Permutations and Combinations

Transcription

1 Massachusetts Istitute of Techology 6.042J/18.062J, Fall 02: Mathematics fo Compute Sciece Pof. Albet Meye ad D. Radhika Nagpal Couse Notes 9 Pemutatios ad Combiatios I Notes 8, we saw a vaiety of techiques fo coutig elemets i a fiite set: the Sum Rule, Iclusio-Exclusio, the Poduct Rule, tee diagams, ad pemutatios. We will ow itoduce yet aothe ule, the Divisio Rule, ad oe moe cocept, combiatios. We will also lea techiques fo coutig elemets of a fiite set whe limited epetitio is allowed. 1 The Divisio Rule The divisio ule is a commo way to igoe uimpotat diffeeces whe you ae coutig thigs. You ca cout distict objects, ad the use the divisio ule to mege the oes that ae ot sigificatly diffeet. We will state the Divisio Rule twice, oce ifomally ad the agai with moe pecise otatio. Theoem 1.1 (Divisio Rule. If B is a fiite set ad f : A B maps pecisely k items of A to evey item of B, the A has k times as may items as B. Fo example, suppose A is a set of studets, B is a set of tutoials, ad f defies the assigmet of studets to tutoials. If 12 studets ae assiged to evey tutoial, the the Divisio Rule says that thee ae 12 times as may studets as tutoials. The followig two defiitios pemit a moe pecise statemet of the Divisio Rule. Defiitio 1.2. If f : A B is a fuctio, the f 1 (b {a A f (a b}. That is, f 1 (b is the set of items i A that ae mapped to the item b B. I the pecedig example, f 1 (b is the set of studets assiged to tutoial b. This otatio ca be cofusig, sice f 1 omally deotes the ivese of the fuctio f. With ou defiitio, howeve, f 1 (b ca be a set, ot just a sigle value. Fo example, if f assigs o items i A to some elemet b B, the f 1 (b is the empty set. I the special case whee f is a bijectio, f 1 (b is always a sigle value, ad so f 1 by ou defiitio is just the odiay ivese of f. Defiitio 1.3. A fuctio f : A B is k-to-1 if fo all b B, f 1 (b k. Fo example, if f assigs exactly 12 studets to each ecitatio, the f is 12-to-1. Assumig k is o-zeo, a k-to-1 fuctio is always a sujectio; evey elemet of the age is mapped to by k > 0 elemets of the domai. We ca ow estate the Divisio Rule moe pecisely: Theoem (Divisio Rule, estatemet. If B is a fiite set ad f : A B is k-to-1, the A k B. Copyight 2002, Pof. Albet R. Meye.

2 2 Couse Notes 9: Pemutatios ad Combiatios Poof. Sice B is fiite, we ca let B ad let B {b 1, b 2,..., b }. The we have: A f 1 (b 1 f 1 (b 2 f 1 (b Equality holds because each side is cotaied i the othe. The ight side is cotaied i the left side because evey set f 1 (b i is cotaied i A by the defiitio of f 1 (b i. The left side is cotaied i the ight, sice evey item i A maps to some b i, ad theefoe is cotaied i the set f 1 (b i. Futhemoe, all of the sets f 1 (b i ae disjoit. The poof is by cotadictio. Assume fo the pupose of cotadictio that a f 1 (b i ad a f 1 (b j fo some i j. The fist iclusio implies f (a b i, but the secod iclusio implies that f (a b j. This is a cotadictio sice f (a deotes a uique item. Sice the sets f 1 (b i ae disjoit, we ca compute the cadiality of thei uio with the Sum Rule. Note that each set f 1 (b i has size k, sice f is k-to-1. A f 1 (b 1 + f 1 (b f 1 (b } k + k + {{ + k } B tems k B Example: Seatig at a Roud Table. I how may ways ca Kig Athu seat kights at his oud table? Two seatigs ae cosideed equivalet if oe ca be obtaied fom the othe by otatio. Fo example, if Athu has oly fou kights, the thee ae six possibilities as show i Figue 1. Heeafte, we deote a seatig aagemet i text by listig kights i squae backets i clockwise ode, statig at a abitay poit. Fo example, [1234] ad [4123] ae equivalet. As a sig that we should big i the divisio ule, the questio poits out that cetai distict seatigs ae equivalet, so should oly be couted oce that is, we wat to cout equivalece classes istead of idividual objects. The divisio ule is a good way to do this. Claim 1.4. Athu ca seat kights at his oud table i ( 1! ways. I paticula, the claim says that fo 4 kights thee ae (4 1! 3! 6 odeigs, which is cosistet with the example i Figue 1. Poof. The poof uses the Divisio Rule. Let A be the set of odeigs of kights i a lie. Let B be the set of odeigs of kights i a ig. Defie f : A B by f ((x 1, x 2,..., x [x 1 x 2... x ]. The fuctio f is -to-1. I paticula: f 1 ([x 1 x 2... x ] { (x 1, x 2, x 3,..., x, (x, x 1, x 2,..., x 1 (x 1, x, x 1,..., x 2,... (x 2, x 3, x 4,..., x 1 }

3 Couse Notes 9: Pemutatios ad Combiatios Figue 1: These ae the 6 diffeet ways that Kig Athu ca seat 4 kights at his oud table. Two seatigs diffeig oly by otatio ae cosideed equivalet. We deote a seatig aagemet i text by listig kights i squae backets i clockwise ode, statig at a abitay poit. Fo example, the seatigs i the top ow ca be deoted [1234], [2413], ad [3421]. Thee ae tuples i the list of tuples, because x 1 ca appeas i diffeet places i a tuple. By the Divisio Rule, A B. This gives: A B! ( 1! The secod equality holds, because the umbe of odeigs of items i a lie is!, as we showed peviously. 2 Combiatios We ow use the Divisio Rule to fid a fomula fo the umbe of combiatios of elemets fom a set. Oe commo pitfall i coutig poblems is mixig up combiatios with pemutatios; ty to keep tack of which is which! Example: Studets ad pizes. How may goups of 5 studets ca be chose fom a 180- studet class to collaboate o a poblem set? Let A be the set of 5-pemutatios of the set of studets. This umbe is P (180, But that s ot eally what we wat ode of the 5 studets i a goup does t matte. So let B be the set of 5-studet goups. This is what we wat. Let s defie a map f : A B by igoig the ode. We claim that f is 5! to 1, i othe wods, 120 to 1. This is based o the umbe of ways that 5 studets could be lied up. So, by the Divisio Rule, A B

4 4 Couse Notes 9: Pemutatios ad Combiatios 2.1 The -Combiatios of a Set Geealizig the example that we just did, the Divisio Rule ca be used to cout the umbe of -combiatios of a set. Defiitio 2.1. A -combiatio of a set is a subset of size. Fo example, fo the set {a, b, c}, we have the followig thee 2-combiatios: {a, b} {a, c} {b, c} Udestadig the distictio betwee -combiatios ad -pemutatios is impotat. The - combiatios of a set ae diffeet fom -pemutatios i that ode does ot matte. Fo example, the above set {a, b, c} has six 2-pemutatios: (a, b (a, c (b, c (b, a (c, a (c, b The collaboatio goups of 5 studets i pevious example ae the 5-combiatios of the set of studets i the class. 2.2 The umbe of -Combiatios (Biomial Coefficiets The umbe of -combiatios of a -elemet set comes up i may poblems. The quatity is impotat eough to meit two otatios. The fist is C(, ; this is aalogous to the otatio P (, fom befoe fo the umbe of ( pemutatios of a -elemet set. The secod otatio is, ead choose. The values o C(, ae ofte called biomial coefficiets because of thei pomiet ole i the Biomial Theoem, which we will cove shotly. The followig theoem gives a closed fom fo. This theoem is used all the time; emembe it! Theoem 2.2.! (!! The theoem says, fo example, that the umbe of 2-combiatios of the thee elemet set {a, b, c} is 3! 3 as we saw above. (3 2! 2! 180! Also, fo example, the umbe of 5-combiatios of the 180 elemet set of studets is 175!5!. Poof. The poof uses the Divisio Rule. Let X be a set with elemets. Let A be the set of - pemutatios of X, ad let B be the set of -combiatios of X. Ou goal is to compute B. To this ed, let f : A B be the fuctio mappig -pemutatios to -combiatios that is defied by: f ( (x } 1, x 2,..., x {{ } { x } 1, x 2,..., x {{ } } ode mattes ode does ot matte (pemutatio (combiatio

5 Couse Notes 9: Pemutatios ad Combiatios 5 The fuctio f is!-to-1. I paticula: f 1 ({x 1, x 2,..., x } {(x 1, x 2, x 3, x 4,..., x, (x 2, x 1, x 3, x 4,..., x,... all! pemutatios... } By the Divisio Rule, A! B. Now we ca compute the umbe of -combiatios of a -elemet set, B, as follows: B A! P (,!! (!! 2.3 Some iteestig special cases Biomial coefficiets ae impotat eough that some special values of ae woth emembe- ig:! 1 Thee is a sigle size 0 subset of a -set. 0! 0!! 1 Thee is a sigle size subset of a -set. 0!!! Thee ae size 1 subsets of a -set. 1 ( 1! 1!! Thee ae size ( 1 subsets of a -set. 1 1! ( 1! Recall that 0! is defied to be 1, ot zeo. Theoem 2.3. Fo all 1, 0,. Theoem 2.3 ca be ead as sayig that the umbe of ways to choose elemets to fom a subset of a set of size is the same as the umbe of ways to choose the elemets to exclude fom the subset. Put that way, it s obvious. This is what is called a combiatoial poof of a idetity: we wite expessios that eflect diffeet ways of coutig o costuctig the same set of objects ad coclude that the expessios ae equal. All sots of complicated-lookig ad appaetly obscue idetities ca be poved ad made clea i this way. Of couse we ca also pove Theoem 2.3 by simple algebaic maipulatio:!!!. (!!! (! ( (! (!

6 6 Couse Notes 9: Pemutatios ad Combiatios 3 Coutig Poke Hads I the poke game Five-Cad Daw, each playe is dealt a had cosistig of 5 cads fom a deck of 52 cads. Each cad i the deck has a suit (clubs, heats, diamods, o spades ad a value (A, 2,..., 10, J, Q, K. The ode i which cads ae dealt does ot matte. We will cout vaious types of hads i Five-Cad Daw to show how to use the fomula fo -combiatios give i Theoem 2.2. It may ot be clea why we wat to cout the umbe of types of hads. The umbe of hads of a paticula type tells us how likely we ae to ecoute such a had i a adom deal of the cads. Hads that ae less likely ae aked highe tha those that ae moe likely, ad so will wi ove the moe likely oes. Kowig the likelihood (o the umbe of vaious hads ca help us to figue out how to bet o the game. 3.1 All Five-Cad Hads How may diffeet hads ae possible i Five-Cad Daw? A had is just a 5-cad subset of the 52-cad deck. Theefoe, the possible hads i Five-Cad Daw ae exactly the 5-combiatios of 52! a 52-elemet set. By Theoem 2.2, thee ae ( ! 5! 2, 598, 960 possible hads. Cofusio betwee combiatios ad pemutatios is a commo souce of eo i coutig poblems. Fo example, oe might eoeously cout the umbe of 5-cad hads as follows. Thee ae 52 choices fo the fist cad, 51 choice fo the secod cad give the fist,..., ad 48 choices fo the fifth cad give the fist fou. This totals possible hads. The poblem with this way of coutig is that a had is couted oce fo evey odeig of the five cads; that is, each had is couted 5! 120 times. We accidetally couted -pemutatios istead of -combiatios! To get the umbe of 5-combiatios, we would have to divide by Hads with Fou-of-a-Kid How may hads ae thee with fou-of-a-kid? Fo example, has a fou-of-akid, because thee ae fou 9 s. Fist we must choose oe value to appea i all fou suits fom the set of 13 possible values (this follows the tee diagam appoach to exploig distict cases. Thee ae 13 choices of this value, ad if we pick oe we ca t have ay othes (that would equie 8 cads. So the choice of value gives 13 disjoit cases to cout. Afte this choice, we have to choose oe moe cad fom the emaiig set of 48. This ca be doe i ( ways. I total, thee ae hads with fou-of-a-kid. The situatio is illustated by the tee diagam i Figue 2. Hee is a way to itepet the esult that uses pobability. If all ( 52 hads ae equally likely, the 5 the pobability that we ae dealt a fou-of-a-kid is ( I othe wods, we ca oly expect to get fou-of-a-kid oly oce i evey 4165 games of Five- Cad Daw! (This is just a passig obsevatio; we will talk about pobability fo eal i a few weeks.

7 Couse Notes 9: Pemutatios ad Combiatios 7 A 2 3 K 2 spades 2 clubs K diamods Hads with 4-of-a-kid 13 x chose 4-tuple i 13 ways chose fifth cad i 48 ways Figue 2: This (icomplete tee diagam couts the umbe of hads with fou-of-a-kid. Thee ae 13 choices fo the value appeaig i all fou suits. The fifth cad i the had ca be ay of the 48 cads emaiig i the deck. This gives a total of possible hads. Mistakes ae easy to make i coutig poblems. It is a good idea to check a esult by coutig the same thig i aothe way. Fo example, we could also cout the umbe of hads with a fou-of-a-kid as follows. Fist, thee ae 52 ways to choose the exta cad. Give this, thee ae 12 ways to choose the value that appeas i all fou suits. (We caot choose the value of the fist cad. The possibilities ae illustated by the tee diagam i Figue 3. By this method, we fid that thee ae fou-of-a-kid hads. This is cosistet with ou fist aswe (which is a good thig. 3.3 Hads with a Full House A full house is a had with both a thee-of-a-kid ad a two-of-a-kid. Fo example, the had 7 7 J J J is a full house because thee ae thee jacks ad two seves. How may full house hads ae thee? We ca choose the value that appeas thee times i 13 ways. (4 The we ca pick ay thee of the fou cads i the deck with this value; this ca be doe i 3 4 ways. Thee ae the 12 emaiig choices fo the value that ( appeas two times. We ca pick ay two of the fou cads 4 with this value; this ca be doe i 2 6 ways. The total umbe of full house hads is theefoe This is 6 times geate tha the umbe of hads with a fou-of-a-kid. Sice a fou-of-a-kid is ae, it is woth moe i poke! 3.4 Hads with Two Pais How may hads ae thee with two pais, but o thee- o fou-of-a-kid?

8 8 Couse Notes 9: Pemutatios ad Combiatios A of diamods 2 of diamods 2 3 K Ca ot choose A K of spades 3 of diamods Hads with 4-of-a-kid 52 x chose fifth cad i 52 ways chose 4-tuple i 12 ways Figue 3: This tee diagam shows aothe way to cout the umbe of hads with a fou-of-a-kid. Thee ae 52 choices fo the exta cad ad the 12 choices fo the value appeaig i all fou suits. This gives hads, the same aswe as befoe. ( 4 False poof. Thee ae 13 choices fo the value of the cads i the fist pai. Thee ae the 2 6 ways to choose two of the fou cads i the deck ( that have this value. Next, thee ae 12 choices 4 fo the value of the cads i the secod pai, ad 2 6 ways to choose two of the fou cads with this value. Fially, the fifth cad ca be ay oe of the 48 cads emaiig i the deck. Altogethe thee ae had with just two pais. Thee ae actually two bugs i this agumet. The fist is that the fifth cad caot be ay oe of the 48 emaiig cads; i paticula, it caot have the same value as a cad aleady selected. (Othewise, thee would be thee of a kid. This ules out 4 of the 48 cads, leavig oly 44 choices fo the fifth cad. The secod bug is that evey had has bee couted twice! Fo example, a pai of kigs ad a pais of quees is couted oce with the kigs as the fist pai ad a secod time with the quees as the fist pai. The efeeces i the agumet to a fist pai ad a secod pai sigal dage; these tems imply a odeig that is ot pat of the poblem. To fix the secod bug, we ca apply the Divisio Rule. Thee is a 2-to-1 mappig fom the set of hads we couted to the set of hads with two pais. Theefoe, the set of hads with two pais is half as lage as ou iitial cout: hads with two pais 2 123, 552 Why did this facto of 2 aise i coutig pais, but ot i coutig full houses? The easo is that pais ae itechageable, but a pai ad a tiple ae ot. Fo example, a pai of 2 s ad a pai of

9 Couse Notes 9: Pemutatios ad Combiatios 9 9 s is the same as a pai of 9 s ad a pai of 2 s. O the othe had, a pai of 2 s ad a tiple of 9 s is diffeet fom a pai of 9 s ad a tiple of 2 s! To check ou esult, we ca cout the umbe of hads with two pais i a completely diffeet way. The umbe of ways to choose the values fo the two pais is ( We ca choose suits fo each pai i ( ways, ad we ca choose the emaiig cad i 44 ways as befoe. This gives: , This is the same aswe as befoe. Two pais tu out to be 33 times moe likely tha a full house; as oe would expect, a full house beats two pais i poke! The followig optioal sectios cotai simila examples of poke had coutig. 3.5 Thee of a kid, two diffeet [Optioal] Thee ae 13 ways to choose the ak fo the thee-of-a-kid, ad oce that is chose, thee ae 4 ways to choose the 1 3 thee cads. Oce we have the thee-of-a-kid, thee ae 12 ways to choose the two aks fo the othe two cads. Ad ( 4 2 the fo each of these aks, thee ae 1 ways to choose the sigle cad of that ak. The total is: , Sice this is fewe tha the hads with two pais, this is aked highe. 3.6 Oe pai, othes all diffeet [Optioal] ( 4 Thee ae 13 ways to choose the ak fo the pai, ad the 2 ways to choose the two cads of that ak. The thee 1 ae 12 ways to choose the aks of the emaiig thee cads, ad fo each of these aks, 4 ways to choose the sigle 3 1 cad of that ak. The total is: , 098, Fou diffeet aks [Optioal] We have aalyzed all the actual poke hads based o the umbes of cads of diffeet aks (4-1, 3-2, 3-1-1, 2-2-1, How may hads have all 5 cads of diffeet aks? That s the total umbe of hads mius the sum of the umbes of hads of these five kids, o 2, 598, 960 1, 098, , , 912 3, , 317, 888. What is wog with the followig coutig agumet? False coutig agumet. Thee ae 52 choices fo the fist cad, 48 choices fo the secod cad give the fist, 44 choices fo the thid give the fist two, etc. Afte each cad, the umbe of emaiig choices dops by fou, sice we caot

10 10 Couse Notes 9: Pemutatios ad Combiatios subsequetly pick the cad just take o ay of the othe thee cads with the same value. This gives hads with o pai. Is this aswe ight? The efeeces to fist cad, secod cad,... ae a waig sigal that we may be iadvetetly coutig a odeed set. To be sue, we ca check how may times a paticula had is couted. Coside the had A K Q J 10. We could have obtaied this had by dawig cads i the ode listed, but we could also have daw them i aothe ode. I fact, we couted this had oce fo evey possible odeig of the five cads, ad i fact, evey had is couted 5! 120 times! We ca coect ou iitial aswe by applyig the Divisio Rule. This gives: hads with o pais , 317, 888 Ad still aothe way: Thee is aothe way to cout the hads with o pais that avoids this cofusio. Sice thee ae o pais, o two cads have the same value. This meas that we could fist choose 5 distict values fom the set of 13 possibilities. The we could choose oe of the fou suits fo each of the five cads. This gives: 13 hads with o pais , 317, 888 This is the same aswe as befoe. Notice that the umbe of hads with o pais is moe tha half of all the hads. This meas that i moe tha half ou games of Five-Cad Daw, we will ot have eve oe pai! 3.8 Hads with Evey Suit [Optioal] How may hads cotai oe cad of evey suit? Such a had must cotai two cads fom oe suit ad oe cad fom each of the othe thee suits. Agai, we ca cout i two ways. Fist cout. Pick oe cad fom each suit. This ca be doe i 13 4 ways. The pick ay oe of the emaiig 48 cads. This gives a total of hads. Howeve, we have ovecouted by two, sice the two cads with the same suit could be picked i eithe ode. Accoutig fo this, thee ae , 464 hads. 2 Secod cout Fist, we ca pick the suit that cotais two cads i fou ways. The we ca pick values fo these two cads i 13 2 ways. Fially, we ca pick a cad fom each of the othe thee suits i 13 ways. This gives: hads with all suits Moe tha a quate of all Five-Cad Daw hads cotai all suits , Remides To make sue that you get the ight aswe whe coutig: 1. Look at a sample 5-cad had (o a sample of whateve you ae coutig ad make sue that it is coveed oce ad oly oce i you cout.

11 Couse Notes 9: Pemutatios ad Combiatios Ty coutig by two diffeet ways ad check that you get the same aswe. Fo example, whe we fist couted the umbe of hads with o pais, we discoveed that we had massively ovecouted; a sample had was actually coveed 120 times by ou cout! We coected this aswe by applyig the Divisio Rule, ad we couted the same thig agai i a diffeet way as a double-check. 4 The Magic Tick Thee is a Magicia ad a Assistat. The Assistat goes ito the audiece with a deck of 52 cads while the Magicia looks away. Five audiece membes ae asked to select oe cad each fom the deck. The Assistat the gathes up the five cads ad aouces fou of them to the Magicia. The Magicia thiks fo a time ad the coectly ames the secet, fifth cad! Oly elemetay combiatoics udelies this tick, but it s ot obvious how. (The eade should take a momet to thik about this; the Assistat does t cheat. 4.1 How the Tick Woks The Assistat has somehow commuicated the secet cad to the Magicia just by amig the othe fou cads. I paticula, the Assistat has two ways to commuicate. Fist, he ca aouce the fou cads i ay ode. The umbe of odeigs of fou cads is 4! 24, so this aloe is isufficiet to idetify which of the emaiig 48 cads is the secet oe. Secod, the Assistat ca choose which fou of the five cads to eveal. The ( amout of ifomatio that ca be coveyed 5 this way is hade to pi dow. The Assistat has 4 5 ways to choose the fou cads evealed, but the Magicia ca ot detemie which of these five possibilities the Assistat selected, sice he does ot kow the secet cad! Nevetheless, these two foms of commuicatio allow the Assistat to covetly eveal the secet cad to the Magicia. Hee is a oveview of how the tick woks. The secet cad has the same suit as the fist cad aouced. Futhemoe, the value of the secet cad is offset fom the value of the fist cad aouced by betwee 1 ad 6. See Figue 4. This offset is commuicated by the ode of the last thee cads. Hee ae the details. The audiece selects five cads ad thee ae oly 4 suits i the deck. Theefoe, by the Pigeohole Piciple, the Assistat ca always pick out two cads with the same suit. Oe of these will become the secet cad, ad the othe will be the fist cad that he aouces. Hee is how he decides which is which. Note that fo ay two cad umbes, oe is at most six clockwise steps away fom the othe i Figue 4. Fo example, if the cad umbes ae 2 ad Q, the 2 is thee clockwise steps away fom Q. The Assistat esues that the value of the secet cad is offset betwee 1 ad 6 clockwise steps fom the fist cad he aouces. The offset is commuicated by the ode that the Assistat aouces the last thee cads. The Magicia ad Assistat agee i advace o a ode fo all 52 cads. Fo example, they might use: A, 2,..., K, A,..., K, A,..., K, A,..., K,

12 12 Couse Notes 9: Pemutatios ad Combiatios Assistat eveals this cad fist. Q J K A Offset passed by odeig of last 3 cads This is the secet cad Figue 4: The 13 possible cad values ca be odeed i a cycle. Fo ay two distict values, oe is offset betwee 1 ad 6 clockwise steps fom the othe. I this diagam, 3 is offset five steps fom J. I the cad tick, the value of the secet cad is offset betwee 1 ad 6 steps fom the fist cad aouced. This offset is commuicated by the ode of the last thee cads amed by the Assistat. though ay othe ode will do as well so log as the Magicia ad his Assistat use the same oe :-. With this ode, oe of the last thee cads aouced is the smallest (S, oe is lagest (L, ad the othe is medium (M. The offset is ecoded by the ode of these thee: SML 1 SLM 2 MSL 3 MLS 4 LSM 5 LMS 6 Fo example, suppose that the audiece selects 3 8 A J 6. The Assistat picks out two cads with the same suit, say 3 ad J. The 3 is five clockwise steps fom J i Figue 4. So the Assistat makes 3 the secet cad, aouces J fist, ad the ecodes the umbe 5 by the ode i which he aouces the last thee cads: A 6 8 LSM Same Tick with 4 Cads? Could the same magic tick wok with just 4 cads? That is, if the audiece picks fou cads, ad the Assistat eveals thee, the ca the Magicia detemie the fouth cad? The aswe tus out to be o. The poof elies o the Pigeohole Piciple. Theoem 4.1. The magic tick is ot possible with 4 cads. Poof. The audiece ca select ay 4-combiatio of cads; let A be the set of all such 4-combiatios. The Assistat ca aouce ay 3-pemutatio of cads; let B be the set of all such 3-pemutatios. The fomulas fo -combiatios ad -pemutatios give the followig sizes fo A ad B:

13 Couse Notes 9: Pemutatios ad Combiatios 13 A C(, 52! 48! 4! 270, 725 B P (, 52! 49! 132, 600 The Assistat sees a 4-combiatio of cads selected by the audiece ad must aouce a 3- pemutatio to the Magicia. Let f : A B be the fuctio that the Assistat uses i mappig oe to the othe. Sice A > B, the Pigeohole Piciple implies that f maps at least two 4- combiatios to the same 3-pemutatio. That is, thee ae two diffeet sets of fou cads that the audiece ca pick fo which the Assistat says exactly the same thig to the Magicia. Fo these two sets of fou cads, thee cads must be the same (sice the Assistat aouces the same thee cads i both cases ad oe cad must be diffeet (sice the two sets ae diffeet. Fo example, these might be the two sets of fou cads fo which the Assistat says exactly the same thig to the Magicia: 3 8 A J 3 8 A K I this case, the Assistat aouces 3 8 A i some ode. The magicia is ow stuck; he ca ot detemie whethe the emaiig cad is J o K. Thee ae may vaiats of the magic tick. Fo example, if the audiece picks 8 cads, the evealig 6 to the Magicia is eough to let him detemie the othe two. This sot of subtle tasmissio of ifomatio is impotat i the secuity busiess whee oe wats to pevet ifomatio fom leakig out i udetected ways. This is actually a whole field of study. 4.3 Hall s Theoem applied to the Magic Tick We kow how the Magic tick woks i which a Assistat eads fou cads fom a five cad had ad the Magicia pedicts the fifth cad. We also kow the tick caot be made to wok if the Assistat oly shows thee cads fom a fou cad had, because thee ae fewe sequeces of thee out of 52 cads tha thee ae 4-cad hads out of 52 cads. So the questio is, whe ca the tick be made to wok? Fo example, what is the lagest size deck fo which ou tick of eadig 4 cads fom a 5-cad had emais possible? A elegat esult kow as Hall s Maiage Theoem allows us to aswe the geeal questio of whe the tick ca be made to wok. Namely, the tick is possible usig h-cad hads chose fom a -cad deck, with the Assistat eadig of the h cads iff P (h, C(, h. (1

14 14 Couse Notes 9: Pemutatios ad Combiatios I paticula, usig h 5-cad hads ad evealig 4 cads, we ca do the tick with a -cad deck as log as 120 P (5, 4 C( 4, 1 4. So we ca do the tick with a deck of up to size 124. Fo example, we could still do the tick, with oom to spae, if we combied a ed deck with a blue deck to obtai a 104-cad deck. Of couse it s ot clea whethe with the double-size deck thee will be a simple ule fo detemiig the hidde cad as thee is with the 52-cad deck, but Hall s Theoem guaatees thee will be some ule. 4.4 Hall s Maiage Theoem The Magicia gets to see a sequece of 4 cads ad has to detemie the 5th cad. He ca do this if ad oly if thee is a way to map evey 5-cad had ito a sequece of 4 cads fom that had so that o two hads map to the same sequece. That is, thee eeds to be a ijectio fom the set of 5-cad hads ito the 4-cad sequeces, subject to the costait that each 5-cad had maps to a 4-cad sequece of cads fom that had. This is a example of a Maiage Poblem. The taditioal way to descibe such a poblem ivolves havig a set of wome ad aothe set of me. Each wome has a list of me she is willig to may ad who ae also willig to may he. The Maiage Poblem is to fid, fo each woma, a husbad she is willig to may. Bigamy is ot allowed: each husbad ca have oly oe wife. Whe such a matchig of wives to husbads is possible, that paticula Maiage Poblem is solvable. Fo ou cad tick, each 5-cad had is a woma ad each 4-cad sequece as a ma. A ma (4-cad sequece ad woma (5-cad had ae willig to may iff the cads i the sequece all occu i the had. Hall s Theoem gives a simple ecessay ad sufficiet coditio fo a Maiage Poblem to be solvable. Theoem 4.2 (Hall s Maiage Theoem 1. Suppose a goup of wome each have a list of the me they would be willig to may. Say that a subset of these wome has eough willig me if the total umbe of distict me o thei lists is at least as lage as the umbe of wome i the subset. The thee is a way to select distict husbads fo each of the wome so that evey husbad is acceptable to his wife iff evey subset of the wome has eough willig me. Hall s Theoem ca also be stated moe fomally i tems of bipatite gaphs. Defiitio. A bipatite gaph, G (V 1, V 2, E, is a simple gaph whose vetices ae the disjoit uio of V 1 ad V 2 ad whose edges go betwee V 1 ad V 2, viz., E {{v 1, v 2 } v 1 V 1 ad v 2 V 2 }. A pefect matchig i G is a ijectio f : V 1 V 2 such that {v, f (v} E fo all v V 1. Fo ay set, A, of vetices, defie the eighbo set, N (A :: {v a A {a, v} E}. A set A V 1 is called a bottleeck if A > N (A.

15 Couse Notes 9: Pemutatios ad Combiatios 15 Theoem (Hall. A bipatite gaph has a pefect matchig iff it has o bottleecks. A simple coditio esuig that a bipatite gaph has o bottleeck is give i the i-class poblems fo Wedesday, Oct. 30, This coditio is easy to veify fo the gaph descibig the cad tick, ad it implies that iequality (1 is ecessay ad sufficiet fo the tick to be possible Othe Examples [Optioal] [Optioal] Hall s Maiage Theoem guaatees the possibility of selectig fou cads of distict suits, oe each fom ay fou sepaate piles of 13 cads fom a stadad deck of 52. Likewise, thee will always be 13 cads of distict deomiatios, oe fom each of 13 piles of fou cads. I this case, thik of each pile as the set of oe to fou the distict cad deomiatios that appea i that pile. Notice that amog ay k piles of cads at least k distict deomiatios appea by the Geealized Pigeoholig Piciple, thee ae at least that may distict deomiatios amog ay 4k cads. Thus the existece of oe cad of each deomiatio spead out acoss 13 piles is assued by the Theoem. Pullig i this big theoem is comfotig but ot completely satisfactoy. Although it assues us that these feats ca always be accomplished, it gives absolutely o idicatio of how to pefom them! Tial ad eo woks well eough fo the simple fou-pile poblem, but ot with 13 piles. Is thee a geeal pocedue fo pullig out a selectio of distict deomiatios fom 13 piles of fou cads? Fotuately thee is: choose a abitay pile cotaiig a Ace, the oe cotaiig a two, ad so o, doig this fo as log as possible util you get stuck. Place these selected piles i a ow. Suppose you have selected a total of te piles, cotaiig a Ace, ad two though te, espectively. This sceaio leaves thee utouched piles. If ay of the 12 cads i those thee piles is a Jack, Quee, o Kig the you ae ot eally stuck; you ca cotiue a little futhe (but pehaps ot i the usual sequetial ode. If you ae tuly stuck, select ay cad i, say, the 11th pile, ad go to oe of the te piles that coespods to the umbe of that cad. Thus if you select a thee you go to the thid pile. Tu a cad i that thid pile ove (to make it cospicuous ad move to the pile that coespods to the umbe of that tued cad. Keep doig this to ceate a chai of piles ad tued cads, util you evetually hit upo a pile that cotais a cad ot i the iitial list; i ou case, util we come upo a Jack, Quee, o Kig. (Note: Oe ca pove that you will ot fall ito a closed loop of choices. Now shift all the piles oe place back alog the chai of piles to obtai a cofiguatio that allows you to add oe moe pile to the iitial list of te. Repeat this pocess util you solve the puzzle completely Poof of Hall s Theoem by Iductio [Optioal] [Optioal] This method of chai shiftig is kow i the liteatue as the techique of augmetig paths ad is the basis fo efficiet ways to fid pefect matchigs ad to solve elated etwok flow poblems. This is descibed i may combiatoics texts 2, but we shall ot develop this method hee. Istead, we ll go back to the ma-woma maiage temiology ad give a simple poof by: Poof. Stog iductio o the umbe,, of wome. Base case ( 1. If thee is just a sigle woma with at least oe ame o he list, the she ca simply may the fist ma o he list. Iductio step ( > 1. Case I. Suppose these wome s lists satisfy the stoge coditio that amog ay lists, 1 <, at least + 1 distict ames ae metioed. Select oe woma. She has at least two ames o he list ( 1 case. Have he may oe of these me (call him Poidexte. This leaves 1 wome to may. 2 Fo example, Ia Adeso pesets a pocedual poof usig augmetig paths i his book A Fist Couse i Combiatoial Mathematics, Oxfod Applied Mathematics ad Compute Sciece Seies, Claedo Pess, Oxfod, 1974.

16 16 Couse Notes 9: Pemutatios ad Combiatios The lists possessed by these 1 wome have the popety that amog ay of them (1 < at least + 1 distict ames ae metioed. Oe of these ames could be Poidexte s who is o loge available fo maiage. But we ca still say that amog ay lists, at least distict ames of available me ae metioed. This is all we eed to ivoke the iductio hypothesis fo these emaiig 1 wome. Thus we have a meas to may all wome i this sceaio. Case II. Suppose this stoge coditio does ot hold. Thus thee is subgoup of 0 wome (1 0 < such that amog thei lists pecisely 0 distict ames ae metioed. By the stog iductio hypothesis fo 0 <, we ca may these wome. This leaves m 0 < wome to coside. Is it tue that amog these m wome s lists ay of them (1 m metio at least distict ames of available me? The aswe is yes! If ot, say some of these wome metio fewe tha available me. The these wome plus the 0 wome above have metioed fewe i 0 + me i total. This cotadicts the popety satisfied by these lists. Thus we ca ivoke the stog iductio hypothesis fo the emaiig m wome ad successfully have them may as well. This completes the poof by iductio. Poblem 1. The iductio poof ca ead as a ecusive pocedue to costuct a maiages fo woma i tems of costuctig maiages fo smalle sets of wome. Why is t this pocedue vey efficiet? Poblem 2. A deck of cads is shuffled ad dealt ito 26 piles of two cads. Is it possible to select a black Ace fom oe pile, a Red ace fom aothe, a black two fom a thid, a ed two fom a fouth, ad so o all the way dow to a black Kig ad a ed Kig fom the two emaiig piles? 5 Popeties of Biomial Coefficiets Thee ae ifiitely may iteestig idetities ivolvig biomial coefficiets. We aleady see a simple example i Theoem 2.3. These idetities ae impotat because of how ofte biomial coefficiets aise i aalysis of algoithms (6.046 ad pobability (ext week. You ca t lea them all, but you ca lea the basic oes ad look othes up/deive them as eeded. Let s begi with a fomula fo the atio betwee successive biomial coefficiets same, successive : Theoem 5.1. Suppose 1. The / Algebaic poof. /! 1! (! ( 1! ( ( 1!! + 1.

17 Couse Notes 9: Pemutatios ad Combiatios 17 Combiatoial poof. Coside choosig a committee of size ad a leade, fom ( people. Oe way is to fist pick the 1 commos ad the thei leade; this ca be doe i 1 ( ( 1 ways. Aothe way is to pick the committee membes fist ad the pick a leade fom amog them; this ca be doe i ways. Thus, ( This atio is geate tha 1 if < +1 ad less tha 1 if > This says that the successive coefficiets icease util eaches +1 2 ad the decease, i.e., they ae uimodal like the cuve i Figue 5. Figue 5: Bell Cuve I fact, the biomial coefficiets fom a eal bell cuve. These aise o examiatios i a atual way. Suppose that a test has questios. The the bell cuve descibes, fo evey, the umbe of ways the studet ca get exactly of the questios ight, amely. If we suppose that each studet takig the test is doig adom guessig, so is equally likely to get each of the questios ight o wog, the these coefficiets tu out to be popotioal to the umbe of studets that get those umbes of questios ight. (We ll see this whe we do pobability. Aothe idetity: Theoem 5.2. Suppose 1. The 1. 1 Algebaic poof.! ( 1!!(! ( 1!(! ( 1! 1. ( 1! (( 1 ( 1! 1

18 18 Couse Notes 9: Pemutatios ad Combiatios Combiatoial poof. Coside choosig a committee of size ad a leade, f ( om people. Oe 1 way is to fist pick the leade ad the his 1 subjects; this ca be doe i 1 ways. Aothe way is to ( pick the committee membes fist ad the pick a leade fom amog them; this ca be doe i ways. Thus, 1. 1 Now a impotat theoem due to Pascal: Theoem 5.3 (Pascal. Suppose 1 1. The This is pobably ou most impotat idetity, sice it gives a kid of ecuece fo biomial coefficiets. Memoize it! Algebaic poof ( 1! ( 1!!( 1! + ( 1!(! ( 1! ( 1! (!(! +!(! ( 1!!(!!!(! Combiatoial poof. We use case aalysis (a tee diagam ad the sum ule. Let S :: {1,..., }. Let A be the set of -elemet subsets of S. Let B be the set of -elemet subsets of S that cotai. Let C be the set of -elemet subsets of S that do t cotai. The A B C, ad B ad C ae disjoit. So A B + C, by the Sum Rule. But ow we ca get expessios fo A, B ad C as umbes of combiatios: A.

19 Couse Notes 9: Pemutatios ad Combiatios 19 1 B. 1 This is because, i additio to, aothe 1 elemets must be chose fom {1,..., 1}. 1 C. This is because elemets must be chose fom {1,..., 1}. So (by the Sum Rule 1 1 A B + C +. 1 Pascal s theoem has a ice pictoial epesetatio: The ow epesets, statig with 0 i the top ow. Successive elemets i the ow epeset, statig with 0 at the left Figue 6: Pascal s tiagle. (Notice that it s just the double-iductio matix eshaped. This tiagle has lots of ice popeties. Expeimet with it. Fo example, what happes if we add the coefficiets i oe ow? Fo the followig two theoems we povide thei combiatoial poofs, oly. The coespodig algebaic poofs ae easy iductive agumets. Theoem 5.4. Suppose is ay atual umbe. The 2. 0

20 20 Couse Notes 9: Pemutatios ad Combiatios Combiatoial poof. Agai, we use case aalysis ad the sum ule fo disjoit uios. Thee ae 2 diffeet subsets of a set of elemets. Decompose this collectio based o the sizes of the subsets. That is, let A be ( the collectio of subsets of size. The the set of all subsets is the disjoit uio A. Thee ae subsets of size, fo 0, 1,...,. Hece the theoem follows. [Optioal] Theoem 5.5 (Vademode. Suppose 0 m,. The m + m k k k0 Combiatoial poof. Agai, we use case ( aalysis with the sum ad poduct ules. Suppose thee ae m ed balls ad m+ blue balls, all distict. Thee ae ways to choose balls fom the two sets combied. That s the LHS. Now decompose this collectio of choices based ( o how may of the chose balls ae blue. Fo ay k, 0 k, ( thee ( ae m m ways to choose k ed balls, ad ways to choose k blue balls. By the poduct ule, that makes k k k k ways to choose balls such that k of them ae blue. Addig up these umbes fo all k gives the RHS. 5.1 The Biomial Theoem A impotat theoem gives the coefficiets i the expasio of powes of biomial expessios. It tus out they ae biomial coefficiets thus the ame! Theoem 5.6. Suppose N. The (x + y x k y k. k k0 Example 5.7. (x + y 4 1x 4 + 4x 3 y + 6x 2 y 2 + 4xy 3 + 1y 4. Algebaic poof. By iductio, usig Pascal s idetity. Combiatoial poof. Coside the poduct (x + y (x + y(x + y(x + y(x + y This poduct has factos. To expad this poduct, we geeate idividual tems of the sums by selectig (i all possible ways oe of the two vaiables iside each facto to get a -vaiable poduct. If we select k x vaiables ad k y vaiables, we get a tem of the fom x k y k. The we gathe all those tems togethe. So how may ways do we get such ( a tem? Each x k y k tem selects x fom k of the factos ad y fom the othe k. So thee ae k ways of choosig which factos have x. The biomial theoem ca be used to give slick poofs of some biomial idetities, fo example: Theoem k k0

21 Couse Notes 9: Pemutatios ad Combiatios 21 Poof. Plug x 1, y 1 ito biomial theoem. Theoem 5.9. ( 1 k 0. k k0 Poof. Plug x 1, y 1 ito biomial theoem. The combiatoial itepetatio is as follows: the umbe of ways of selectig a eve umbe of elemets fom a set of equals the umbe of ways of selectig a odd umbe. This woks fo both eve ad odd. Example Coside 5. We ca choose a odd-size set ( 5 ( 5 ( ways. We ca choose a eve-size set ( 5 ( 5 ( ways. Example : Thee ae odd sets. Thee ae eve sets. Fo odd, this is ituitive (choosig a eve set leaves a odd set. Fo eve, it may be somewhat supisig thee is o obvious bijectio betwee the eve ad odd sets. Theoem 5.9 ca also be pove usig iclusio/exclusio i a ticky way. 6 Piciple of Iclusio ad Exclusio Now we etu to the Piciple of Iclusio-Exclusio fom last week. Ou biomial coefficiet idetities give us a way to pove this theoem, via ou alteatig sum idetity (poved by biomial theoem. Recall the Iclusio-Exclusio theoem: Theoem 6.1. i A i i A i i<j A i A j + i<j<k A i A j A k The theoem ca be poved by iductio o the umbe of sets, but this appoach is messy. Istead we will give a combiatoial poof usig biomial coefficiets. Poof. Each tem i the summatio couts cetai elemets i i A i. We pove that evey elemet of this uio is couted exactly oce. So, coside ay paticula elemet a, ad suppose it is i exactly of the sets A i. We see how may times a is couted by each tem. I the fist tem, a is couted 1 times, oce fo each of the sets that cotais it. I the secod tem, a is couted 2 times, oce fo each of the pais of sets that cotai it. I the thid tem, a is

22 22 Couse Notes 9: Pemutatios ad Combiatios couted 3 times,..., ad i the th tem, a is couted 1 times. The tems alteate sigs, so the total umbe of times a is couted is: ( 1. 3 But Theoem 5.9 implies that ( 1 k 0. k k0 This implies that the sum above is equal to 0 1. That is, a is couted exactly oce, as eeded. 7 -Pemutatios with Repetitio We ow tu to pemutatios ad combiatios whee elemets ae allowed to epeat. A - pemutatio with epetitio of a set S is the umbe of ways to choose elemets fom S with epetitio allowed ad whee ode mattes. Fo example, thee ae ie 2-pemutatios with epetitio of the set S {A, B, C}, which ae listed below. (A, A (A, B (A, C (B, A (B, B (B, C (C, A (C, B (C, C 7.1 Coutig -Pemutatios with Repetitio Fotuately, -pemutatios with epetitio ae easy to cout. It is the same as usig the poduct ule to cout the umbe of stigs of legth fom a alphabet with lettes. Theoem 7.1. The umbe of -pemutatios with epetitio of a -elemet set is. Fo example, the theoem says that the umbe of 2-pemutatios with epetitio of the 3-elemet set S {A, B, C} is 3 2 9, which checks with ou pevious aswe. Poof. Let S be a set with elemets. The -pemutatios with epetitio of S ae pecisely the elemets of: S } S {{ S } tems By the Poduct Rule, this set has cadiality S.

23 Couse Notes 9: Pemutatios ad Combiatios Pemutatios with Limited Repetitio We might wat to cout the umbe of -pemutatios whee each elemet ca be epeated a limited umbe of times. I geeal, this leads to some haiy aalysis ad o closed-fom aswe. Theefoe, we will coside oly a special case, the umbe of -pemutatios whee each elemet is epeated a pecisely specified umbe of times. Fo example, i how may ways ca we aage the lettes i the wod P EP P ER? This is equal to the umbe of 6-pemutatios of the set {P, E, R} whee P is epeated 3 times, E is epeated 2 times, ad R is epeated 1 time. (Sice the total umbe of epetitios defies, we will use the tem pemutatios i place of -pemutatios fo the emaide of the sectio. Iitially, suppose that we make all the lettes distict by addig subscipts. That is, we wat the umbe of ways to aage the lettes P 1 E 1 P 2 P 3 E 2 R. I this case, thee ae 6! 720 aagemets because thee ae six choices fo the fist lette, five choices fo the secod lette, etc. Next, suppose that we ease the subscipts o the E s. This maps each aagemet of the lettes P 1 E 1 P 2 P 3 E 2 R to a aagemet of the lettes P 1 EP 2 P 3 ER. Sice E 1 ad E 2 could be odeed i 2! ways befoe the easue, the mappig is 2!-to-1. Fo example, we have: P 1 E 1 P 2 P 3 E 2 R P 1 EP 2 P 3 ER ad P 1 E 2 P 2 P 3 E 1 R P 1 EP 2 P 3 ER Theefoe, by the Divisio Rule thee ae 6!/2! 360 aagemet of the lettes i P 1 EP 2 P 3 ER. Fially, suppose that we ease the subscipts o the P s. This maps each aagemet of the lettes P 1 EP 2 P 3 ER to a aagemet of the lettes P EP P ER. Sice P 1, P 2, ad P 3 could be odeed i 3! ways befoe the easue, this mappig is 3!-to-1. Theefoe, by the Divisio Rule, the umbe of aagemets of the lettes P EP P ER is 6! 2! 3! 60. We ca pove a geeal theoem usig the same agumet as i the P EP P ER poblem. Theoem 7.2. Let A be the set {a 1, a 2,..., a }, ad let 1, 2,..., be o-egative iteges. The umbe of pemutatios of the set A whee each elemet a i is epeated exactly i times is: ( ! 1! 2!...! Fo example, the theoem says that the umbe of pemutatios of {P, E, R} whee P is epeated 3 times, E is epeated 2 times, ad R is epeated 1 time is This is the aswe we foud befoe. ( ! 60. 3! 2! 1!

24 24 Couse Notes 9: Pemutatios ad Combiatios Poof sketch. We iitially make i distict copies of each elemet a i. The the umbe of pemutatios whee each elemet is epeated exactly oce is ( !. We the ease the subscipts o the distict copies of elemet a 1. This defies a 1!-to-1 mappig fom old pemutatios to pemutatios whee a 1 is epeated 1 times ad all othe elemets ae epeated oce. Theefoe the umbe of ew pemutatios is ( !. 1! Cotiuig this way with a 2, a 3,..., we fid that the umbe of pemutatios whee each elemet a i is epeated exactly i times is. ( ! 1! 2!...! 7.3 Multiomial Coefficiets What is the umbe of pemutatios of the set A {a 1, a 2 } whee a 1 is epeated 1 times ad a 2 is epeated 2 times? Accodig to Theoem 7.2, the aswe is: ( 1 + 2! ! 2! 1 We ca estate the questio as follows. How may stigs cotai 1 copies of the symbol a 1 ad 2 copies of the symbol a 2? This is equal to the umbe ( of ways to choose 1 distict positios fo the a 1 s fom the set of all positios, which is This is the same as the pevious aswe. By shiftig fom pemutatios to stigs i this way, we ca give a alteative poof of Theoem 7.2. Alteative poof of Theoem 7.2. Thee is a bijectio betwee pemutatios of the set A {a 1, a 2,..., a } whee each elemet a i appeas exactly i times ad stigs whee each symbol a i appeas exactly i times. Theefoe, we ca cout pemutatios by coutig stigs as follows. The umbe ( of ways to choose 1 distict positios fo the a 1 s fom the set of all positios is The the umbe of ways to choose 2 distict positios fo the a 2 s fom the set of all emaiig positios is , ad so foth. The total umbe of stigs is theefoe: ( ! ( 2 + +! ( 3 + +!! 1! ( 2 + +! 2! ( 3 + +! 3! ( 4 + +!! 0! ( ! 1! 2!...!

25 Couse Notes 9: Pemutatios ad Combiatios 25 The fist equality uses the defiitio of biomial coefficiets, ad the secod follows by cacellig tems. The quatity ( ! 1! 2!...! is called a multiomial coefficiet ad is deoted , 2,..., Give a set with elemets, the multiomial coefficiet , 2,..., epesets the umbe of ways to choose 1 elemets, the 2 of the emaiig elemets, ad so foth. Multiomial coefficiets also aise i the Multiomial Theoem, a geealizatio of the Biomial Theoem. The esult is stated below, but ot poved. Theoem 7.3 (Multiomial Theoem. (x 1 + x x x 1 x 2... x 1, 2,..., Combiatios with Repetitio Now we move fom pemutatios with epetitio to combiatios with epetitio. Let S be the set {A, B, C}. As we saw i the pevious lectue, this set has thee 2-combiatios. That is, thee ae thee ways to choose two distict elemets of S whee ode does ot matte. The thee 2- combiatios of S ae show below. {A, B} {A, C} {B, C} Suppose that we ae ot equied to choose distict elemets of S, but athe ca choose the same elemet epeatedly. The esultig sets ae called the -combiatios with epetitio of the set S. Listed below ae the six 2-combiatios with epetitio of S. {A, B} {A, C} {B, C} {A, A} {B, B} {C, C} Stictly speakig, these ae multisets (bags, ot sets, sice a elemet may appea multiple times. 8.1 Coutig -Combiatios with Repetitio The followig theoem gives a ice fomula fo the umbe of -combiatios with epetitio of a -elemet set. Theoem 8.1. The umbe of -combiatios with epetitio of a -elemet set is + 1.

26 26 Couse Notes 9: Pemutatios ad Combiatios I the example above, we foud six ways to choose two elemets fom the set S {A, B, C} with epetitio allowed. Sue eough, the theoem says that the umbe of 2-combiatios of a 3-elemet set is ( Fo compaiso, ecall that the umbe of odiay -combiatios of a -elemet set is. Evey odiay -combiatio is also a valid -combiatio with epetitio. So, as oe would expect, the umbe of -combiatios with epetitio is geate if > 1. The poof of this theoem uses a impotat tick called stas ad bas. Poof. Let S be a set with elemets that ae odeed i some way. We will establish a bijectio betwee -combiatios with epetitio of the set S ad stigs of stas ad bas. Let R be a paticula -combiatio with epetitio of S. Wite dow 1 bas. These 1 bas divide the lie ito egios. } {{... } 1 bas defie egios Put oe sta i the i-th egio fo each time that the i-th elemet of S appeas i R. This pocedue maps each -combiatio with epetitio to a stig with stas ad 1 bas. (Fo example, let S be the set {A, B, C, D, E}, with elemets odeed alphabetically. Let R be the 7-combiatio with epetitio {A, B, B, B, D, E, E}. The stas-ad-bas stig coespodig to R is show below. }{{} } {{ } }{{} }{{} A B,B,B D E,E The two bas with o stas betwee idicate that elemet C eve appeas i R. This mappig is a bijectio because it has a ivese. That is, give ay stas-ad-bas stig, we ca costuct the coespodig -combiatio with epetitio. The umbe of stas i the fist egio detemies the umbe of times that the fist elemet of S appeas i the -combiatio, the stas i the secod egio detemie the umbe of times that the secod elemet appeas, etc. Sice the mappig is a bijectio, the umbe of -combiatios with epetitio of a -elemet set is equal to the umbe of stigs cotaiig 1 bas ad stas. The umbe of such stigs is equal to the umbe of ways to choose distict positios fo the stas i a stig of + 1 stas ( ad bas. This is the umbe of odiay -combiatios of a set with + 1 elemets, which is Tiple-Scoop Ice Ceam Coes Baski-Robbis is a ice ceam stoe that has 31 diffeet flavos. How may diffeet tiple-scoop ice ceam coes ae possible at Baski-Robbis? Two ice ceam coes ae cosideed the same if oe ca be obtaied fom the othe by eodeig the scoops. Of couse, we ae pemitted to have two o eve thee scoops of the same flavo. Of couse, the best solutio to this poblem is to go to the Baski-Robbis i Havad Squae ad explicitly costuct all possible combiatios. This is called the cosumptio method. Howeve, thee is also a puely mathematical appoach.