Notes on Combinatorics. Peter J. Cameron


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1 Notes o Combiatorics Peter J Camero
2 ii Preface: What is Combiatorics? Combiatorics, the mathematics of patters,, helps us desig computer etwors, crac security codes, or solve sudous Ursula Marti, VicePricipal (Sciece ad Egieerig), Quee Mary, Uiversity of Lodo These otes accompaied the course MAS219, Combiatorics, at Quee Mary, Uiversity of Lodo, i the Autum semester 2007 It is impossible to defie combiatorics, but a approximate descriptio would go lie this We are give the job of arragig certai objects or items accordig to a specified patter Some of the questios that arise iclude: Is the arragemet possible? I how may ways ca the arragemet be made? How do we go about fidig such a arragemet? This is best illustrated by examples Example 1: Sudou You are give a 9 9 grid, divided ito ie 3 3 squares Your job is to put the umbers 1,2,,9 ito the cells of the grid i such a way that each umber occurs just oce i each row, oce i each colum, ad oce i each 3 3 subsquare It is ot hard to see that the arragemet is ideed possible A heroic calculatio by Bertram Felgehauer ad Frazer Jarvis i 2005 showed that there are 6,670,903,752,021,072,936,960 differet ways of fillig the grid Now suppose that someoe has complicated the problem by writig some umbers ito the grid already I geeral it may or may ot be possible to complete the grid; ad eve if it is, it may be very difficult to fid a solutio Nevertheless, may people aroud the world ejoy egagig with this combiatorial problem every day Example 2: Euler s officers 1782: The great mathematicia Leohard Euler ased i
3 Six differet regimets have six officers, each oe holdig a differet ra (of six differet ras altogether) Ca these 36 officers be arraged i a square formatio so that each row ad colum cotais oe officer of each ra ad oe from each regimet? Euler cojectured that the aswer is o, ad this guess was evetually proved correct i 1900 However Euler also cojectured that the aswer is o if six is replaced by 10, 14, or ay umber cogruet to 2 mod 4 He was completely wrog about this, but this was ot discovered util the 1960s Example 3: Kirma s schoolgirls I 1843, Thomas Kirma ased: Fiftee schoolgirls go for a wal every day for a wee i five rows of three Is it possible to arrage the wals so that every two girls wal together exactly oce durig the wee? This is certaily plausible Each girl has to wal with fourtee others; every day there are two other girls i her row, so seve days would be the right umber for the schedule However, this does ot prove that the arragemet is possible I fact, it ca be doe; Kirma himself foud a schedule satisfyig the coditios Examples ad reality The examples may give you the impressio that combiatorics is a collectio of charmig puzzles of little relevace to our moder techological world I fact this is completely wrog The course is ot really about applicatios, but i the digital world this subject is of eormous sigificace People (ad computers!) sped a lot of time sortig data, sedig messages through etwors, correctig faulty data or ecodig data to eep it safe from uauthorised access, desigig better etwors, looig for ew combiatios of atoms to form molecules which will provide us with better drugs, ad so o We eed to decide whe such a problem has a solutio, ad to fid the solutio efficietly These otes These otes reflect the cotets of the course i 2007 I have added a couple of proofs of major theorems ot covered i the course The otes have bee provided with exercises (some of them with wored solutios) ad a idex The recommeded textboo for the course was my ow boo Combiatorics: Topics, Techiques, Algorithms, first published i 1994; but rather tha followig the boo I have writte everythig aew The course covers roughly the first half of the boo; if you ejoyed this, you may wat to read more, or to loo at my Notes o coutig o the Web I am grateful to Vola Yildiz who spotted a umber of misprits i a prelimiary versio of the otes iii
4 iv Further readig Either of the two level 4 courses at Quee Mary ca be tae by studets who have doe the Combiatorics course: MAS408: Graphs, Colourigs ad Desig MAS439: Eumerative ad Asymptotic Combiatorics I metioed above my Notes o coutig which are o the web i the same place as these otes Some other boos which cotai further material (icludig the recommeded course text) are: Marti Aiger, Combiatorial Theory, Spriger, 1979 Norma Biggs, Discrete Mathematics (2d editio), Oxford Uiversity Press, 2002 Peter J Camero, Combiatorics: Topics, Techiques, Algorithms (2d editio), Cambridge Uiversity Press, 1996 J H va Lit ad R M Wilso, A Course i Combiatorics, Cambridge Uiversity Press, 1992 Jiri Matouse ad Jaroslav Nešetřil, Ivitatio to Discrete Mathematics, Oxford Uiversity Press, 1998
5 Cotets Preface ii 1 Subsets ad biomial coefficiets 1 11 Subsets 1 12 Subsets of fixed size 2 13 Properties of biomial coefficiets 3 14 The Biomial Theorem 6 15 Further properties of biomial coefficiets 7 16 Appedix: Proof of Lucas Theorem 10 2 Selectios ad arragemets The formulae Proofs Balls i urs Maig words from letters 18 3 Power series Power series Operatios o power series The Biomial Theorem Other power series 28 4 Recurrece relatios Fiboacci umbers Liear recurreces with costat coefficiets Liear recurreces with ocostat coefficiets Noliear recurreces 41 5 Partitios ad permutatios Partitios: Bell umbers Partitios: Stirlig umbers 49 v
6 vi CONTENTS 53 Permutatios: cycle decompositio Permutatios: Stirlig umbers 52 6 The Priciple of Iclusio ad Exclusio PIE Surjectios ad Stirlig umbers Deragemets 61 7 Families of sets Sperer s theorem Itersectig families The de Bruij Erdős theorem Fiite fields ad projective plaes Appedix: Proof of the Erdős Ko Rado Theorem 72 8 Systems of distict represetatives Hall s Theorem How may SDRs? Sudou 81 9 Lati squares Row by row Youde squares Orthogoal Lati squares Sets of mutually orthogoal Lati squares Appedix: Proof of Bose s Theorem Steier triple systems Existece of STS() Kirma s schoolgirls Appedix: Proof of Kirma s Theorem 101 Solutios to oddumbered exercises 107 Miscellaeous problems 119 Idex 123
7 Chapter 1 Subsets ad biomial coefficiets Oe of the features of combiatorics is that there are usually several differet ways to prove somethig: typically, by a coutig argumet, or by aalytic methods There are lots of examples below If two proofs are give, study them both Combiatorics is about techiques as much as, or eve more tha, theorems 11 Subsets Let be a oegative iteger, ad let X be a set with elemets How may subsets does X have? Propositio 11 The umber of subsets of a elemet set is 2 First proof We ecode subsets by sequeces (e 1,e 2,,e ), where each e i is either 0 or 1 There are 2 choices for e 1, 2 choices for e 2,, 2 choices for e ; so altogether 2 sequeces So we are doe if we ca establish a bijectio betwee subsets ad sequeces To each subset Y of X, we associate the sequece (e 1,e 2,,e ) where e i = { 1 if i Y, 0 if i / Y It is easy to see that each sequece arises from a subset, ad distict sequeces arise from distict subsets; so the correspodece is a bijectio Secod proof This is a proof by iductio Let f () be the umber of subsets of {1,2,,} We see that f (0) = 1 (the empty set has just oe subset, amely itself) Also, f ( + 1) = 2 f (); for each subset Y of {1,2,,} ca be exteded i two ways to a subset of {1,2,, + 1}: we ca choose whether or ot to 1
8 2 CHAPTER 1 SUBSETS AND BINOMIAL COEFFICIENTS iclude + 1 i the subset Now we ca easily prove by iductio that f () = 2 The iductio starts because f (0) = 1 = 2 0 For the iductive step, assume that f () = 2 ; the f ( + 1) = 2 f () = 2 2 = 2 +1 So the iductio goes through, ad the proof is complete 12 Subsets of fixed size If ad are itegers satisfyig 0, how may elemet subsets does a elemet set X have? ( ) Defie the biomial coefficiet by ( ) = ( 1) ( + 1) ( 1) 1 (There are factors i both the umerator ad the deomiator, the ith factors beig i + 1 ad i + 1) For 0, the umber of elemet subsets of a elemet set is ( ) Proof We choose distict elemets of the elemet set X There are choices for the first elemet; 1 choices for the secod; i+1 choices for the ith; ad + 1 choices for the th Multiply these umbers ( together ) to get that the total umber of choices is the umerator of the fractio This is ot the aswer, sice choosig the same elemets i a differet order would give the same subset for example, 1, the 4, the 3 would be the same as 3, the 1, the 4 So we have to divide by the umber of differet orders i which we could choose the elemets There are choices for the first; 1 for the secod; i + 1 for the ith; ad + 1 = 1 choice (really o choice at all!) for the last Multiplyig these umbers gives the deomiator of the fractio So the result is proved ( ) It will sometimes be coveiet to give a meaig to the symbol eve if is greater tha We specify: ( ) If >, the = 0
9 13 PROPERTIES OF BINOMIAL COEFFICIENTS 3 This is a reasoable choice sice, if >, there are( o) elemet subsets of a elemet set You should chec that our formula for remais correct i this case: if >, the oe of the factors i the umerator is equal to 0 13 Properties of biomial coefficiets 131 Sum of biomial coefficiets The total umber of subsets of a elemet set is 2 We ow the umber of subsets of size, for each value of : addig these up must give the total I other words, =0 ( ) = Biomial coefficiets ad factorials Here is a alterative formula for the biomial coefficiets This uses the factorial fuctio, defied by! = ( 1)( 2) 1, the product of all the itegers from 1 to iclusive Now we have ( )! =!( )! For if we tae the defiitio of the biomial coefficiet, ad multiply top ad bottom by ( )!, the i the umerator we have the product of all the itegers from 1 to, that is,!; the deomiator is!( )! I order to mae this formula valid i the limitig cases = 0 ad =, we have to adopt the covetio that 0! = 1 This may seem strage, but if we wat the recurrece! = ( ( ) 1)! ( ) to hold for = 1, the it ( is) forced upo us! This 0 the correctly gives = = 1, ad i particular = However, the formula does ot wor if >, sice the < 0 ad we caot defie factorials of egative umbers 133 A recurrece relatio There is a simple recurrece relatio for the biomial coefficiets, which eables big oes to be calculated from smaller oes by additio: ( ) ( ) ( ) = 1
10 4 CHAPTER 1 SUBSETS AND BINOMIAL COEFFICIENTS First proof Cosider the problem of coutig the elemet subsets of a  elemet set X, which cotais oe special elemet called x First we cout the sets which cotai x ( Each ) of these must have 1 out of 1 the remaiig 1 elemets So there are such sets 1 Next we cout the sets which do ot cotai x Each of these must ecessarily have ( ) elemets chose from the 1 elemets differet from x; so there are 1 such sets Addig these umbers together gives all the Secod proof ( ) sets We ca prove the result by calculatio, usig our formula: ( ) ( ) = = = = ( 1)! ( 1)!( )! ( 1)!( )! ( 1)!!( )! ( ), + ( 1)!!( 1)! + ( 1)! ( )!( )! where we have used the facts that! = ( 1)!,! = ( 1)!, ad ( )! = ( ) ( 1)! I mae o secret of the fact that I lie the first proof better! 134 Symmetry We have ( ) ( ) = For the first proof, we fid a bijective correspodece betwee the elemet sets ad the ( )elemet sets i a set of size ; this is easily doe by simply matchig each set with its complemet The secod proof, usig the formula i 2 above, is a simple exercise for the reader
11 13 PROPERTIES OF BINOMIAL COEFFICIENTS Pascal s Triagle It is possible to arrage the biomial coefficiets i a symmetrical ( ) triagular ( ) patter, i which the ( + 1)st row cotais the + 1 umbers,, 0 The triagle begis as follows: Although we call this Pascal s Triagle, Pascal was ot the first perso to write it dow Below is a versio due to ChuShiChieh (Zhu Shijie), tae from wor of Yag Hui, i his boo Ssu Yua Yü Chie, dated 1303 Jia Xia ew it about 250 years earlier Other people who ew about it at roughly the same time iclude Halayudha i Idia, ad AlKaraji ad Omar Khayyam i Ira We do t ow who iveted it! The property i 134 above shows that the triagle has leftright symmetry The recurrece relatio 133 shows that each etry of the triagle is the sum of the two etries immediately above it This gives a very quic method to geerate as much of the triagle as required
12 6 CHAPTER 1 SUBSETS AND BINOMIAL COEFFICIENTS 14 The Biomial Theorem We ow come to the Biomial Theorem, a geeralisatio of the property 1 of the precedig paragraph (put x = y = 1 to see this) Theorem 12 (Biomial Theorem) (x + y) = =0 ( ) x y First proof We have (x + y) = (x + y)(x + y) (x + y), where there are factors o the righthad side of the equatio If all the bracets are expaded, we get a sum of very may terms; but each term is obtaied by choosig x from some of the bracets ad y from the remaiig oes If we choose x from bracets ad y from the remaiig, we obtai a term x y So the coefficiet of this term is the umber of ways we ca do this, i other words, the umber ( of) choices of out of the bracets from which x is selected This umber is So the theorem is proved Secod proof We prove the theorem by iductio o For = 0, the lefthad side ( is )(x+y) 0 = 1, while the righthad side has just the sigle term = 0, which 0 is x 0 y 0 = 1 So the iductio starts 0 Suppose that the Biomial Theorem holds for a value The (x + y) +1 = (x + y)(x + y) = x ( =0x y ) + y ( =0x y ) ( ) For = m, secod term gives us a cotributio x m y +1 m What is the cotributio to of the first term to the coefficiet ( of x) m y +1 m? To get this term, we m must put = m 1, ad the coefficiet is m 1 So the coefficiet of x m y +1 m i (x + y) +1 is ( ) ( ) ( ) =, m 1 m m
13 15 FURTHER PROPERTIES OF BINOMIAL COEFFICIENTS 7 which is just what we require to mae the iductio wor So the proof is complete Sometimes it is coveiet to have a oevariable form of the Biomial Theorem Puttig y = 1, we obtai ( ) (1 + x) = x =0 15 Further properties of biomial coefficiets 151 Eve ad odd ( ) We ow that, for fixed, the sum of the biomial coefficiets over all values of from 0 to is 2 What if we add them up just for eve, or just for odd? For > 0, /2 i=0 ( ) = 2i ( 1)/2 ( ) = 2 1 i=0 2i + 1 Proof Let S e ad S o be the sums of the eve ad odd biomial coefficiets respectively The S e +S o is the sum of all the biomial coefficiets; i other words, S e + S o = 2 If we put x = 1 i the oevariable Biomial Theorem, we obtai =0( 1) ( ) = ( 1 + 1) = 0 Now i this sum, the eve biomial coefficiets have coefficiet +1 ad the odd oes have coefficiet 1; so the equatio says that S e S o = 0 The two displayed equatios show that S e = S o = 2 /2 = Biomial idetities There are a huge umber of other equatios coectig biomial coefficiets Here is oe Let m,, be positive itegers The i=0 ( i )( m i ) ( m + = (This result is sometimes called the Vadermode covolutio) )
14 8 CHAPTER 1 SUBSETS AND BINOMIAL COEFFICIENTS First proof Suppose a school class cosists of m girls ad boys, ad we eed to choose a team of childre I how may ways ( ca ) this be doe? We cout m the umber of teams cotaiig i girls: there are ways to choose the girls, ( ) i ad ways of choosig the remaiig i team members from the boys i Multiplyig these umbers gives us the umber of possible teams cotaiig i girls, ad ( summig ) over i gives the total umber of teams But we ow that the m + total is Secod proof Cosider the equatio (1 + x) m (1 + x) = (1 + x) m+ ( ) m + What is the term i x? O the right, it is, by the Biomial Theorem O the left, we could choose the term x i from the first factor ad( x i ) from( the secod ) m ad multiply them The coefficiets of these two terms are ad ; so i i we multiply these umbers, ad the sum over i ( ) Puttig m = =, ad otig that = i i=0 153 Sum of sizes of sets ( ) 2 = i ( i ( ) 2 Here are a some further results ad proof techiques ( ) ( ) 1 First result: = 1 ), the equatio reduces to First proof From a class of childre, ( we ) have to choose a team of members, ad a captai for the team There are teams, ad choices of a captai for ( ) ay team; altogether choices But we could proceed differetly: we could choose the captai first (i ways), ad ( the) the remaiig ( 1 team) members 1 1 from the remaiig 1 childre (i ways), givig i all 1 1
15 15 FURTHER PROPERTIES OF BINOMIAL COEFFICIENTS 9 Secod proof ( )! =!( )! ( 1)! = ( 1)!( )! ( ) 1 = 1 ( ) Secod result: = 2 1 =1 First proof ( ) =1 = = =1 1 l=0 = 2 1 ( ) 1 1 ( ) 1 l Secod proof We have Differetiatig gives (1 + x) = =0 ( ) x (1 + x) 1 ( ) = x 1 =1 (We omit the = 0 term sice it is zero) Now put x = 1 to get the result ( ) Third proof There are subsets of size of a elemet set X; so the sum o the left simply adds up the sizes of all these subsets But we ca calculate this sum aother way Pair up each subset A with its complemet X \ A; these two sets cotai elemets betwee them There are 2 subsets, ad so they fall ito 2 /2 = 2 1 pairs Thus the value of the sum is Cogrueces Here is a picture of part of Pascal s triagle I have put to mea that the etry is odd, ad left a bla if the etry is eve Notice the fractal structure of the
16 10 CHAPTER 1 SUBSETS AND BINOMIAL COEFFICIENTS diagram: If we ow the triagle formed from the first 2 rows, we obtai the first 2 +1 rows by puttig two copies of the triagle side by side below the first oe, ad leavig the positios i the middle triagle bla This is explaied by a result called Lucas Theorem: Theorem 13 (Lucas Theorem) Let p be a prime umber Write ad to the base p: = a 0 + a 1 p + a 2 p a d p d, = b 0 + b 1 p + b 2 p b d p d, where 0 a i,b i p 1 The ( ) I particular, d i=0 ( ai b i ) (mod p) ( ) is divisible by p if ad oly if a i < b i for some value of i The proof of the theorem is i the ext sectio You should try to explai how this justifies the fractal shape of the diagram showig the parities i Pascal s triagle 16 Appedix: Proof of Lucas Theorem Recall the statemet of Lucas Theorem: Theorem (Lucas Theorem) base p: Let p be a prime umber Write ad to the = a 0 + a 1 p + a 2 p a d p d, = b 0 + b 1 p + b 2 p b d p d, where 0 a i,b i p 1 The ( ) d i=0 ( ai The proof comes from the followig lemma: b i ) (mod p)
17 16 APPENDIX: PROOF OF LUCAS THEOREM 11 Lemma Let p be prime, ad let = cp + a, = d p + b, with 0 a,b p 1 The ( ) ( )( ) c a (mod p) d b Proof Here is a short proof usig the Biomial Theorem The ey is the fact that, if p is prime, the (1 + x) p 1 + x p (mod p) ( ) p For each biomial coefficiet, for 1 i p 1, is a multiple of p, so all i ( ) p itermediate terms i the Biomial Theorem vaish mod p (We have = i p!/i!(p i)!, ad p divides the umerator but ot the deomiator) Thus (cogruece mod p): (1 + x) = (1 + x) cp (1 + x) a = (1 + x p c (1 + x) a ( ) c x pi i c i=0 a j=0 ( a j ) x j Sice 0 a,b < p, the oly way to obtai a term i t = t d p+b i this expressio is to tae the term i = d i the first sum ad the term j = b i the secod; this gives as required ( ) ( c d )( ) a b (mod p), Proof of the theorem The proof is by iductio o d The iductio starts with d = 1 sice, the = a 0, = b 0, ad there is othig to prove Suppose that the theorem holds with d 1 replacig d As i the statemet of the theorem, let = a 0 + a 1 p + a 2 p a d p d, = b 0 + b 1 p + b 2 p b d p d, where 0 a i,b i p 1 Put a = a 0, c = a 1 +a 2 p+ +a d p d 1, b = b 0, d = b 1 + b 2 p + + b d p d 1 The = cp + a, = d p + b, ad we have (with cogrueces mod p): ( ) ( )( ) c a (by the Lemma) d b
18 12 CHAPTER 1 SUBSETS AND BINOMIAL COEFFICIENTS Corollary = ( d ( ai i=1 b i d ( ) ai i=0 b i ) ) ( a0 b 0 With the hypotheses of the theorem, if a i < b i for some i with 0 i d Proof If a i < b i, the ( ai b i ) (by the iductio hypothesis) ( ) is divisible by p if ad oly ) = 0 So oe of the factors o the righthad side of the theorem is zero, whece the product is zero ( ) ai If a i b i, the the biomial coefficiet is ot divisible by p (it is ozero ad there are o factors p i the umerator sice a i p 1 Now a product of d umbers ot divisible by p is itself ot divisible by p Example Let = 2 m 1 The all the digits a i of i base 2 are equal to 1, so we have b i a i for ay This meas that every etry i row of Pascal s triagle is odd Exercises 1 Write 1001 as a biomial coefficiet b i ( ) with 20 2 If X is a set of 8 elemets, the the umber of 3elemet subsets of X is twice the umber of 2elemet subsets Is there ay other size of the set X for which this holds? 3 Calculate =0 2( ) 4 Let X be a elemet set Fid a bijectio F betwee the set of elemet subsets of X ad the set of all ( )elemet subsets of X Deduce that ( ) ( ) = 5 This exercise exteds the result about summatio of eve ad odd biomial coefficiets i 151 Similar methods ca deal with sums of biomial coefficiets where lies i ay fixed cogruece class of positive itegers
19 16 APPENDIX: PROOF OF LUCAS THEOREM 13 Let i deote the square root of 1, ad ote that 1+i = 2e π/4 Hece fid the real ad imagiary parts of (1 + i) for ay atural umber (You will probably fid it coveiet to cosider the differet cogruece classes mod 8 separately) Expadig (1 + i) by the Biomial Theorem, fid expressios for ( t)/4 ( ), j=0 4 j +t for t = 0, 1, 2, 3, agai separatig the cogruece classes mod 8 (This ivolves a lot of repetitious wor You should at least do all the calculatios for 0 (mod 8)) 6 By calculatig the coefficiet of x o the two sides of the idetity or otherwise, prove that 7 (1 + x) (1 x) = (1 x 2 ), =0( 1) ( ) { 2 0 if is odd, = ( ) 2m ( 1) m if = 2m m (a) Prove that ( ) = ( ) ( ) ( ) (b) Prove that, if > 2+1, the > ; if = 2+1, the ( ) ( + 1 ) ( ) ; ad if < 2 + 1, the < + 1 ( ) = + 1 (c) Hece show that, for fixed ad = 0,1,,, the biomial coefficiets icrease, the remai costat for oe step (if is odd), the decrease (Such a sequece is said to be uimodal) ( ) 2m (d) Show further that the largest biomial coefficiet is ( ) 2m + 1 while if = 2m+1 is odd, the ad m (e) Deduce that, if = 2m, the 2 2m 2m + 1 ( ) 2m 2 2m m ( 2m + 1 m + 1 if = 2m is eve, m ) are equal largest
20 14 CHAPTER 1 SUBSETS AND BINOMIAL COEFFICIENTS 8 (a) Show that the biomial coefficiet ( ) 2m is divisible by every prime p sat m isfyig m + 1 p 2m (b) Usig the estimate o Problem Sheet 1, Questio 2, show that the umber 2m of primes betwee m + 1 ad 2m is at most log 2 m Remar: This is a wea versio of the famous Prime Number Theorem, which says that the umber of prime umbers p satisfyig 1 p is asymptotically log
21 Chapter 2 Selectios ad arragemets 21 The formulae We have a hat cotaiig ames, ad we are goig to draw out ames I how may ways ca we do this? To aswer the questio, we have to clarify the strategy a bit First, do we care about the order i which the ames are draw, or ot? Secod, whe we have draw a ame, do we put it bac i the hat ad shae it up before the ext draw, or do we discard it? The aswers to this questio correspod to samplyig with order sigificat or ot, ad with repetitio allowed or ot allowed If the order is sigificat, we have a tuple of ames; if ot, we have a set (if repetitio is ot allowed), or what might be called a multiset if repetitio is allowed We will write multisets i square bracets to distiguish them from sets For example, if the ames are a,b, ad we draw two of them, the order importat, repetitio allowed: there are four possibilities, (a, a), (a, b), (b,a) ad (b,b) order importat, repetitio ot allowed: there are two possibilities, (a, b) ad (b,a) order uimportat, repetitio allowed: there are three possibilities, [a, a], [a,b], ad [b,b] (Choosig a the b is the same as choosig b the a) order uimportat, repetitio ot allowed: just oe possibility, amely {a, b} I geeral, the umbers of selectios are give by the etries i the followig table We use the otatio () ( for ) the umber ( 1) ( + 1) This is the umerator i our defiitio of, ad is ofte called the fallig factorial 15
22 16 CHAPTER 2 SELECTIONS AND ARRANGEMENTS Repetitio allowed Repetitio ot allowed Order Order ot sigificat sigificat ( ) + 1 ( ) () Note that the umerator i the top right etry is ( + 1) ( + 1), the socalled risig factorial 22 Proofs Order sigificat, repetitio allowed: We get to mae choices, ad there are ames to choose at each step So there are possibilities Order sigificat, repetitio ot allowed: This time, there are ames to choose at the first step; 1 at the secod step (sice we discarded the first ame after we chose it); 2 at the third step; ad + 1 at the th step Multiplyig these umbers gives the aswer Order ot sigificat, repetitio ot allowed: We simply choose a set with ( ) elemets from the elemets i the hat The umber of ways of doig this is, by defiitio Alteratively choose with order sigificat, ad repetitio allowed, ad ote that each uordered sample has! differet orderigs; so the aswer is () /! Order ot sigificat, repetitio allowed: This case is the most difficult But ote, before we begi, that we caot just use the argumet i the precedig paragraph to get /! [WHY NOT?] Step 1: The umber of choices of objects from, with order ot sigificat ad repetitio allowed, is equal to the umber of ways of choosig oegative itegers x 1,,x satisfyig x x = For give the selectio, we ca let x i be the umber of times that the ith ame was selected; clearly x 1,,x satisfy the stated coditios Coversely, give x 1,,x satisfyig the coditios, form a selectio i which the ith ame is chose x i times Thus, for example, suppose that = 3 ad = 6 If the ames are a,b,c, the the selectio [a,a,b,b,b,c] correspods to x 1 = 2, x 2 = 3, x 3 = 1
23 23 BALLS IN URNS 17 Step 2: So we have to cout the umber of choices of oegative itegers x 1,,x with sum To do this, tae a row( of + ) 1 cells; ( choose ) 1 of them ad put marers i them There are = ways of 1 maig this choice Havig made the choice, defie x 1,,x as follows: Let x 1 be the umber of cells before the first mared cell Let x 2 be the umber of cells betwee the first ad secod mared cell Let x 1 be the umber of cells betwee the 1st ad th mared cell Let x be the umber of cells after the th mared cell The clearly the umbers x 1,,x are oegative itegers; they add up to the umber of umared cells, which is ( + 1) ( 1) = Moreover, every way of choosig oegative itegers addig up to is represeted uiquely by such a marig of 1 out of + 1 boxes So the result is proved For example, our choice x 1 = 2, x 2 = 3, x 3 = 1 would come from a marig of the followig 1 = 2 out of + 1 = 8 boxes: To mae this clearer, here is a table which gives both steps i the case = 3, = 2 Let the ames i the hat be a,b,c The first colum gives a selectio of two ames (with repetitio allowed ad order uimportat) The secod gives three umbers addig up to 2 The third gives four boxes with a choice of two of them mared aa (2, 0, 0) ab (1, 1, 0) ac (1, 0, 1) bb (0, 2, 0) bc (0, 1, 1) cc (0, 0, 2) 23 Balls i urs There is aother way to loo at the mai result of the last sectio Suppose that we have urs, or vases, U 1,,U We have idistiguishable balls How may ways ca we put the balls i the urs? [Of course this problem ca be put ito
24 18 CHAPTER 2 SELECTIONS AND ARRANGEMENTS may disguises I have idetical sweets I how may ways ca I distribute them to a class of childre?] If x i is the umber of balls I put ito the ith ur [or the umber of sweets I give to the ith child], the x 1,,x are oegative itegers ( which add ) up to So + 1 the umber of ways of puttig the balls ito the urs is The coditios ca be varied i may ways Suppose, for example, that I have to distribute balls amog urs as above, but with the requiremet that o ur should be empty This ass that x i 1 for all i If we defie ew variables y 1,,y by y i = x i 1, the the sum of the ys is ; so the umber of choices of the y s is ( ) ( ) + ( ) 1 1 = 1 The simple way to thi about this is: Suppose each ur is to be oempty The I first tae balls ad put oe i each ur The I distribute the remaiig balls ito the urs i ay way This gives the same result as above Example How may ways ca I distribute 100 sweets to a class of 30 boys ad 20 girls, if it is required that each boy has at least oe sweet ad each girl has at least two sweets? To solve this, I first give oe sweet to each boy ad two to each girl, usig up = 70 sweets The I( distribute the) remaiig ( ) 30 sweets amog the childre, which ca be doe i = ways Maig words from letters How may ways ca we arrage distict objects i order? By the formula i the bottom left of the box, the aswer is simply () =! Aother way of seeig this is as follows Let F() be this umber The F(0) = 1, F() = F( 1) for > 0 (We tae F(0) = 1 because there is just oe list with o etries, the empty or bla list To get the secod equatio, we choose oe of the objects to be first o the list (there are ways of doig this), ad the we have to put the remaiig 1 i order after the first oe) Now a easy iductio argumet shows that F() =! for all Now we mae the questio a bit harder How may ways of arragig some (possibly all) of the objects i a list?
25 24 MAKING WORDS FROM LETTERS 19 Example How may words ca I mae from the letters of the word FACE TIOUS? (A word is simply a strig of letters chose from those i the word; we do ot require that it maes sese i Eglish or ay other laguage The order of the origial letters i the word is irrelevat; a better aalogy is that you are playig Scrabble ad you have these letters By covetio we iclude the empty word, which is the strig cotaiig o letters) I the case give, the letters are all distict; this maes life easier, so we start with this case Suppose that we are give letters, all differet How may  letter words ca we costruct? These words are just selectios of letters from the give, with order importat ad repetitio ot allowed; so the umber is () = ( 1) ( + 1) So the total umber W() of words is W() = =0 () We ca express this aother way Note that () =!/( )! So W() =! ( =0 Now recall from calculus that ) ( 1 =! ( )! m=0 m=0 1 m! = e ) 1 m! Iside the bracets of the formula for W(), we see the sum of the reciprocals of the factorials from 0 to, i other words, the sum of the first + 1 terms of the ifiite series So we see that W() is approximately e! We ca be more precise: e! W() = = <! m=+1 m! ( + 1)( + 2) + 1 ( + 1)( + 2)( + 3) ( + 1) ( + 1) 3 + = 1 (I the last term we summed a geometric series)
26 20 CHAPTER 2 SELECTIONS AND ARRANGEMENTS I other words, e! is bigger tha the iteger W() but smaller tha W() + 1/; so we get W() by calculatig e! ad roudig dow to the iteger below So fially we coclude, usig the floor or iteger part fuctio, that W() = e! For the word FACETIOUS, we have = 9, ad W() = e 9! = Remar Although W() = e! is a beautiful simple formula, ad gives us a very good estimate for the size of W(), it is ot so good for the purpose of calculatio For example, 70! is a umber with about 100 digits, so i order to decide whether W(70) is odd or eve we would eed to ow e to 100 places of decimals (at least) For exact calculatio it is better to use the formula W() = () = ( 1) + ( 1)( 2) + =0 We ca also fid W() by a recurrece method We have W(0) = 1, W() = 1 + W( 1) for > 0 (The coditio W(0) = 1 is because of the empty word I geeral, to form a word of from letters, we choose oe letter to go first (i ways), ad mae a word from the remaiig 1 letters (i W( 1) ways) to follow it; but we have missed out oe word, amely the empty word, so we eed to add 1) A easy iductio ow gives the formula for W() If the letters we are give cotai repetitios, it is more difficult to write dow a formula Here, we will simply do a example Example How may words ca be made from the letters of SYZYGY? For the case whe we use all the letters, the aswer is ot too hard There are 6! ways of arragig the six letters, but ay rearragemet of the three Ys will give the same word So the umber of arragemets is 6!/3! = 120 If we allow words of arbitrary legth, it is a bit more difficult To solve it, we subdivide the words accordig to the umber of occurreces of the letter Y At most oe Y We have to mae words out of the four letters S, Z, G ad Y This ca be doe i W(4) = = 65 ways
27 24 MAKING WORDS FROM LETTERS 21 Two Ys Temporarily label the Ys as Y 1 ad Y 2 so we ca distiguish them Now we have five letters S, Z, G, Y 1 ad Y 2, but we must use the two Ys Choose some of the other three letters, order all letters icludig the two Ys i ay way, ad add up all possibilities; fially divide by 2 sice the Ys are really idistiguishable We get (( ) ( ) ( ) ( ) / 2! + 3! + 4! + 5!) 2 = All three Ys Similarly the total for this case is (( ) ( ) ( ) ( ) / 3! + 4! + 5! + 6!) 3! = So the total is = 364 Exercises 1 (a) How may ordered sequeces of legth 5 ca be made usig the elemets {1,2,3,4,5,6,7} if repetitios are allowed? How may of these cotai exactly two of the umbers 1,2,3? I how may of them do eve ad odd umbers alterate? (b) What are the aswers to these questios if repetitios are ot allowed? 2 How may words ca be made usig the letters of the word STARTS? How may of these are palidromes (that is, read the same bacward as forward)? 3 Let X ad Y be sets with X = ad Y = m (a) Determie the umber of fuctios f mappig X ito Y (b) How may of these fuctios are ijectios, ie oetooe? (c) How may of these fuctios are bijectios, ie oetooe ad oto? (d) (much harder) How may of these fuctios are surjectios, ie oto? 4 How may permutatios of the set {1,2,,} are there? How may of these are cyclic permutatios, that is, their cycle decompositio cosists of a sigle cycle of legth? 5 For which values of is W() odd?
28 22 CHAPTER 2 SELECTIONS AND ARRANGEMENTS
29 Chapter 3 Power series A lot of combiatorics is about sequeces of umbers: We ll see such sequeces as (Fiboacci umbers), or (a 0,a 1,a 2,) (1,1,2,3,5,8,13,21,34,) (1,1,2,6,24,120,720,5040,) (factorials) A very useful device to represet such a sequece of umbers is to tae the umbers to be the coefficiet i a power series a x = a 0 + a 1 x + a 2 x 2 + a 3 x We call this power series the geeratig series or geeratig fuctio for the sequece of umbers I this chapter we loo at power series ad some of their uses i combiatorics Example We saw that the umber of subsets of a elemet set is 2 This gives us a sequece of umbers, amely whose geeratig fuctio is (2 0 = 1,2 1 = 2,2 2 = 4,2 3 = 8,) 2 x = x usig the formula for the sum of a geometric series 23
30 24 CHAPTER 3 POWER SERIES 31 Power series You ve met power series i calculus, ad maybe i aalysis also So how do they compare with combiatorics: First, the good ews We are ot doig calculus here, so we do t have to worry whether the sequeces coverge or ot For us, a power series is just a booeepig device, to wrap up ifiitely may terms ito a sigle mathematical object For example, if our sequece is the factorials above, the the power series is!x 0 ad if you remember the ratio test from calculus, you should be able to show that this series ever coverges uless x = 0 (The ratio of successive terms is ( + 1)!x +1 /!x = ( + 1)x, which teds to ifiity as ) But this power series might still be useful! Secod, the good ews If a power series does coverge, ad if you ow somethig about the properties of the fuctio A(x) it defies, the you ca use those properties i combiatorics also! We ll see some examples later I the example above, the sum of the series is 1/(1 2x); the series coverges if x < 1/2 We deote the set of all power series with iteger coefficiets by Z[[x]] This should remid you of the otatio Z[x] for the set of polyomials with iteger coefficiets; power series are very similar to polyomials, but ca have ifiitely may coefficiets Similarly, if we wat the coefficiets to be real umbers, we write R[[x]], with similar modificatios for the other umber systems 32 Operatios o power series There are various operatios that ca be doe to power series If you studied Algebra, you have met the idea of a rig; the first two operatios below (additio ad multiplicatio) mae R[[x]] ito a rig, for ay rig R (though we wo t stop to prove this) Additio We add two power series term by term: ( ) ( a x ) b x = (a + b )x 0
31 33 THE BINOMIAL THEOREM 25 Multiplicatio We multiply power series i the same way as we multiply polyomials To get a term i x i the product, we multiply the term i x i the first factor by the term i x i the secod, ad sum over all values of from 1 to Thus Substitutio ( ) ( ) a x b x 0 0 where c = = c x, 0 a b =0 Let A(x) = a x ad B(x) = b x Suppose that a 0 = The we ca substitute A(x) for x i the secod series: B(A(x)) = b (A(x)), 0 where A(x) is calculated usig the multiplicatio rule Why do we eed the costat term of A(x) to be zero? Cosider the costat term of the series B(A(x)) It would be b 0 +b 1 a 0 +b 2 a 2 0 +, ad we would have a ifiite series of umbers, ad would have to worry about covergece But if a 0 = 0, the the smallest power of x occurrig i A(x) is at least x ; so whe we come to calculate the coefficiet of x i B(A(x)), we oly have to cosider fiitely may terms b A(x) for 0 I other words, we oly eed fiitely may additios ad multiplicatios to wor out ay term Differetiatio the We ca also differetiate power series If A(x) = a x, 0 d dx A(x) = a x 1 = (m + 1)a m+1 x 1 m 0 Notice what has happeed here The term = 0 is zero, so we leave it out i the first step; the we use a ew summatio variable m = 1, so that as rus from 1 to ifiity, m rus from 0 to ifiity 33 The Biomial Theorem We saw the Biomial Theorem, a formula for (1+x) for positive itegers Here is a geeralisatio of it, first proved by Isaac Newto
32 26 CHAPTER 3 POWER SERIES We eed to geeralise the defiitio of biomial coefficiets first Let a be ay umber, positive or egative, ratioal or irratioal, real or complex Let be a atural umber (a positive iteger or zero) Defie ( a ) = a(a 1) (a + 1) ( 1) 1 This has the properties ( ) a if a is a atural umber, the = 0 for ; ( ) a otherwise, 0 for all a ( ) a For the oly way we ca have = 0 is for oe of the factors i the umerator to be zero, that is, a i = 0 (that is, a = i) for some i 1 Now we have: Theorem 31 (The Biomial Theorem) For ay complex umber a, ( ) a (1 + x) a = x 0 There are two ways to iterpret this theorem I terms of calculus: the series o the right coverges for x < 1, ad its sum is (1 + x) a Secod, i terms of combiatorics: The usual rules of expoets hold A calculus proof of the Biomial Theorem (without all the tricy details about covergece) is give i a appedix Example 1 The first law of expoets says that By the Biomial Theorem, ( ( ) a )x 0 (1 + x) a (1 + x) b = (1 + x) a+b ( 0 ( ) b )x = 0 Now by the rule for multiplicatio of power series, ( )( ) ( a b a + b = i i i=0 ( a + b ) ) x This is the Vadermode covolutio We saw it for atural umbers a ad b i Sectio 152; but ow we ow that it holds for ay a ad b at all
33 33 THE BINOMIAL THEOREM 27 Example 2 We get some iterestig examples by choosig expoets which are ot atural umbers so Case a = 1 We have ( ) 1 = ( 1)( 2) ( ) ( 1) 1 ( 1 (1 x) 1 = 0 ) = ( 1), ( x) = x, 0 so we have the formula for the sum of a geometric series We already used this i calculatig the geeratig fuctio for the powers of 2 Case a = 1/2 We have ( ) 1/2 = = = = ( 1/2)( 3/2) ( (2 1)/2) ( 1) 1 ( ) 1 (2 1)(2 3) 1 2 ( 1) 1 ( ) 1 1 2(2 1) 1 4!! ( ) 1 ( ) 2, 4 where we have used the fact that 2(2 2) 2 = 2! Thus ( 1/2 (1 4x) 1/2 = 0 ) ( 4x) = 0 ( 1 4 ) ( ) 2 ( 4x) = 0 So the geeratig fuctio for the cetral biomial coefficiets =0 ( ) 2 x ( ) 2 is 1/ 1 4x Exampe 2, cotiued We ca use what we just leared to prove the followig idetity for the cetral biomial coefficiets: ( )( ) 2 2( ) = 4
34 28 CHAPTER 3 POWER SERIES Proof We start from the idetity (1 4x) 1/2 (1 4x) 1/2 = (1 4x) 1 Now the coefficiet of x o the left is obtaied by taig the coefficiet of x i the first factor (1 4x) 1/2, multiplyig by the coefficiet of x i the secod factor, ad summig over from 0 to This gives precisely the lefthad side of the result we are provig O the right, (1 4x) 1 = 4 x, 0 so the coefficiet of x is 4, ad we are doe Example 3 Here is a simple example of the use of power series to solve a recurrece We will have more complicated examples later Suppose that a sequece of umbers a 0,a 1,a 2 satisfy a 0 = 1 ad a = 2a 1 for 1 Of course it is clear that these umbers are the powers of 2 But let us see this aother way The geeratig fuctio is A(x) = a x 0 = 1 + 2a 1 x 1 = 1 + 2a m x m+1 m 0 = 1 + 2xA(x) (Chec that you ca follow all these steps I the third step we have used a ew summatio variable m = 1) This equatio ca be rearraged to give A(x) = 1 1 2x = (2x) = 2 x 0 Now if two power series are equal the their coefficiets must be the same; so we have a = 2 for all 0 34 Other power series Apart from the Biomial Theorem, there are a couple of other famous power series which crop up from time to time:
35 34 OTHER POWER SERIES 29 The expoetial fuctio I calculus this is usually writte as e x I will usually write it as exp(x); this meas the same thig The power series is The most importat properties are x exp(x) = 0! d exp(x) = exp(x) This is easy to prove from the power series sice dx d dx x! = x 1 ( 1)! exp(x + y) = exp(x) exp(y) (We prove this below) The logarithm fuctio The fuctio log(x) is ot defied at x = 0 so we caot write it as a power series Istead, we have log(1 + x) = 1 If we differetiate term by term we get ( 1) 1 x d dx log(x) = ( x) 1 = (1 + x) 1 1 The logarithm is the iverse of the expoetial: exp(log(1 + x)) = 1 + x, log(exp(x)) = x (Remember that we ca substitute oe power series i aother if the first oe has costat term zero This is OK for the first result above I the secod case, it is really log(1 + y), where y = exp(x) 1, which does ideed have costat term zero) Example is Cosider the equatio exp(x + y) = exp(x) exp(y) The left had side (x + y) 0! = 0 1! =0 ( )( x = 0! l 0 = exp(x) exp(y)!!( )! x y ) y l l!
36 30 CHAPTER 3 POWER SERIES (I the secod lie we used a dummy variable l = We have to chec the rages of summatio: taig all values ad ruig from 0 to is the same as ad l idepedetly taig all oegative values) We could have reversed the procedure ad derived the Biomial Theorem from the property of the expoetial fuctio Actually there is a lot of very iterestig combiatorics hidde i the power series for the expoetial ad logarithm fuctios If you are iterested i this, see my Notes o Coutig o the Web 341 Appedix: Proof of the Biomial Theorem This proof is a bit of a cheat, sice all the hard wor is i the calculus Suppose we have a power series a x whose sum is a ow fuctio f (x) 0 How do we wor out the coefficiets a? If we differetiate the series times, we get d dx f (x) = a ( 1) ( + 1)x (We start the sum at = because the th derivative of ay smaller power of x is zero) The if we put x = 0, we fid [ d ] dx f (x) =!a, x=0 so that a = [(d /dx ) f (x)] x=0 /! Taig f (x) = (1 + x) a, whe we differetiate times we get Puttig x = 0, we get [ d d dx (1 + x)a = a(a 1) (a + 1)(1 + x) a (1 + x)a dx ] x=0 = a(a 1) (a + 1) So the coefficiet of x i the power series for (1 + x) a is ( ) a(a 1) (a + 1) a =,! ( ) a so that (1 + x) a = x 0
37 34 OTHER POWER SERIES 31 Exercises 1 The purpose of this exercise is to show you that, eve whe a power series fails to coverge, algebraic maipulatios o it ca still give us somethig iterestig (a) Let π be a permutatio of the set {1,,} We say that π is decomposable if there is a umber, with 1 1, such that π maps the umbers 1,, to themselves If o such exists the π is idecomposable There are! permutatios of the set {1,,} Suppose that g() of them are idecomposable (By covetio we tae 0! = 1 but we do ot defie g(0)) For ay permutatio π, let be the smallest umber such that π maps 1,, to themselves (so that = if π is idecomposable) Show that there are g()( )! permutatios with ay give value of Hece show that g()( )! =! =1 Now let F(x) =!x ad G(x) = g()x be the geeratig fuctios 0 1 for the factorial umbers ad the umbers g() respectively Note that G(x) has costat term zero sice we start at 1 Prove that F(x)(1 G(x)) = 1 Note that this equatio maes sese eve though the power series do ot coverge for ay ozero value of x
38 32 CHAPTER 3 POWER SERIES
39 Chapter 4 Recurrece relatios Recurrece relatios are a very powerful method of calculatig combiatorial umbers But there are ot may geeral methods for dealig with them, so mostly we will just loo at a few importat examples The mai idea is that we ca tur a recurrece relatio for a sequece of umbers ito a equatio (algebraic or differetial) for the geeratig fuctio 41 Fiboacci umbers Leoardo Fiboacci was a Italia mathematicia of the 13th cetury His most importat wor was the itroductio of the Arabic umerals 0,1,2,3,4,5,6,7,8,9 to Europe I order to show how much easier it is to calculate with these tha with the Roma umerals previously used, he posed the followig problem as a exercise i his boo Liber Abaci (The Boo of Calculatio): A pair of rabbits do ot breed i their first moth of life, but at the ed of the secod ad every subsequet moth they produce oe pair of offsprig If I acquire a ewbor pair of rabbits at the begiig of the year, how may pairs of rabbits will I have at the ed of the year? Uder these coditios, the umber of pairs of rabbits after moths is called the th Fiboacci umber F How do we calculate these umbers? First, we have F 0 = 1, F 1 = 1 For we are give that we have oe pair of rabbits at the start of moth 0, ad they do ot produce offsprig i moth 1 Next, F = F 1 + F 2 for 2 33
40 34 CHAPTER 4 RECURRENCE RELATIONS To show this, let G be the umber of pairs of rabbits which are old eough to breed at the ed of moth Now by the coditios of the problem, we have G = F 2 (sice the rabbits breedig i moth are all those bor i moth 2 or earlier) Also, F F 1 = G, sice G pairs are bor i moth ad are those cotributig to F but ot to F 1 Elimiatig G from these two equatios gives the result So the aswer to Fiboacci s exercise ca be foud by a doze additios, a simple job usig Arabic umerals Fiboacci did ot ivet these umbers, which had bee ow to Idia mathematicias icludig Pigala, Virahaa ad Hemachadra for early 1500 years whe he wrote his boo The coditio F = F 1 + F 2 is a example of a recurrece relatio This is a relatio which eables ay term of the sequece to be calculated if the earlier terms are ow I this case we oly eed to ow the two precedig terms Usually, a recurrece relatio eeds to be supplemeted with iitial coditios, tellig us how the sequece starts I this case the recurrece relatio oly applies for 2, so we eed to be give the values of F 0 ad F 1 separately I the ext sectio, we will solve this recurrece relatio to fid a explicit formula for the th Fiboacci umber First, though, we give a couple more coutig problems for which the Fiboacci umbers are the solutio Example I have a staircase with steps At a sigle stride, I ca go up either oe or two of the steps I how may differet ways ca I wal up the staircase? Let a be this umber The a 0 = 1 (sice if there are o steps, the there is oly oe way to do othig!) ad a 1 = 1 (obviously) We claim that a = a 1 +a 2 for 2 For let S be the set of all ways of walig up the steps The last step we use before we reach the top is either umber 1 or umber 2 (sice we asced either oe or two steps i the last stride); so let S 1 be the set of ways i which the peultimate step is umber 1, ad S 2 those i which it is umber 2 The S 1 ad S 2 are disjoit ad have uio S Moreover, clearly we have S 1 = a 1 while S 2 = a 2, ad S = a So the recurrece relatio holds Now a straightforward iductio shows that a = F for all atural umbers This represetatio of the Fiboacci umbers was discussed by Virahaa i the 6th cetury, i coectio with Sasrit poetry A vowel i Sasrit ca be log or short If we assume that a log vowel is twice as log as a short vowel, i how may ways ca we mae a lie of poetry of legth out of log ad short vowels? Clearly this is the same problem, ad the aswer is the th Fiboacci umber F
41 41 FIBONACCI NUMBERS 35 From this theorem we get a curious formula for F : F = /2 =0 ( For aother way of statig our result is that the umber of ways of writig as a ordered sum of oes ad twos is F Now we ca cout these expressios aother way Suppose that we have twos i the sum The we must have 2 oes, so there are terms altogether (ad we see that /2 ) So the umber of expressios with twos is the umber of selectios of elemets from ( ) (the positios i the sequece where the 2s occur), of which there are Summig over gives the result For example, whe = 4, we have ( ) 4 = 1 (correspodig to ); 0 ( ) 3 = 3 (correspodig to , ad ); 1 ( ) 2 = 1 (correspodig to 2 + 2) 2 Summig, we have F 4 = 5 Example How may sequeces of legth are there cosistsig of zeros ad oes with o two cosecutive oes? (Call such a sequece admissible) Let b be this umber Clearly b 0 = 1 (oly the empty sequece), ad b 1 = 2 (the sequeces 0 ad 1 are both admissible) Partitio the set T of all admissible sequeces ito two subsets T 0 ad T 1, where T 0 is the set of sequeces edig i 0, ad T 1 is the set of sequeces edig i 1 Now give ay admissible sequece of legth 1, we ca add a zero to it to get a admissible sequece of legth ; so T 0 = b 1 But we may oly add a 1 to a admissible sequece if it eds i zero; so T 1 is the umber of admissible sequeces of legth 1 edig i zero, which by the precedig argumet is b 2 Thus, b = b 1 + b 2 We have the same recurrece relatio as for the Fiboacci umbers, but differet iitial coditios However, we have b = F +1 for all The proof is by iductio We have b 0 = 1 = F 1, b 1 = 2 = F 2, ad for 2, ) b = b 1 + b 2 = F + F 1 = F +1
42 36 CHAPTER 4 RECURRENCE RELATIONS 42 Liear recurreces with costat coefficiets I this sectio we will fid a formula for the th Fiboacci umber The two methods we use ca be exteded to a wider class of recurrece relatios Method 1 We are tryig to solve the recurrece relatio with iitial coditios F 0 = 1, F 1 = 1, F = F 1 + F 2 for 2 We begi by observig that there is a uique solutio For F 0 ad F 1 are give, ad the the recurrece determies F 2,F 3, (This is really a argumet by iductio!) So, if we ca fid by ay method at all a solutio, the we ow it is the uique solutio We will cosider just the recurrece relatio a = a 1 + a 2, ad worry about the iitial coditios later The ext observatio that we mae is that the recurrece relatio is liear That meas that, if two sequeces (a ) ad (b ) satisfy it, the so does ay liear combiatio (c ) with c = pa + qb for ay umbers p ad q So we cocetrate o fidig specific solutios We try a solutio of the form a = α for some umber α [Why? Oe aswer is that it wors, as we will see A better aswer is that, if you cosider a oeterm recurrece relatio lie a = αa 1, it is obvious that there will be a solutio a = α ] Now a = α will satisfy the recurrece relatio if ad oly if α = α 1 + α 2 for 2 This will be the case if ad oly if α 2 = α + 1 The quadratic equatio x 2 = x + 1 has two solutios α = , β = So a = α ad a = β both satisfy our recurrece relatio, ad by the liearity priciple, so does a = pα + qβ for ay p ad q Fially, we try to choose p ad q such that this solutio also satisfies the iitial coditios a 0 = a 1 = 1 This gives us two equatios p + q = 1, pα + qβ = 1
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