is a -sample without repetitio of elemets, that is, a -permutatio is a sample without repetitio of elemets which cotais all the elemets I other words, permutatios of elemets are all the possible -arragemets each of which cotais every elemet oce, with two such arragemets differig oly i the order of their elemets The umber of -permutatios is deoted by P P V, (! Defiitio 4 A multiset is a geeralizatio of the otio of set i which elemets are allowed to appear more tha oce Defiitio 4 Let M be a multiset with elemets of the first type, elemets of the secod type,, elemets of the th type The differet arragemets of M are the permutatios with repetitio I other words, a permutatio with elemets of the first type, elemets of the secod type,, elemets of the th type is a sequece formed with elemets of the first type, elemets of the secod type,, elemets of the th type P (,,, deotes the umber of such permutatios How much is P (,,,? Example 43 Calculate the umber of sequeces that ca be formed with p zeros ad q oes p + q Choose where to place the zeros: C p+q,p ( p+q p We get the same choosig the places for the oes: C p+q,q ( p+q q Example 44 Calculate the umber of sequeces that ca be formed with p zeros, q oes, ad r twos Choose the places for the zeros ad the choose the places for the oes: C p+q+r,p C q+r,q ( p+q+r p ( q+r (p+q+r! q (q+r! (p+q+r! (p+q+r p!p! (q+r q!q! (q+r!p! (q+r! r!q! (p+q+r! p!q!r!

19 4 Tree diagrams 9 Notice that P (,,, is the umber of sequeces that ca be formed with elemets of the first type, elemets of the secod type,, elemets of the th type Let ( ( ( ( P (,,, 3! (! (!!(!!(! 3!( 3! (!!(!!!!!! 45 Combiatios with repetitios Defiitio 45 A -combiatio of elemets with repetitio is a multiset of elemets amog the The umber of -combiatios with repetitio of elemets is deoted by the symbol CR, How much is CR,? Example 46 How much is CR 5,3? Let Ω {a, a, a 3, a 4, a 5 } be a set of five elemets We wat to form multisets of 3 elemets Defie A {a α a α a α3 : α, α, α 3 {,, 3, 4, 5}, α α α 3 } This is the set of multisets of 3 elemets of Ω Thus, CR 5,3 A Defie Ω {b, b, b 3, b 4, b 5, b 6, b 7 } ad B {b β b β b β3 : β {,, 3, 4, 5}, β {, 3, 4, 5, 6}, β 3 {3, 4, 5, 6, 7}, β < β < β 3 } Defie the map φ : A B a α a α a α3 b α b α b α3 This is a bijectio Ijective: Let φ(a α a α a α3 φ(a γ a γ a γ3 The b α b α +b α3 + b γ b γ +b γ3 + α γ, α + γ +, α 3 + γ 3 + α γ, α γ, α 3 γ 3 a α a α a α3 a γ a γ a γ3 Surjective: Let b β b β b β3 with β {,, 3, 4, 5}, β {, 3, 4, 5, 6}, β 3 {3, 4, 5, 6, 7} ad β < β < β 3 Let α β, α β, α 3 β 3 Thus, α {,, 3, 4, 5}, α {,, 3, 4, 5}, α 3 {,, 3, 4, 5} ad α α α 3 Hece, there exists a α a α a α3 A such that φ(a α a α a α3 b β b β b β3 Therefore, the sets A ad B both have the same umber of elemets Sice B C 7,3, CR 5,3 A B C 7,3

20 0 Basic combiatorics Example 47 Let Ω {a, b, c, d} Therefore 4 Let 3 {a, a, a} 000 {a, a, b} 000 {a, a, c} 000 {a, b, b} 000 {a, b, c} 000 {c, c, d} 000 Therefore, CR 4,3 umber of sequeces that ca be formed with 3 zeros ad 3 oes This ca be calculated choosig three places out of 6, which is ( 6 3 For each multiset of elemets there is a sequece formed with zeros ad oes, ad coversely Therefore, CR, is the umber of sequeces formed with zeros ad oes Thus, CR, C +, V +,! ( +! ( +!! ( +! (!! ( + Example 48 Calculate the umber of sequeces that ca be formed with p zeros ad q oes CR q+,p ( ( q++p p p+q This is aother way of calculatig example 43 p Example 49 Calculate the umber of sequeces that ca be formed with zeros ad 3 oes such that there are ot cosecutive zeros Choose of the 4 possible places: C 4, ( 4 4! 6!! Example 430 (Geeralizatio Calculate the umber of sequeces that ca be formed with p zeros ad q oes such that there are ot two cosecutive zeros

21 5 Factorial powers Choose p of the possible q + places: C q+,p ( q+ p Example 43 Calculate the umber of sequeces that ca be formed with zeros, 3 oes ad 4 twos such that the zeros are ot cosecutive Choose a sequece formed by s ad s (place the s ad the isert the zeros (place the 0s i the holes: C 7,3 C 8, ( 7 3( 8 Example 43 (Geeralizatio Calculate the umber of sequeces that ca be formed with p zeros, q oes ad r twos such that there are ot two cosecutive zeros Choose a sequece of s ad s (place the s ad the isert the zeros (place the 0s i the holes: C q+r,q C q+r+,p ( ( q+r q+r+ q p 5 Factorial powers Defiitio 5 Let a C Its descedig or lower or fallig factorial (power of order, N, is { a(a (a + a 0 Defiitio 5 Let a C Its ascedig or upper or risig factorial (power of order, N, is { a(a + (a + a 0 Propositio 53 Some properties are the followig: (a ( a ( a (b ( a ( a Proof- ( a ( a( a ( a + ( a( (a + ( (a + ( a

22 Basic combiatorics (c ( a ( a Proof- ( a ( a( a + ( a + ( a( (a ( (a + ( a (a a +l a a l (b a +l a (a l Proof- a +l a(a (a + (a (a (a l + a (a l (c a +l a (a + l 3 a N {0} (a If 0 ( ( + ( ( +( ( ( (! (! (b If > 0 Proof- ( ( ( ( Classificatios Rule of sum: Let Ω be a set If {A,, A } is a partitio of Ω, that is, ia i Ω ad A i A j for all i j the Ω A + + A Classificatio is useful: sometimes there is o other solutio but to classify, we may fid iterestig idetities, 3 classificatio is the mai idea to get recurrece relatioships Example 6 We have already doe this A ways of placig 5 distiguishable balls i 3 umbered boxes such that oe of them is empty Defie A ways i which there are two boxes with two balls each, A ways i which there is a box with three balls Therefore, A A + A Example 6 Ω {a,, a } P(Ω collectio of subsets of Ω power set of Ω Subsets of Ω:

23 6 Classificatios 3 0 elemets: elemet: {a }, {a },, {a } elemets elemets: Ω P(Ω C,0 + C, + + C, i0 C,i i0 ( i Example 63 Let Ω {a,, a } How much is P(Ω? Set x P(Ω Subsets that cotai a : there are x P({a,, a } Subsets that do ot cotai a : there are x P({a,, a } Therefore, x x + x x x 3 x 3 x By example 6, i0 ( i Example 64 a umber of sequeces of legth formed with zeros ad oes i which there are ot two cosecutive zeros a (0, a 3 (0, 0, a 3 5 (00, 0, 0, 0, We classify by the first elemet: Startig with zero: the secod umber must be so there are a Startig with oe: there are a Hece, a a + a, 3 Example 65 (Recurrece relatioship of the biomial coefficiets

24 4 Basic combiatorics Let Ω {a,, a } ad Recall that -combiatios without repetitio of elemets are the subsets of Ω with elemets First type: those that cotai a There are C, Secod type: those that do ot cotai a There are C, Therefore, C, C, + C, ( ( ( + We get Pascal s triagle of the biomial coefficiets or combiatios: ( 0 ( 0 ( ( 0 ( ( ( 0 3 ( 3 ( 3 ( 3 ( 0 4 ( 4 ( 4 ( 4 3 ( Thus, Example 66 a umber of sequeces of legth formed with zeros, oes ad twos i which there are ot two cosecutive oes a 3 (0,, a 8 (00, 0, 0, 0,, 0,, For 3: Startig with 0: there are a Startig with 0: there are a Startig with : there are a Startig with : there are a Therefore, a a + a, 3 Example 67 There are two types of floor tiles: x, x Let a be the umber of ways to tile a rectagular floor of dimesio x

25 7 The priciple of iclusio ad exclusio 5 a a 3 We classify the ways to tile by the firs elemet: First elemet x: there are a First elemets x, x: there are a First elemet x: there are a Therefore, a a + a 3 7 The priciple of iclusio ad exclusio Notatio: A B AB for simplicity Example 7 Let A { : ad is divisible by or 3} { :, 0 (mod } { :, 0 (mod 3} Deote B { :, 0 (mod } ad C { :, 0 (mod 3} Is {B, C } a partitio of A? It is true that B C A but B C For example 6 B C So we caot deduce that A B + C But, is there ay relatioship? Let us cosider the Ve diagram Figura A B + C B C B, C 3, B C 6 Therefore, A Example 7 A { :, 0 (mod } B { :, 0 (mod 3} C { :, 0 (mod 5}

26 6 Basic combiatorics How much is D A B C? Ve diagram: Figura D A B C (A B C (A B + C (A B C A + B A B + C A C B C A + B A B + C ( A C + B C A B C A + B + C A B A C B C + A B C Thus, D Theorem 73 (The Priciple of Iclusio-Exclusio Let A,, A be subsets of a fiite set Ω The i A i A i A i A j + A i A j A + +( r i i <i < <i r i<j i<j< A i A i A ir + + ( i A i Proof- By iductio o We have already see it for Suppose it is true for ia i i A i A i A i + A ( i A ia i A i + A i A ia A i i ( i<j A i A i A i A j + +( r i<j i <i < <i r + A A i A j A + + ( r A i A i A ir + +( i A i + i <i < <i r + + ( i A ia A i A i A j + A i A j A + i +( r i<j i <i < <i r How may quatities appear i the formula? i<j< A i A i A ir + + ( i A i A i A i A ir A + ++C, +C,3 + +C,r + +C, 0 C, (Notice that the correspods to the quatity o the left

27 7 The priciple of iclusio ad exclusio 7 Example 74 Suppose that we have married couples (a ma ad a woma i a ball: (M, W,, (M, W I how may differet ways ca we pair the me off with the wome so that o ma daces with his wife? Let Ω {all the couples that ca be doe}, Ω {couples i which o ma daces with his wife} Notice that Ω P (M :, M :,, M or it ca be see as a arragemet of {,, } We wat to calculate Ω but it is easier to calculate Ω c Ω c {couples i which there is at least oe that is a married couple} Hece, Ω Ω Ω c Defie A {married couple daces together},,, Fix married couple i A i umber of ways to arrage the remaider couples P (! Fix married couples i, j A i A j umber of ways to arrage the remaider couples P (! Fix married couple i,, i r A i A ir umber of ways to arrage the remaider couples P r ( r! A A P 0 Thus, Ω c i A i i,j A ia j + + ( r i < <i r A i A ir + + ( A A (! C, (! + + ( r C,r ( r! + + ( i ( i C,i ( i! Therefore, Ω Ω Ω c! i ( i C,i ( i! i0 ( i C,i ( i! Remar 75 A simpler expressio for the priciple of iclusio-exclusio: Give I {,, } defie A I i I A i The, i A i ( I A I I {,,}

28 8 Basic combiatorics 7 Euler s totiet or phi (φ fuctio Let N Defie A { {,, } : is prime with } { {,, } : gcd(, } Euler s totiet or phi fuctio is defied as φ( A If, φ( If > ad p α p α p α l l is its prime decompositio the φ( ( ( p p ( pl l i ( pi Example 76 φ( φ( 3 ( ( Remar 77 Notice that φ( is a iteger Proof- We ca write A { {,, } : is ot divisible by ay p i } The A c { {,, } : is divisible by at least oe of the p i } Let B i { {,, } : is divisible by p i }, i,, l Thus, ( l B i i l i p i + i<j l i<j l φ( A A c l i B i B i B j + + ( r p i p j + + ( r i < <i r l i < <i r l B i B ir + + ( l B B l p i p ir + + ( l p p l

29 8 Traslatios 9 ( l i p i + i<j l p i p j + + ( r i < <i r l ( ( p p ( pl + + ( l p i p ir p p l l i ( pi 8 Traslatios Defiitio 8 To traslate a problem is to state a equivalet problem i other terms Example 8 Poits i R (x, y pairs of poits i R Lie ax + by c I this way we traslate a geometrical problem i a algebraic oe Example 83 Placemet of m distiguishable balls i umbered boxes Placemet without exclusio (several balls ca be placed i the same box mappig from {,, m} to {,, } Placemet with exclusio (each box has at most oe ball ijective mappig from {,, m} to {,, } Example 84 Placemet of m udistiguishable balls i umbered boxes Placemet without exclusio (several balls ca be placed i the same box sequece formed by m zeros ad oes Placemet with exclusio (each box has at most oe ball sequece formed by m zeros ad oes i which there are ot two cosecutive zeros Example 85 Let Ω {,, }

30 30 Basic combiatorics If Ω {,, 3, 4, 5}: {} 0000 {, 5} 000 Therefore, subset of Ω sequece of legth formed by zeros ad oes Example 86 Let Ω {,, } Subset of elemets of Ω sequece of legth formed by oes ad zeros Example 87 H-V trajectory a trajectory formed by flights of the followig types: (x, y + (x, y (x +, y, Therefore, this is a H-V trajectory: (x, y Figura Let T (p,q (0,0 set of H-V trajectories from (0, 0 to (p, q, with p, q N Put 0 for H ad for V, so we get a sequece of zeros ad oes H-V trajectory from (0, 0 to (p, q sequece of zeros ad oes Example 88 U-D trajectory a trajectory formed by flights of the followig types: (x, y (x +, y +, (x, y (x +, y Let Θ (p,q (0,0 set of U-D trajectories from (0, 0 to (p, q, with p, q N Let x umber of flights U, y umber of flights D Thus, x + y p ad x y q Thus, x p+q, y p q If p < q the y < 0, so there is t ay U-D trajectory from (0, 0 to (p, q If p ad q have differet parity the x ad y will ot be itegers, so there is t ay U-D trajectory from (0, 0 to (p, q If p q 0 ad p ad q have the same parity the U-D trajectory from (0, 0 to (p, q sequece formed by p+q zeros ad p q oes

31 9 The Dirichlet pigeohole priciple ad the salutes lemma 3 9 The Dirichlet pigeohole priciple ad the salutes lemma The Dirichlet Pigeohole Priciple: If m pigeos occupy pigeoholes ad m > the at least oe will house at least two pigeos I mathematical terms, if m > there is ot ay ijective mappig from {a,, a m } to {b,, b } Example 9 Let s prove that i New Yor there are at least two persos with the same umber of hairs o the head Suppose that a perso has at most hairs o the head ad that i New Yor more tha persos live By the Dirichlet Pigeohole Priciple it follows The Salutes Lemma: Let be the umber of guests i a party Suppose that if a perso says hello to aother perso, the latter always says hello to the former Thus, the umber of persos that say hello to a odd umber of persos is eve Therefore, if the umber of guests is odd the there is at least oe perso that says hello to a eve umber of guests

32 3 Basic combiatorics

33 Chapter Combiatorial idetities Combiatorial idetities Combiatorial idetities ad proofs A combiatorial idetity is a equality where there are combiatorial umbers For example, ( ( ( + + +, 0 0 A idetity ca be proved i differet ways: Aalytically: by iductio, reductio to aother ow equality, Combiatorially: by fidig a combiatorial problem ad two differet coutig methods whose results are each side of the equality Example ( + 0 ( + + (, 0

34 34 Combiatorial idetities Combiatorial proof: Let Ω be a set of elemets First coutig method: Subsets of Ω sequeces of legth formed with zeros ad oes There are Secod coutig method: Subsets of Ω of elemets: subsets of Ω (, 0,,, Therefore, there are 0 Aalytical proof: By iductio o : For 0, ( Let ad assume that the formula is true for Havig i mid that ( ( ( +, ( ( 0 [( 0 + ( + ] [( 0 + ( + ] [( + + ( + ( ] + ( ( 0 + ( + ( 0 + ( + ( + + ( + ( + ( ( ( ( + ( 0 ( Example ( 0 ( + ( ( + ( 0, Aalytical proof: Sice ( ( ( + (, ( 0 ( + ( + ( ( ( 0 ( ( 0 + ( + ( ( ( +( +( +( 0 Combiatorial proof: We have to prove ( 0 + ( + ( + ( 3 +, that is, the umber of subsets of {,, } with a eve umber of elemets is equal to the umber of subsets of {,, } with a odd umber of elemets Let P set of subsets of {,, } with a eve umber of elemets ad I set of subsets of {,, } with a odd umber of elemets Defie the map Sice it is a bijectio P I P { I A \ {} if A A A {} if / A Example 3 a ( + 0 ( +, b ( + ( + 3 By the two previous examples a + b ad a b Therefore a ad a

35 Combiatorial idetities 35 Basic idetities of combiatorial umbers Recall Propositio 49: (!, 0,,!(! ( (, 0,, ( ( 0 ( ( ( +,, 0,, ( (,,,,,,,, ( ( 3 Pascal s triagle Recall Pascal s triagle: ( 0 ( 0 ( ( 0 ( ( ( 0 3 ( 3 ( 3 ( 3 ( 0 4 ( 4 ( 4 ( 4 3 ( 4 ( ( 5 ( 5 ( 5 ( 5 4 ( 5 ( ( 6 ( 6 ( 6 ( 6 ( 6 5 ( 6 ( ( 7 ( 7 ( 7 ( 7 ( 7 ( 7 6 ( Thus, From the triagle we see, for example, that ( ( ( + 5 ( + 6 ( I geeral, ( ( ( ( ( r + r ,, r 0 0 r r

36 36 Combiatorial idetities Proof : ( ( ( ( ( ( + r + + r + r + r + r + r r r r r r r ( ( ( ( ( + r + r + r r r r 0 ( ( ( ( ( + r + r + r r r r 0 Proof (iductio o r: For r 0 ( ( Assume that it is true for r ad let s see it for r: ( ( ( ( ( ( ( r + r + r + r , 0 r r r r where the first equality is true by the iductio hypothesis ad the secod by oe of the basic combiatorial idetities 4 Vadermode s formula ad other idetities That is, ( ( m 0 r + ( ( m + + r r 0 Proof- (by iductio o ( ( m r ( r ( m 0 ( + m For 0, ( 0 0( m r ( m r, for all r, m N {0} r ( + m r,, m, r N {0},, m, r N {0} Assume that the formula is true for for ay r, m The ( ( ( ( ( ( m m m r r r 0 ( ( (( ( ( (( ( ( m m m r 0 r r r 0 (( ( ( ( ( ( m m m r r r 0 (( ( ( ( ( ( m m m r r r 0

37 Combiatorial idetities 37 ( ( ( + m + m + m + r r r Proof- (combiatorial Let Ω {a,, a, b,, b m }, A {a,, a }, B {b,, b m } The umber of subsets of Ω with r elemets is ( +m r ( O the other had, the umber of subsets of Ω with r elemets, of which are i A is m ( r, for 0,, r Thus, r ( ( m ( 0 r +m r Other idetities ( + 0 ( + + ( Proof- By Vadermode s formula whe m r: ( ( 0 + ( ( + + ( ( ( 0 ( Proof- (aalytical ( ( r r ( ( r, r 0! r! r!( r!!(r!! (!!( r!(r! (! ( ( r Proof- (combiatorial Let Ω {,, } umber of pairs (A, B where A is a subset of Ω with r elemets ad B is a ( ( r r subset of A with elemets Aother strategy is to choose B as a subset of Ω with elemets ad add r elemets i order to build A: ( ( r

38 38 Combiatorial idetities Biomial formula Theorem Let x, y C ad 0,,, The (x + y 0 ( x y ( Proof- (by iductio o For 0, (x + y 0 ad ( 0 0 x 0 y 0 For, (x + y x + y ad ( 0 x 0 y + x y 0 y + x Suppose it is true for ad let s see it for ( x 0 y + 0 ( x 0 y + 0 (x + y (x + y (x + y x(x + y + y(x + y ( ( x x y + y x y 0 ( ( x + y + x y 0 0 ( ( ( xy + x y + + x y ( ( ( + x 0 y + xy + + x y 0 (( ( (( ( + xy (( ( ( + + x y + x y 0 ( ( ( ( xy + x y + + x y + x y 0 0 x y ( x y Proof- (combiatorial Notice that (x + y(x + y xx + xy + yx + yy, (x+y(x+y(x+y (xx+xy+yx+yy(x+y xxx+xxy+xyx+xyy+yxx+yxy+yyx+yyy

39 Biomial formula 39 For : (x + y (x + y(x + y (x + y xx x + xx xy + sequeces of legth that ca be formed with x ad y 0 ( x y Corolary ( + x 0 ( x ( + 0 ( x + + ( x Corolary 3 ( x 0 ( ( x 0 ( ( x ( 0 ( ( x + + ( x Corolary 4 ( + 0 ( x + ( x 4 + ( + x + ( x 4 Proof- ( + x + ( x (( ( 0 + x + ( 4 x 4 + Corolary 5 ( x + ( x ( x 5 + ( + x ( x 5 Proof- ( + x ( x (( ( x + 3 x 3 + ( 5 x 5 + Some combiatorial idetities we obtai from the biomial formula: x : ( ( 0 + ( ( ( 0 ( + + (,

40 40 Combiatorial idetities ( 0 + ( + ( 4 + ( + ( 3 + ( 5 + x i: ( ( ( ( ( ( + i i + i + i + + i 0 0 ( ( ( ( ( ( r r (( ( ( ( ( +i ( r + A + Bi, r + ( ( ( ( (( ( (( ( A ( ( ( ( (( ( (( ( B Put ( ( S 0 + +, 0 4 ( ( S + +, 6 ( ( S + +, 5 ( ( S , + Notice that S j is the umber of subsets of {,,, } whose umber of elemets is j ( (mod 4, for j 0,,, 3 Therefore, ( + i (S 0 S + (S S 3 i We call r α re iα rcosα + ir si α With this otatio + i π e i π 4 ad 4 ( + i ( e i π 4 cos π + i si π Hece, 4 4 S 0 S cos π 4, S S 3 si π 4 Moreover, we ow that S 0 + S S + S 3 Thus, S 0 + cos π 4,

41 Biomial formula 4 S + si π 4 S cos π 4,, Derivig: For x : For x : Itegratig: t 0 For t : For t : ( + x ( + x dx Itegratig agai: x S 3 si π 4 ( x ( + x 0 (, 0 ( (, 0 ( t x dx 0 ( + t ( + t + dt + [ ( + x [ ( + t + ( + ( + t ] x ( x, ] t 0 ( t + + ( + ( + ( ( + 0 x 0 0 ( [ ] x + t + 0 t + dt ( [ t + ( + ( + ] x 0

42 4 Combiatorial idetities For x : For x : Other operatios: ( + x + ( + ( + x + + ( + ( + + ( + ( ( x + ( + ( + ( ( + ( + ( ( + ( + ( + ( + x ( + x m ( + x +m is a polyomial idetity, so both sides of the equality have the same coefficiets Therefore, r ( ( ( m + m, i r i r i0 which is Vadermode s formula A little bit more: x y (x + ( y ( (x + y x ( y For x y : ( ( ( 0 3 Multiomial coefficiets 3 Multiomial coefficiets Defiitio 3 Let N {0} ad r, r,, r m N {0} such that r +r + +r m (! r, r,, r m r!r! r m! Example 3 ( 6 6! 3,, 3!!!

43 3 Multiomial coefficiets 43 Propositio 33 ( ( r, r r ( ( ( ( ( r r r r r m r, r,, r m r r r 3 r m Proof- ( ( ( ( r r r r r m r r r 3 r m! ( r! ( r r! r!( r! r!( r r! r 3!( r r r 3! ( r r m! r m!r m!! r!r! r m! 3 Combiatorial meaig ( r,r,,r m! r!r! r m! is the umber of sequeces of legth that ca be formed with r symbols α,, r m symbols α m, which is also the umber of ways to place distiguishable balls i m umbered boxes so that there are r balls i the first box, r i the secod,, ad r m i the last oe Strategy: choose the places for the α s or choose the balls i the first box: ( r choose the places for the α s or choose the balls i the secod box: ( r r 3 4 choose the places for the α m s or choose the balls i the peultimate box: ( r r m r m 5 choose the places for the α m s or choose the balls i the last box:

44 44 Combiatorial idetities 33 Multiomial formula Theorem 34 Let x, x,, x m C ad 0 The, ( (x + x + + x m x r x r x rm m r, r,, r m r i N {0} r + r + + r m Proof- (x +x + +x m (x +x + +x m (x +x + +x m m m i i ( umber of sequeces x i x i x i i which there are r symbols x,, r m symbols x m x r ( x r x r x rm m r, r,, r m r i N {0} r + r + + r m m x i x i x i i x r x rm m 34 Some applicatios m r i N {0} r + r + + r m Proof- Multiomial formula whe x x m r i N {0} r + r + r 3 + r 4 ( r, r,, r m ( ( r 3+r 4 0, r, r, r 3, r 4 Proof- By the multiomial formula whe m 4 ( (a + b + c + d r, r, r 3, r 4 r i N {0} r + r + r 3 + r 4 a r b r c r 3 d r 4 Put a b, c d

45 4 Geeralized biomial formula 45 3 If r + + r m the r! r m! is a divisor of! Proof- Sice ( r,,r m is a iteger, r! r m! is a divisor of! 4 [[(m!]!] m [m!]! Proof- Put r r r m (m! ad r + + r m m(m! m! Thus r! r m! [[(m!]!] m divides [m!]! Defiitio 35 m!! is called semifactorial of m { m(m (m 4 if m 0 (mod m(m (m 4 if m (mod 4 Geeralized biomial formula Defiitio 4 Let α R, N {0} ( { α α! 0 α(α (α +! Propositio 4 If α N {0} the ( C, ; otherwise, if α / N {0} the ( α has o combiatorial meaig ( α + ( α ( α +, α R, N {0} 3 ( ( α α + ( Proof- ( ( α + α (α+ ( α+! α (! + α! α (! [ ] + α + α α+ (! (α+α(α (α +! 3 ( α ( α ( α ( α(α+ (α+ ( (α+ ( ( α+!!!! ad CR, ( + Notice that C, ( Recall ow Taylor s formula for a polyomial: ( (

46 46 Combiatorial idetities Theorem 43 Give a polyomial of degree at most, p(x c 0 +c x+c x + +c x 0 c x, it turs out that c p( (0, 0,, ad p(x p ( (0! 0 x! Proof- Sice p(x 0 c x, p(0 c 0 ad p (r (x r ( ( r + c x r r r c x r for r,, Therefore, p (r (0 r r c r r!c r ad c r p(r (0 r! Let s apply this theorem to p(x ( + x Note that p (r (x ( ( r + ( + x r r ( + x r Thus, c r p(r (0 r ( r! r! r Therefore, ( ( + x x r r This is a ew proof of Newto s biomial formula We are goig to geeralize this taig f(x ( + x α Notice that r0 If α the f(x +x ad D f R {} If α the f(x + x ad D f [, + If α the f(x +x ad D f (, + What is the domai of f? Lemma 44 Let f(x ( + x α For every α R, (, + D f Proof- Let x (, + Thus, x + > 0, x + e l(x+ ad ( + x α e α l(+x exist ad is fiite Therefore x D f Moreover, f(x is derivable i (, + It turs out that f (r α r ( + x α r, which is cotiuous for x (, +, ad f (r (0 α r By Taylor-Maclauri s Theorem f(x 0 Therefore, f(x lim 0 f ( (0 x + R (x with lim R (x 0 whe x <! f ( (0 x f ( (0! 0 ( + x α There are some especial cases: 0 ( α! x I our case, x for x <

47 4 Geeralized biomial formula 47 α N {0} ( + x is a polyomial of degree ad the geeralized biomial formula is already ow sice for > ( 0 α if x < Sice (+x 0 Thus, Cosequetly, α, x < ( + x Thus, Cosequetly, (+x α, N {0} ( + x (+x ( x for x <, +x 0 ( ( + x 0 ( x + x x + x x 3 + x 4 x + x + x + x 3 + x 4 + x + x + x 4 + x 6 + x 8 + x r + xr + x r + x 3r + x 4r + 0 ( x 0 ( ( + x 0 ( ( + x ( + x x + 3x 4x 3 + 5x 4 ( x + x + 3x + 4x 3 + 5x 4 + ( x r + xr + 3x r + 4x 3r + 5x 4r + 0 ( x 0 ( ( + x 0 CR,x Notice that C, is the coefficiet of x i the series ( + x 0 C,x ad CR, is the coefficiet of x i the series ( x 0 CR,x α ( + x +x ( 3 5 ( 0 ( 3 5 ( (!! (! x Now, ( ( ( ( +( + ( +! (!!! ( 3 ( ( (! 0! 4 6 ( (!( ( ( 3 ( ( (! ( ( ( (! 4!! 4 Thus, ( + x ( ( x + x 4!

48 48 Combiatorial idetities Some applicatios of the geeralized biomial formula are: Sice for α, β R ad x <, ( + x α ( + x β ( + x α+β the [( ( ( ] [( ( ( ] α α α β β β + x + x + + x + x [( ( ( ] α + β α + β α + β + x + x + 0 ( The coefficiets of x r o the left ad right had sides must be equal Therefore, α ( β ( 0 r + α β ( ( r + + α β ( r( 0 α+β r ad ( α + β r r 0 which is the geeralized Vadermode formula ( ( α β r Sice for α R ad x <, ( + x α ( x α ( x α the [( ( ( ] [( ( ( ] α α α α α α + x + x + x + x 0 0 [( ( ( ] α α α x + x 4 0 Therefore, ( ( ( α 0 ( r α r + α ( { ( ( r α r + + α α 0 r (mod r( 0 ( ( r α r r 0 (mod Thus, r ( ( { α α ( r r 0 0 r (mod r 0 (mod ( r ( α r

49 Chapter 3 Geeratig fuctios ad recurrece relatios 3 Geeratig fuctios 3 Power series Theorem 3 (Covergece Let (a 0 be a sequece of real or complex umbers, 0 a x a power series ad ρ lim sup a The, If ρ 0 the series coverges oly for x 0 If ρ > 0: For x < ρ the series is absolutely coverget For x > ρ the series diverges For x ρ othig ca be said (there are cases i which coverges ad cases i which it does t

50 50 Geeratig fuctios ad recurrece relatios ρ is called the radius of covergece of the series Let r (r 0 be a sequece of real umbers ad S r collectio of subsequeces of r with limit (fiite or ifiite Let L collectio of the limits of the subsequeces i S r It turs out that lim sup r max L ad lim if r mi L Example 3 r 0,, 0,,, r { 0 if eve if odd L {0, } ad lim sup r, lim if r 0 Propositio 33 Assume that r (r 0 has limit, l lim r The every subsequece has limit l Therefore L {l} ad lim sup r l, lim if r l lim if( r lim sup r ad lim sup( r lim if r If r r for all the lim if r lim if r ad lim sup r lim sup r If lim a a l the lim a l Example 34 Let a α, 0,,,, α C Let 0 (αx ρ lim sup α α lim sup α - If α the a for all ad the series is + x + x + which is the geometric series of ratio (ρ 3- If α 0 the the series is + 0x + 0x +, which coverges for all x (ρ 0 Example 35 Let a ( α, 0,,,, α R Notice that a ( a α a lim a α + lim ρ Hece, ( α α + ad, therefore, lim a Thus, lim sup a ad 0 ( α x coverges for x < Furthermore, if α is a positive iteger, α m: ( m 0, ( m,, ( m m, 0, 0, lim a 0 ad ρ m ( m 0 x coverges for all x

51 3 Geeratig fuctios 5 3 Geeratig fuctio of a sequece of umbers Defiitio 36 Let a (a 0 (a 0, a, a, be a sequece of real (complex umbers The geeratig fuctio of a is g a (x a x a 0 + a x + a x + 0 i the real (complex x i which the series coverges 33 Examples (direct problems Example 37 a (C, 0 (C,0, C,, C,,, C,, C,+, C,+, (( ( ( ( ( (,,,,,,, The geeratig fuctio is g a (x C, x 0 C, x 0 0 ( x ( + x, x R Example 38 g a (x CR, x 0 a (CR, 0 ( ( x ( x, x < 0 Example 39 g a (x 0 (( α a 0 ( α x ( + x α, x < Example 30 g a (x a,,, a x + x + x +, x < x 0

52 5 Geeratig fuctios ad recurrece relatios Example 3 a, 0,, 0,, 0, a { 0 (mod 0 (mod a ( + g a (x + x + x 4 + x, x < ( x < Example 3 a, 0, 0,, 0, 0,, 0, 0, a { 0 (mod 3 0 otherwise g a (x + x 3 + x 6 + x 3, x3 < ( x < Example 33 a, 5, 5, 5 3, 5 4, a 5 V R 5, g a (x +5x+5 x +5 3 x x+(5x +(5x 3 + 5x, 5x < ( x < 5 Example 34 a, r, r, r 3, r 4, a r V R r,, r 0 g a (x +rx+r x +r 3 x 3 + +rx+(rx +(rx 3 + rx, rx < ( x < r Example 35 a a 0, a,, a, 0, 0, g a (x a 0 +a x+a x + +a x +0x + +0x + + a 0 +a x+a x + +a x, x R Example 36 a 0!,!,!, 3!, a! g a (x 0! +! x +! x + 3! x3 + e x, x R

53 3 Geeratig fuctios Examples (iverse problems Example 37 g a (x ( + x m α, x < ( ( ( α α α g a (x ( + x m α + x m + 0 Thus, a x m + { ( α 0 (mod m m 0 otherwise ( α x 3m + 3 Example 38 g a (x e x ( x 0!! x +! x4 3! x6 + Thus, a { ( 0 (mod (! 0 otherwise Falta ejemplos 35 Operatios with geeratig fuctios Propositio 39 g ca (x cg a (x, ca (ca 0, ca, g a+b (x g a (x + g b (x 3 g a (xg b (x g c (x, c a 0 b + a b + + a b 0, 0,,, Proof- g ca (x 0 (ca x c 0 a x cg a (x g a (x + g b (x 0 a x + 0 b x 0 (a + b x g a+b (x Defiitio 30 I the last item of the previous propositio c is the covolutio of a ad b, which is writte c a b

54 54 Geeratig fuctios ad recurrece relatios Example 3 Let a (a 0 with geeratig fuctio g a (x defied for x < ρ Let c a 0 + a + + a, 0,,, The, c a b with b (,,, Thus, g c (x g a (xg b (x g a (x( + x + x + ga(x x Example 3 Let d a 0 + a a, 0,,, Let d a b with b ( 0,,, Thus, g b (x + x + (x + ( x ad g d (x ga(x x Example 33 Let r ( ( α 0 + α ( + + α This is a particular case of Example 3 whe a ( α Hece, gr (x ga(x (+xα x x Example 34 (Derivatives Let ρ > 0 ad g a (x 0 a x, x < ρ (radius of covergece The g a is idefiitely derivable i the iterval of covergece ad g a(x a x, g a(x ( a x,, g a (r (0 r r a r ad a r g(r a (0 g (r a r! (x r r a x r Taig x 0, Remar 35 If g a is the geeratig fuctio of a (a 0, a, a, the g (r a (x r r a r + (r + r a r+ x + (r + r a r+ x + is the geeratig fuctio of a (r (a (r 0, a (r, with a (r (r + r a r+ 3 Geeratig fuctios ad combiatorial problems 3 Number of solutios of a equatio Theorem 3 Let B {b, b, } N {0} Z +, C {c, c, } N {0} Z +,, H {h, h, } N {0} Z + be subsets of N {0} Z + Let a umber of solutios of x + x + + x where x B, x C,, x H The the geeratig fuctio of (a 0 is g a (x a x (x b + x b + (x c + x c + (x h + x h + Proof- 0 (x b +x b + (x c +x c + (x h +x h + r B x r x r x r r C r H a x 0

55 3 Geeratig fuctios ad combiatorial problems 55 Remar 3 If x < x b + x b + + x + x + + x b + x b x b + x b + + x Hece, x b + x b + is absolutely coverget I the same way, x c + x c + ad x h + x h + are absolutely coverget Thus, g a is well defied for x < 3 Applicatios ad examples Example 33 Let a umber of solutios of x + +x where x {0,,, },, x {0,,, } The geeratig fuctio of a (a 0 is g a (x (x 0 +x +x + (x 0 +x +x + (x 0 +x +x + (x 0 +x +x + Thus, ( ( x x 0 ( ( ( ( + a ( ( ( x ( + Example 34 Let a umber of ways of gatherig euros with cois or bills of euro, euros ad 5 euros Put x umber of cois of euro, x umber of cois of euros ad x 3 umber of bills of 5 euros The a umber of solutios of x + x + 5x 3 where x, x, x 3 N {0} umber of solutios of y + y + y 3 where y {0,,, }, y {0,, 4, }, y 3 {0, 5, 0, } The geeratig fuctio of (a 0 is g a (x (x 0 +x +x + (x 0 +x +x 4 + (x 0 +x 5 +x 0 + x x x 5 ( + x( x ( x 5 A + x + B x + C ( x + Dx4 + Ex 3 + F x + Gx + H x 5 Now it ca be wored out i order to obtai a Example 35 A die is rolled 0 times Determie the probability the outcome is 30 poits Let x i be the outcome i the i-th time, i 0 Possible outcomes: (x,, x 0 where x i {,, 6}, i 6 There are 6 0

56 56 Geeratig fuctios ad recurrece relatios Favorable outcomes: (x,, x 0 where x + + x 0 30 Set a 30 umber of favorable outcomesumber of solutios of x + + x 0 30 where x i {,, 6}, i 6 g a (x (x +x + +x 6 (0 (x +x + +x 6 (x +x + +x 6 0 x 0 (+x+ +x 5 0 a 30 is the coefficiet of x 30 i g a (x, that is, the coefficiet { of x 0 i ( + x + + x 5 0 Let x 6 S + x + + x 5 The ( xs x 6 x ad S x 6 x Thus, ( 0 a 30 0 ( ( x ( + x + + x ( x 6 0 ( x 0 x ( ( ( 0 ( 0 0 ( x ( 6 l x l l 0 l0 ( ( ( ( 0 + ( ( ( ( ( ( ( ( ( ( ( 0 ( 0 ( 3 ( 0 ( 0 ( 8 ( ( Example 36 I how may ways ca a studet get a total of 8 poits with 6 tass where studets are awarded 0, or 4 poits for each tas? Let T i be the tas i, let x i be the score obtaied i T i ad a the umber of solutios of x + + x 6 where x i {0,, 4}, i 6 We wat to fid a 8 The geeratig fuctio of (a 0 is Three ways of fidig a 8 : (+x +x 4 6 g a (x (x 0 + x + x 4 (6 (x 0 + x + x 4 ( + x + x 4 6 ( x 6 ( + x + x 4 6 r +r +r 3 6 a 8 ( ( 6,, ,3,3 6! 6 x ( x 6 6 ( x 6 ad goig o as i the previous example ( 6 r,r,r 3 r x r x 4r 3 ( 6 r +r +r 3 6 r,r,r 3 x r +4r 3 Thus ! + 6! 3!3! ( + x + x 4 6 ( + x ( + x 6 6 ( x 6 ( 6 0 l0 ( l 0 ( 6 x l 6 0 l0 ( 6 ( l 6! 3!3! + 6 5! 4! (x ( + x 6 ( 6 0 x ( + x x (+l Hece, a 8 ( 6 6 ( 6( ( 4

57 33 Recurrece relatios Recurrece relatios 33 Combiatorial problems ad recurrece relatios Examples Example 33 Let a umber of subsets of {,,, } a 0, a a, Example 33 Let a umber of sequeces of legth formed with 0s ad s where there are ot two cosecutive 0s a 0, a, a a + a, Example 333 Let a umber of ways of tilig a floor with ad tiles a 0, a, a a + a, Example 334 Let a cosecutive itegers umber of subsets of {,,, } i which there are ot two a 0, a, a a + a, Example 335 Let a umber of ways of climbig a stair of steps if i each move we climb or steps a 0, a, a a + a, Example 336 (Tower of Brahma or Tower of Haoi The Tower of Haoi (also called the Tower of Brahma or Lucas Tower is a mathematical game or puzzle It cosists of three rods, ad a umber of diss of differet sizes which ca slide oto ay rod The puzzle starts with the diss i a eat stac i ascedig order of size o oe rod, the smallest at the top, thus maig a coical shape The objective of the puzzle is to move the etire stac to aother rod, obeyig the followig rules: Oly oe dis may be moved at a time Each move cosists of taig the upper dis from oe of the rods ad slidig it oto aother rod, o top of the other diss that may already be preset o that rod

58 58 Geeratig fuctios ad recurrece relatios No dis may be placed o top of a smaller dis Let a the miimum umber of moves to solve the problem whe there are diss a, a a +, Example 337 Let a umber of regios i the plae that are determied by lies that are cocurret i pairs but there are ot three that itersect at a sigle poit a 0, a, a 4, a a +, 3 Example 338 Let T umber of poits that are ecessary to build a triagle of rows as the followig: T T 3 T 3 6 T 4 0 T 5 5 T 6 T 0 0, T T +, Example 339 Let a umber of maps f : {,, } {,, } such that f f idetity Notice that: f(, there are a f( f(, there are a f( f(, there are a Therefore, a, a, a a + ( a, Whe we have a recurrece relatio ad eough iitial values we ca virtually get ay term a of the sequece But, ca we give a explicit formula for a i terms of?

59 33 Recurrece relatios 59 Example 330 Related to Example 338, we saw that T 0 0, T T +, Therefore, T 0 0, T 0 +, T , T , T , Thus, T ( + Sice this ca be writte also as T + ( + ( , T ( + ad T (+ Example 33 Suppose we are give the followig recurrece relatio a 0 a, a αa + β, If α the a a +β, is a arithmetic progressio with commo differece β If β 0 the a αa is a geometric progressio with commo ratio α a 0 a, a αa + β, a α(αa + β + β α a + β( + α, a 3 α(α a + β( + α + β α 3 a + β( + α + α By iductio o, a α + β( + α + α + + α { α + β α α α a + β α Example 33 Let a 0 a, a b, a αa + βa,, α, β costats ad β 0 a αa + βa 0 αb + βa, a 3 αa + βa α(αb + βa + βb b(α + β + αβa, a 4 αa 3 + βa α(b(α + β + αβa + β(αb + βa b(α 3 + βα + a(α β + β It is ot as simple as the previous example We have to fid aother method 33 Geeratig fuctios ad recurrece relatios The method is, give the recurrece relatio fid the geeratig fuctio i order to obtai a We see it with some examples:

60 60 Geeratig fuctios ad recurrece relatios Example 333 Let a 0 a, a αa +β, Let g(x a 0 +a x+a x +a 3 x 3 + be the geeratig fuctio of (a 0 Suppose x < Thus, g(x a 0 + (αa 0 + βx + (αa + βx + (αa + βx 3 + a 0 + αx(a 0 + a x + a x + + βx( + x + x + a 0 + αxg(x + βx x g(x a 0 + β x x αx a αx + βx ( x( αx βx If α, βx( x βx ( x 0 βx a Moreover, a a + β, 0 If α, β A( α ad A had, βx ( x( αx a αx x βx ( x( αx A x + β 0 a 0 x Thus, g(x B αx ( ( x 0 ( + βx+ 0 (a + βx Therefore, Thus, βx A( αx + B( x For x, For x, β B( Thus, B β β O the other α α α α α A + B A α x αx 0 x +B 0 α x 0 (A+Bα x Moreover, 0 aα x Hece, g(x 0 (A + (B + aα x ad a A + (B + aα β α + ( β α + aα aα + β( α α α Example 334 Let a 0 a, a b, a αa + βa,, α, β costats ad β 0 Let g(x a 0 + a x + a x + a 3 x 3 + be the geeratig fuctio of (a 0, x Dg Thus, g(x a 0 + a x + (αa + βa 0 x + (αa + βa x 3 + a 0 +a x+αx(a x+a x + +βx (a 0 +a x+ a 0 +a x+αx(g(x a 0 +βx g(x Thus, g(x a 0 + (a αa 0 x αx βx If α, β, a 0, a the a a + a umber of ways to tile a floor with ad tiles The geeratig fuctio is g(x x + x ( + x( x A + x + B x A ( x + B 0 x 0 (A( + B x 0

61 33 Recurrece relatios 6 O the oe had, A( x + B( + x so 3B ad B 3, ad 3A, A 3 Thus, g(x 0 3 (( + + x Thus, a ( Example 335 a 0 a, a b, a c, a αa + βa + γa 3, 3, α, β, γ Z, γ 0 g(x a 0 +a x+a x + a 0 +a x+a x +(αa +βa +γa 0 x 3 +(αa 3 +βa +γa x 4 + a 0 + a x + a x + αa x 3 + αa 3 x βa x 3 + βa x γa 0 x 3 + γa x 4 + Thus, a 0 + a x + a x + αx(g(x a 0 a x + βx (g(x a 0 + γx 3 g(x g(x a 0 + a x + a x + αx( a 0 a x + βx ( a 0 αx βx γx 3 a + (b αax + (c bα aβx αx βx γx 3 Example 336 a 0 a, a b, a αa + βa + γ, g(x a 0 + a x + a x + a 0 + a x + (αa + βa 0 + γx + (αa + βa + γx 3 + Thus, a 0 + a x + αx(a x + a x + + βx (a 0 + a x + + γx ( + x + If γ (γ 0 the a 0 + a x + αx(g(x a 0 + βx g(x + γx x g(x a 0 + (a αa 0 x + γ x x αx βx g(x a 0 + (a αa 0 x + x g γ (x αx βx 333 Other methods (algebraic method Let a 0 a, a b, a αa + βa, S {(x 0 : x αx + βx } S is a vector space over C