we have the probability for a given event to occur equals 1 the event will not occur. probability that

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1 76 Chapter 6 Buckets & Balls Suppose you walk to a room wth people. How may people do there eed to be the room before you bet that there are two people wth the same brthday? Of course f there were a thousad people, you would bet. It s a sure thg. But probably obody would bet agast you. How about f there are oly 100 people? 50? Before we fd the surprsg aswer, let's look at a smple prcple from probablty. Sce the sze of a set plus the sze of ts complemet s always the sze of the uverse (frst coutg prcple), or equvaletly, symbols, A U A U U U, we have the probablty for a gve evet to occur equals 1 the evet wll ot occur. probablty that Example 1. Let A be the set of people the room. We wat to compute the probablty that at least two of them have the same brthday. What s the expermet? We go aroud the room askg for people's brthdays. If A m, the there are 365 m (we have m decsos or stages ad 365 choces for each oe of them) outcomes to our expermet. It seems trcky to compute the sze of the outcomes that gve at least two wth the same brthday, but the complemet of ths set s that o two of them have the same brthday. How may ways ca that occur? For the frst perso we ask we have 365 choces, but for the secod oe we have oly 364, ad for the thrd oe 363, etc. ad ths kd of reasog should soud famlar. So we ca buld a table of probabltes. Let's say p deotes the probablty of o two persos havg the same brthday whe there are people the room. We ca wrte dow a aswer by smple coutg, ad we obta 365! p. (365 )!365 Ufortuately, ths may ot be easy to compute sce 365! has more tha 500 dgts. A much better way to do the computato s recursvely. Namely, at stage 1 we have 365, for 365, ad for the ext, prevous result the ext oe we have our prevous result tmes tmes 363, etc. I short, 365

2 p 1 p. 365 The table below gves the values recursvely computed. # of All NOT All people Thus, amazgly, you should be ready to bet whe there are oly 23 people the room! I a typcal classroom of 35 studets, the odds that at least two have the same brthday are better tha 4 to 1; ad at 40 people they are better tha 8 to 1 thus, we should have ot bee surprsed whe two Presdets had the same brthday. Wth 50 people the odds are better tha 32 to 1, ad f there are as may as 100 people the room your odds are astroomcal, better tha 3,000,000 to 1. I our ow Mathematcs Departmet of 40 faculty members, the odds were close to 10-to-1 tha 2 of us would have the same brthday. Ad that s the case. What was ot lkely s that we would be ext to each other alphabetcally, Mea ad Merryfeld. At oe tme, there were 3 of us wth the same brthday! Ad that was ot very lkely ether. A more sophstcated, but ot ay more elucdatg, way to vew the prevous problem s to realze that all we are askg for s the probablty of the brthday fucto from the set A of people to the set of dates of the year to be oe-to-oe. Ths way of vewg leads to may other terestg coutg problems some of whch we wll tackle the ext examples. Example 2. Suppose you are to dstrbute some moey amog your favorte relatves: Alphose, Bertrad ad Costace. I the frst stuato we are gog to look at, you have 5 blls: $1, $5, $10, $20 ad $50. I how may ways ca you dstrbute the moey? Ths s very easy: we are lookg at fuctos from the set 1,5,10, 20,50 to the set abc,,. The aswer to the questo s the Suppose we vary the questo a lttle bt, ad we ask how may ways ca we dstrbute the moey so that everybody gets somethg. Equvaletly, we could have started by askg what s the probablty that everybody wll receve

3 78 somethg gve that the moey wll be dstrbuted at radom. What we are coutg ow s the umber of oto fuctos from {1,5,10,20,50} to the set { a, b, c }. There are several ways to accomplsh ths coutg. Oe of them s to dvde the blls to three ples ad the assg the ples to the three people. To cout the last stage s easy, sce we have three ples of moey ad we are to dstrbute t to three people, there are 3! 6 ways to do that part. But how ca we dvde the blls to three oempty ples? Dog t by brute force, we see from the table below that there are 25 ways to break 5 thgs to 3 oempty ples. Sce each set of three ples gves rse to 6 oto fuctos (as we remarked the

4 79 prevous paragraph), we have a total of 150 oto fuctos from a 5-set to a 3-set. So the probablty of everybody gettg somethg s 150 =62% (whether Alphose should be paraod f he receves othg s a questo for moralsts to decde.) Observe also that we have grouped these 25 objects the table specal ways, frst havg grouped 7 of them, ad the the remag 18 were separated to 6 groups of 3. Frst we lsted the 7 whe was by tself. Where s the umber 7 comg from? It s the umber of ways of dvdg 4 objects to two ples, ad we wll call ths umber S (4,2). But because every tme we break up 4 objects to two ples we ca just break up 5 to 3 ples by makg the ew, ffth object to a ple of ts ow. What about the remag 18 ways? By ecessty, was ot by tself. So what we dd s take ay of the S (4,3) 6 ways of dvdg 4 objects to 3 ples, ad the added to ay of the 3 ples, gvg us a total of (see the vertcal groupgs the lst above). I other words, we have derved the fact that S(5,3) S(4,2) 3 S (4,3). Suppose you had ow te dfferet blls ad sx relatves. As usual we could get a mache to do the work, but we ca perhaps see a recursve relato comg o. It s clear that f we cout the umber of ways of breakg up the te blls to sx (udfferetated) ples, the every set of ples gves rse to 6!=720 oto fuctos, so the total umber of oto fuctos wll be 720 tmes the umber of ways of breakg up 10 thgs to 6 udfferetated ples. As above, let s call ths umber S (10,6). What about S (10,6)? Take the bll. It s ether by tself a ple or ot (we are parttog the collecto of ples to two peces.) How may sets of ples are there where t s by tself? A lttle reflecto shows there are S (9,5). If s ot by tself, what must we have doe? We took ay of the S (9,6) sets of ples, ad added to oe of the sx ples (ad there are 6 ways to do ths) so we have 6 S (9,6) sets of ples. I short, we have S(10,6) S(9,5) 6 S (9,6). Note that ths recurso s somewhat smlar to Pascal's recurso, ad the tme has come to get geeral. The umber of ways of dvdg m dfferet objects to exactly (oempty) udfferetated ples s called a Strlg umber of the secod kd, ad s deoted by S( m, ). We wll have o use for Strlg umbers of the frst kd (they are smlar), ad these umbers are amed after the Brtsh mathematca of the 18 th cetury (a studet ad colleague of Newto's). If we just abstract the argumet for 10 ad 6 above for ay m ad, we get 243

5 80 Theorem. Strlg Recurso. Let m 2. The S( m, ) S( m 1, 1) S( m 1, ). We should remark that clearly S( m, ) 0 f m, ad also order to start buldg our table, we eed Sm (,1) 1 ad S( m, m ) 1 (are these clear?). I the table we compute the frst twelve rows of the Strlg umbers of the secod kd usg the recurso: Hece S (10,6) Multplyg by 720, we get that there are 16,435,440 oto m/ Strlg Numbers of the Secod Kd fuctos from a 10-set to a 6-set, ad sce there are 60,466,176 fuctos, the probablty of a fucto beg oto s roughly 27% (Alphose should't get paraod uder these crcumstaces.) Corollary. Oto Fuctos. Let A have m elemets ad let B have elemets. The the umber of oto fuctos from A to B s! S( m, ). But we ca cout the umber of oto fuctos aother way. Let's go back to our three favorte relatves ad our 5 dfferet blls. Let's cosder our uverse of 243 fuctos, ad let A be the set of them whch Alphose receves othg, B the oes whch Bertrad receves othg ad the same wth C ad Costace. The what we are tryg to do s fd A B C ad we ca use the cluso-excluso prcple. We eed to cout some thgs. Frst A 32, sce t s the umber of fuctos from a 5-set to a 2- set. Smlarly, B C 32. Trvally, A B 1 (f ether Alphose receves moey, or Bertrad..., or equvaletly, how may fuctos from a 5-set to a 1-set), A C B C 1 also. Fally, A B C 0, so A B C

6 81 Hece, A B C Just as before. Ths type of reasog ca obvously be exteded, ad we wll ask you to do some of that the homework. Let's vary Example 2 a lttle bt. Example 3. Oto Fuctos Aga. Suppose we wated to cout the umber of oto fuctos from a 8 set to a 5 set. The aswer s of course 5! S(8,5) ,000. But attackg the problem from the cluso-excluso pot of vew, we let the codoma be {1,2,3,4,5}, for ay subset S of ths set, we let A S be the set of fuctos whose rage cotas o elemet of S. Our uverse cossts of all 8 fuctos from a 8 set to a 5 set so t has 5 390,625 elemets. Clearly, the sze of 8 A S s 5 S. We are terested coutg A1 A2 A3 A4 A 5. By clusoexcluso: A1 A2 A3 A4 A 5 5(4 ) 10(3 ) 10(2 ) 5 264,625 so our aswer s the complemet so t s as promsed. Example 4. Suppose ow we are stll gog to dstrbute some moey to Alphose, Bertrad ad Costace. But ow what we have s 5 crsp, ew $20 blls. How may ways do we have of dog ths? As usual, frst brute force. Here what matters s how may blls each Al, Bert ad Coe receve. We are gog to let ( x, y, z ) stad for a dstrbuto where x s the umber of blls Al receved, y the umber Bert got ad z how may Coe receved. So x, y ad z are oegatve whole umbers that add up to 5. The possbltes are the (5,0,0) (4,1,0) (4,0,1) (3,2,0) (3,0,2) (3,1,1) (2,2,1) (0,5,0) (0,4,1) (1,4,0) (0,3,2) (2,3,0) (1,3,1) (1,2,2) (0,0,5) (1,0,4) (0,1,4) (2,0,3) (0,2,3) (1,1,3) (2,1,2) So there are 21 ways. If we just try to buld a tree we wll soo otce that the umber of braches at a stage does deped o prevous choces. However, whe vewed the rght way, ths problem s ot very hard. Thk of leavg structos for somebody to delver your presets. Oe easy way s havg two peces of whte paper ad puttg them amog your blls. The you smply leave the structos: Al gets all the blls utl the frst pece of paper, Bert gets the ext group ad fally Coe gets the remag blls. Thk of the peces of whte paper as separators, so total we have seve peces of paper, ad of the seve postos where oe ca put the separators, oe has to choose two for them, so the aswer s Whch s correct! So smple t s deceptve. Let's look at a few examples. Of the umbers {1,2,3,4,5,6,7} we are gog to choose two to correspod to the postos for the separators (thk of the blls ad separators as occupyg those 7 postos.) Thus, {1,2} correspods to the frst two postos, hece Al gets 0, Bert gets 0 ad Coe gets 5. What about the subset {3,6}? Here the frst two 2

7 82 postos your stack are occuped by blls, so Al s gog to get 2, the ext stack also cotas 2 blls ad that s Bert's share ad fally Coe gets 1. Gog reverse, suppose you wated Al to get 1, Bert to get 1 ad Coe to get 3. Where are the separators? {2,4}. Ad so we have a correspodece betwee the places for the separators ad the dstrbutos. Suppose that t s requred that everybody gets at least oe bll. How may ways ca we do t ow? Ths s easy. Gve everybody a bll. We have two left. Use the same scheme wth the separators (we stll have two), so ow we have a total of four postos (2 for blls & 2 for separators), ad we are to choose 2 for the separators, so we ca do t ways: (3,1,1), (1,3,1), (1,1,3), (2,2,1), (2,1,2) ad (1,2,2). What the prevous examples all have commo s the dea of dstrbutg balls buckets. I some stuatos, the balls all have dfferet colors, so they are dfferetated (Examples 1 ad 2); others they are all the same color (Example 3). The same apples for the buckets where, for example, for the Strlg umbers they were ot dfferetated. Fally, sometmes we wated at most oe ball ay bucket (Example 1) ad sometmes we wated at least oe ball every bucket (Example 2). Ths leads to a potetal 16 dfferet stuatos, ot all of whch are terestg. Below we wll buld a table hghlghtg the most mportat oes. We wll be usg the followg otato: d or D for dfferetated, u or U for udfferetated. We wll be usg a captal letter for the balls whe we wat at most oe ball ay bucket, ad a captal letter for the buckets meas we wat at least oe ball ay bucket. But before we go to our table, ad before we look at the (u,u) stuato, we should ecourage the reader to realze that ths cosderato of dstrbutg balls to buckets s more tha just a gmmck, t s a actual physcal model of the stuatos we wll ecouter, ad that two essetal gredets the physcal model should be kept md at all tmes: frst, that a ball caot go to two buckets (but a bucket ca receve two balls), ad secod, that oe must dstrbute all the balls (but that a bucket may rema empty). Example 5. Suppose we have 5 black balls, ad we wat to dstrbute them 3 gray buckets. Our acroym hece would be ( uu, ) where m 5 ad 3. How may ways ca ths be doe? Oly 5 ways: 5 oe bucket; 4 oe, 1 aother; 3 ad 2; 3 ad 1 ad 1; 2 ad 2 ad 1. It s clear that what we have here are the parttos of 5 at most 3 peces. Suppose we had bee after ( uu,, ) so that every bucket had to have at least oe ball? The the aswer would have bee 2: ad The aswer to ( Uu, ) s 0 sce there s o way to dstrbute 5 balls to 3 buckets wth at most oe ball per bucket. We wll retur to parttos a latter secto, but a way, they are behd a lot of the coutg, ad oe ca always use them. However, they ofte lead to tractable ways of coutg due to ther large umber. Hece we suggest you do't get too fascated by ths way of coutg, but sometmes t s uavodable. Let's revst the examples of ths secto by vewg them terms of parttos.

8 83 I Example 5, we had udfferetated balls ad udfferetated buckets. Each partto just gave us oe arragemet. Acroym ( uu, ) : I Example 3, we had udfferetated balls, but dfferetated buckets. Hece a partto of the balls, depedg o ts ature, wll gve a dfferet umber of arragemets. For example, the partto gves 3 arragemets sce there are 3 ways of choosg the bucket that gets oly oe ball. Acroym ( ud, ) : 5 = = = = = = = = = = 3 21 I the latter part of Example 2, the balls were dfferetated but the buckets were't. I addto we requred that every bucket have at least oe ball. Thus oly the parttos to exactly three parts are relevat: ad Each partto, aga, by ts ature, gves us a dfferet umber of arragemets. For example, gves 15 dfferet arragemets: 5 3 where 5 s the umber of ways of choosg the ball that s gog to be aloe, ad 3 s the umber of ways of dvdg the remag 4 balls to two groups of 2. Acroym (d,u): = = Fally the begg of Example 2 we were terested a stuato where the balls ad the buckets were dfferetated (ths s the oly case whe you are dealg wth fuctos). Ad aga every partto gves us a dfferet umber. For example, gves 10*3*2=60 arragemets where 10 s the umber of ways of choosg the 3 balls that wll be together, 3 s the umber of ways of choosg the bucket they wll go to, ad 2 s the umber of ways of dstrbutg the other 2 balls. Acroym (d,d): 5 = = = = =

9 84 Acroy m (d,d) (D,d) (d,u) Summary Table We wll assume we have m balls to be dstrbuted amog buckets. Objects that we are coutg Fuctos from the set of balls to the set of buckets Oe-to-oe fuctos from the balls to the buckets Ways of dvdg m balls to oempty udfferetated ples Cout ( m! m)! S( m, ) (d,d) Oto fuctos! S( m, ) (u,d) Sometmes called selectos: ways of dstrbutg m detcal objects to (at most) dfferetated peces m 1 m 1 The oly oe we should clarfy further s the last oe. It s Example 3. We have detcal blls to be dstrbuted amog relatves. I the geeral stuato we have m blls ad relatves. Usg separators, the oly questo s how may of them do we eed? A lttle thought wll covce that we eed oe less separator tha relatves, therefore 1. So total we have m 1 postos for all the peces of paper, ad we have to choose 1 of them for the separators. Equvaletly we could have chose the m postos for the blls. Ths table should ot be memorzed. What s mportat s the reasog behd t, ad what s sometmes hard s to decde whch are the balls ad whch are the buckets, ad whether they are dfferetated, or somethg else. We wll fll some of the holes the table the exercses. I a prevous chapter we studed the bomal theorem. The Bomal Theorem. ( x y) x y Let us revst t. The followg specal case of the theorem deserves recogto: Corollary. (1 x) x. 0 Ths equato s, aga, a algebrac detty so t leds tself to all kds of mapulatos ad substtutos. For example, we ca dfferetate (wth respect to x) to obta a ew detty: thus f we let 6, we have 1 1 (1 x) x, m

10 x 3 x 4 x 5 x 6 x x 60x 60x 30x 6x 6( 1 x). We wll be seeg some more of these substtutos ad mapulatos the exercses. 6 Suppose we ow we wat to compute ( x y z ) stead. So we would be lookg at ( x y z)( x y z)( x y z)( x y z)( x y z)( x y z ). Here the expaso would be real tedous. But the reasog we dd the bomal theorem s stll vald. Namely as we expad ths product, order to buld a term we get j oe summad out of each of the factors, thus our terms are of the form x y z k where, 2 3 j ad k are oegatve tegers ad j k 6. For example, x y z ad x 4 z 2 are both terms. How may terms wll we have? 6 stages, 3 optos for each decso gve us a 6 total of terms! O the other had, there are very few types of terms: x 6, y 6, z 6, x 5 y, x 5 z,. How may types are there? Let's cout them smartly: how may ways ca we partto 6 to at most 3 peces: 6+0+0, 5+1+0, 4+2+0, 4+1+1, 3+3+0, 3+2+1, ad The partto gves rse to 3 types x 6, y 6, ad z 6. I cotrast, gves rse to 6 types: x 5 y, x 5 z, xy 5, xz 5, y 5 z ad yz 5. The dfferece stems from the fact that for the frst partto all we had to do s assg a varable to the 6 ad the the other two varables would get a 0 (1 decso, 3 optos for that decso). But for 5+1+0, we had two decsos: whch varable gets the 5 ad whch varable gets the 1: 3 optos for the frst oe, 2 for the secod oe gve us 6 types. Sce ths le of reasog has become route, we just proceed by lstg the umber of types each of the parttos wll gve us: 4+2+0, 6 types; 4+1+1, 3 types; 3+3+0, 3 types; 3+2+1, 6 types; 2+2+2, 1 type. For a total of 28 types, All we eed ow s the coeffcet for each of the types. Take for example, x y z. How ca such a term arse? It must have bee that three of the factors cotrbuted x s, two cotrbuted y s ad the remag factor cotrbuted a z. We have three decsos: frst to choose the three factors that cotrbute x s we have 6 3 The we have to pck the y s: there are three factors remag: 3 =20 ways of dog that stage. 2 =3 ad fally the last stage s for the z, but there s oly oe opto left. So we have , so that s the coeffcet of x y z. It should be clear that x yz 3 has also 60 for ts coeffcet, or ay of the types stemmg from the partto j k Let s reaso t geeral: take x y z cotrbute x's ad we have 6 where j k 6. The factors have to ways of dog that choosg. The from the remag

11 86 factors we have to decde for the y's: 6 j sce j k 6. Hece the coeffcet of x y z k s j ways, ad fally, for the z's: 6 j k j 6! ( 6 )! ( 6 j)! j k!(6 )! j!(6 j)! k!0! by cacelg. Ths s a very satsfyg expresso: 6!! j! k! 1 These coeffcets are called multomal coeffcets. Ad we mmedately test whether our deas geeralze to ay expaso. Suppose we are 11 j k dog ( x y z w ). What s the coeffcet of x y z w m? If the expoets do ot add up to 11, the coeffcet s 0. If j k 11, the we have 11 factors that wll cotrbute the x's, the 11 ways for the y's, ad 11 j j k ad the remag factors all have to cotrbute w's. So total we have, ways of choosg the for the z's, j 11 j k 11! (11 )! (11 j)! (11 j k)! j k m!(11 )! j!(11 j)! k!(11 j k)! m! whch reduces cely to 11!! j! k! m! by cacellato ad the fact that j k m 11. As a llustrato we fsh computg ( x y z) x y z 6 ( x y z ) : x y 6x z 6xy 6xz 6y z 6yz x y 15x z 15x y 15x z 15y z 15y z x yz 30xy z 30xyz 20x y 20x z 20y z x y z 60x yz 60x y z 60x yz 60xy z 60xy z 90x 2 y 2 z 2. Wth exactly =729 terms as expected. I geeral, f m1 m2 mt s ay partto of, the we ca defe the multomal coeffcet of that partto by! m1, m2,, mt m1! m2! mt!. These are atural geeralzatos of the bomal coeffcets, ad deed f we let t 2, so m1 m2, the

12 87 m, m! m! m! m m Ad combatorally, they are also very close. We developed the bomal coeffcets by choosg a gve umber of freds for ether of two dfferet actvtes: comg to the party or ot comg to the party. What we are dog the multomal coeffcets s choosg specfc umbers of freds for each of several, dfferet actvtes. We have motvated (although ot formally proved too borg) Theorem. Multomal Theorem. x x x x x x. m1 m2 1 2 t 1 2 m,,, 1 m2 mt m m m 1 2 t As before, the otato ca be tmdatg, but we ca always just wrte oe term at a tme, as we dd above. Example 6. A aagram of a word s a rearragemet of ts letters. Thus, my ame MENA has 24 aagrams sce there are 4! arragemets of 4 objects. Not all are meagful words Eglsh, but that s rrelevat. BOB o the other had does ot have 3!=6 aagrams, oly 3: OBB, BOB ad BBO. The reaso beg that whe we compute the 6 ways we are coutg the B's as dfferet but they are ot. Smlarly, my frst ame ROBERT does ot have 6! aagrams because of double coutg: there are two R's that are beg couted as dfferet the 6!, but do ot gve dfferet words. Let's do t aother way. Thk of the sx blaks that we are gog to fll wth the letters:{r, O, B, E, R, T}: _. Of those 6 blaks, 2 have to go for R's. Choose those 2, for whch we have ways of dog. The choose the spot for the O's : ay of 4 ways. The we have 3 spots left to put the B, 2 for the E ad 1 for the T. Thus the umber of aagrams s: You ca thk of obtag the result terms of actvtes: you have sx blaks: 2 of them are gog for R, 1 for O, 1 for B, 1 for E ad 1 for T. By the same reckog, MISSISSIPPI has 11 4, 4, 2, 1 34, 650 aagrams. Of course, oe ca have varatos o the questos. How may aagrams of ROBERT are there where the vowels are together? Put the vowels together: there are 2 ways of dog that, ad the thk of them as oe letter: V ad we have aagrams of RBRTV, of whch there are 60, so our aswer s 120. Aother varato: how may aagrams are there of ROBERT where the vowels come alphabetcal order (ot ecessarly together). There are 360 aagrams, ad they come par where the vowels occur the same locatos, oly oe of those two has the vowels mt t

13 88 order, so the aswer s There s yet aother mportat way to thk of the multomal coeffcets. We have bee dscussg the umber of ways of dstrbutg balls buckets uder dfferet crcumstaces depedg o whether the balls (or the buckets) were all dfferet colors or all the same. What we have ow s the possblty of repeated colors. Namely, for aagrams, thk of the letters as the balls, the alphabet beg the colors, ad the blaks are dfferetated buckets. So we are dstrbutg balls (some the same, some ot) to buckets of dfferet colors, wth at most oe ball per bucket. I ths stuato, the multomal coeffcets are relevat. Example 7. I how may ways ca we break a group of 10 people to two groups of 2 ad 2 groups of 3. We ca use the aagram model to start ths problem. We have two A s, 2 E s, 3 B s ad 3 C s. So we could mpulsvely aswer 10 10!. But ths s 2,2,3,3 2!2!3!3! ot qute correct because the groups of two are ot dstgushable from each other, ad ether are the groups of three, so we have to dvde our aswer by 2 2 4, so the evetual aswer s 6,300. To dvde 12 people to four groups of 3, we have 12! 3!3!3!3!4! 15,400.