Explorig Pascal s Triagle Tom Davis tomrdavis@earthlik.et http://www.geometer.org/mathcircles November 7, 006 Abstract This article provides material to help a teacher lead a class i a adveture of mathematical discovery usig Pascal s triagle ad various related ideas as the topic. There is plety of mathematical cotet here, so it ca certaily be used by ayoe who wats to explore the subject, but pedagogical advice is mixed i with the mathematics. Geeral Hits for Leadig the Discussio The material here should ot be preseted as a lecture. Begi with a simple defiitio of the triagle ad have the studets look for patters. Whe they otice patters, get them to fid proofs, whe possible. By proof we do ot ecessarily mea a rigorous mathematical proof, but at least eough of a argumet that it is covicig ad that could, i priciple, be exteded to a rigorous proof. Some sample argumets/proofs are preseted below, but they represet oly oe approach; try to help the studets fid their ow way, if possible. 3 3 6 5 0 0 5 It is ot critical to cover all the topics here, or to cover them i ay particular order, although the order below is reasoable. It is importat to let the Figure : Pascal s Triagle ivestigatio cotiue i its ow directio, with perhaps a little steerig if the class is ear somethig very iterestig, but ot quite there. The umbers i Pascal s triagle provide a woderful example of how may areas of mathematics are itertwied, ad how a uderstadig of oe area ca shed light o other areas. The proposed order of presetatio below shows how real mathematics research is doe: it is ot a straight lie; oe bouces back ad forth amog ideas, applyig ew ideas back to areas that were already covered, sheddig ew light o them, ad possibly allowig ew discoveries to be made i those old areas. Fially, the material here does ot have to be preseted i a sigle sessio, ad i fact, multiple sessios might be the most effective presetatio techique. That way there s some review, ad the amout of ew material i each sessio will ot be overwhelmig. Basic Defiitio of Pascal s Triagle Most people are itroduced to Pascal s triagle by meas of a arbitrary-seemig set of rules. Begi with a o the top ad with s ruig dow the two sides of a triagle as i figure. Each additioal umber lies betwee
two umbers ad below them, ad its value is the sum of the two umbers above it. The theoretical triagle is ifiite ad cotiues dowward forever, but oly the first 6 lies appear i figure. I the figure, each umber has arrows poitig to it from the umbers whose sum it is. More rows of Pascal s triagle are listed o the fial page of this article. A differet way to describe the triagle is to view the first lie is a ifiite sequece of zeros except for a sigle. To obtai successive lies, add every adjacet pair of umbers ad write the sum betwee ad below them. The o-zero part is Pascal s triagle. 3 Some Simple Observatios Now look for patters i the triagle. We re iterested i everythig, eve the most obvious facts. Whe it s easy to do, try to fid a proof (or at least a covicig argumet) that the fact is true. There are probably a ifiite umber of possible results here, but let s just look at a few, icludig some that seem completely trivial. I the examples below, some typical observatios are i bold-face type, ad a idicatio of a proof, possibly together with additioal commets, appears afterwards i the stadard fot. All the umbers are positive. We begi with oly a positive, ad we ca oly geerate umbers by icludig additioal s, or by addig existig positive umbers. (Note that this is really a iductive proof, if writte out formally.) The umbers are symmetric about a vertical lie through the apex of the triagle. The iitial row with a sigle o it is symmetric, ad we do the same thigs o both sides, so however a umber was geerated o the left, the same thig was doe to obtai the correspodig umber o the right. This is a fudametal idea i mathematics: if you do the same thig to the same objects, you get the same result. Look at the patters i lies parallel to the edges of the triagle. There are ice patters. The oe that is perhaps the icest example is the oe that goes:, 3, 6, 0, 5,,... These are just the sums: (), ( + ), ( + + 3), ( + + 3 + ), et cetera. A quick examiatio shows why the triagle geerates these umbers. Note that they are sometimes called triagular umbers sice if you make a equilateral triagle of cois, for example, these umbers cout the total umber of cois i the triagle. I fact, the ext row:,, 0, 0, 35,... are called the pyramidal umbers. They would cout the umber of, say, caoballs that are stacked i triagular pyramids of various sizes. Is it clear why addig triagular umbers together give the pyramidal umbers? Is it clear how Pascal s triagle succeeds i addig the triagular umbers i this way? I the same vei, if those rows represet similar couts i ad 3 dimesios, should t the first two rows somehow represet couts of somethig i 0 ad dimesios? They do ad this is could be a ice segue ito the behavior of patters i ad higher dimesios. If you add the umbers i a row, they add to powers of. If we thik about the rows as beig geerated from a iitial row that cotais a sigle ad a ifiite umber of zeroes o each side, the each umber i a give row adds its value dow both to the right ad to the left, so effectively two copies of it appear. This meas that whatever sum you have i a row, the ext row will have a sum that is double the previous. It s also good to ote that if we umber the rows begiig with row 0 istead of row, the row sums to. This serves as a ice remider that x 0, for positive umbers x.
If you alterate the sigs of the umbers i ay row ad the add them together, the sum is 0. This is easy to see for the rows with a eve umber of terms, sice some quick experimets will show that if a umber o the left is positive, the the symmetric umber o the right will be egative, as i: 5 + 0 0 + 5. Oe way to see this is that the two equal umbers i the middle will have opposite sigs, ad the it s easy to trace forward ad back ad coclude that every symmetric pair will have opposite sigs. It s worth messig aroud a bit to try to see why this might work for rows with a odd umber. There are probably lots of ways to do it, but here s a suggestio. Look at a typical row, like the fifth: + 5 + 0 0 + 5. We d like the ext row (the sixth, i this case) to look like this: + 6 + 5 0 + 5 6 +. If we give letter ames to the umbers i the row above it: a +; b 5; c +0; d 0; e +5; f, the how ca we write the elemets i row 6 i terms of those i row 5? Here s oe ice way to do it: + a 0; 6 b a; +5 c b; 0 d c; +5 e d; 6 f e; + 0 f. Now just add the terms: a 0 + b a + c b + d c + e d + f e + 0 f, ad the sum is obviously zero sice each term appears twice, but with opposite sigs. The hockey-stick rule : Begi from ay o the right edge of the triagle ad follow the umbers left ad dow for ay umber of steps. As you go, add the umbers you ecouter. Whe you stop, you ca fid the sum by takig a 90-degree tur o your path to the right ad steppig dow oe. It is called the hockey-stick rule sice the umbers ivolved form a log straight lie like the hadle of a hockeystick, ad the quick tur at the ed where the sum appears is like the part that cotacts the puck. Figure illustrates two of them. The upper oe adds + + + + to obtai 5, ad the other adds + + 0 + 0 to obtai 35. (Because of the symmetry of Pascal s triagle, the hockey sticks could start from the left edge as well.) 3 3 6 5 0 0 5 6 5 0 5 6 7 35 35 7 Figure : The Hockey Stick To see why this always works, ote that whichever you start with ad begi to head ito the triagle, there is a i the other directio, so the sum starts out correctly. The ote that the umber that sits i the positio of the sum of the lie is always created from the previous sum plus the ew umber. Note how this relates to the triagular ad pyramidal umbers. If we thik of pyramids as three-dimesioal triagles ad of lies with,, 3,,... items i them as oe dimesioal triagles, ad sigle items as a zerodimesioal triagle, the the sum of zero-dimesioal triagles make the oe dimesioal triagles, the sum of the oe-dimesioal triagles make the two-dimesioal triagles ad so o. With this iterpretatio, look at the diagoals of Pascal s triagle as zero, oe, two, three,... dimesioal triagles, ad see how the hockey-stick rule adds the items i each diagoal to form the ext diagoal i exactly the maer described above. 3
There are iterestig patters if we simply cosider whether the terms are odd or eve. See figure 3. I the figure, i place of the usual umbers i Pascal s triagle we have circles that are either black or white, depedig upo whether the umber i that positio is odd or eve, respectively. Look at the geeral patter, but it is also iterestig to ote that certai rows are completely black. What are those row umbers? They are rows 0,, 3, 7, 5, 3, ad each of those umbers is oe less tha a perfect power of. How could you possibly prove this? Well, oe approach is basically recursive: Notice the triagles of eve umbers with their tips dow. Clearly, sice addig eves yields a eve, the iteriors will remai eve, but at the edges where they re up agaist a odd umber, the width will gradually decrease to a poit. Now look at the little triagle made from the four rows 0 through 3. At the Figure 3: Odd-Eve Pascal s Triagle bottom, you ve got all odd umbers, so the ext lie will be all eve, except for the other edges. The outer edges must look like two copies of the iitial triagle util they meet. Oce you ve got all odd, we ow have the shape of the triagle made of the first 8 rows, ad the ext step is two odds at the ed, with eves solidly betwee them. The argumet repeats, but with triagles of twice the size, et cetera. There s othig special about odd-eve; the same sorts of ivestigatios ca be made lookig for multiples of other umbers. 3 3 6 5 0 0 5 3 5 8 The Fiboacci sequece is hidde i Pascal s triagle. See figure. If we take Pascal s triagle ad draw the slatig lies as show, ad add the umbers that itersect each lie, the sums tur out to be the values i the Fiboacci series:,,, 3, 5, 8, 3,, 3, 55, 89,, 33, 377,.... The first two umbers are ad every umber after that is simply the sum of the two previous umbers. Oe argumet to covice yourself that this is true Figure : Fiboacci Series is to ote that the first two lies are OK, ad the to ote that each successive lie is made by combiig exactly oce, each of the umbers o the previous two lies. I other words, ote that the sums satisfy exactly the same rules that the Fiboacci sequece does: the first two sums are oe, ad after that, each sum ca be iterpreted as the sum of the two previous sums.
Pascal s Triagle ad the Biomial Theorem Most people kow what happes whe you raise a biomial to iteger powers. The table below is slightly uusual i that coefficiets of are icluded sice it will be the coefficiets that are of primary importace i what follows: (L + R) 0 (L + R) L + R (L + R) L + LR + R (L + R) 3 L 3 + 3L R + 3LR + R 3 (L + R) L + L 3 R + 6L R + LR 3 + R (L + R) 5 L 5 + 5L R + 0L 3 R + 0L R 3 + 5LR + R 5 A quick glace shows that the coefficiets above are exactly the same as the umbers i Pascal s triagle. If this is geerally true, it is easy to expad a biomial raised to a arbitrary power. If we wat to deal with (L + R), we use as coefficiets the umbers i row of Pascal s triagle. (Note agai why it is coveiet to assig the first row the umber zero.) To the first coefficiet, we assig L, ad for each successive coefficiet, we lower the expoet o L ad raise the expoet o R. (Note that we could have said, assig L R 0 to the first coefficiet.) The expoet o L will reach 0 ad the expoet o R will reach just as we arrive at the last coefficiet i row of Pascal s triagle. OK, but why does it work? The easiest way to see your way through to a proof is to look at a couple of cases that are ot too complex, but have eough terms that it s easy to see patters. For the example here, we ll assume that we ve successfully arrived at the expasio of (L + R) ad we wat to use that to compute the expasio of (L + R) 5. The brute-force method of multiplicatio from the algebra class is probably the easiest way to see what s goig o. To obtai (L + R) 5 from (L + R), we simply eed to multiply the latter by (L + R): L + L 3 R + 6L R + LR 3 + R L + R L R + L 3 R + 6L R 3 + LR + R 5 L 5 + L R + 6L 3 R + L R 3 + LR L 5 + 5L R + 0L 3 R + 0L R 3 + 5LR + R 5 () I the multiplicatio illustrated i equatio () we see that the expasio for (L+R) is multiplied first by R, the by L, ad the those two results are added together. Multiplicatio by R simply icreases the expoet o R by oe i each term ad similarly for multiplicatio by L. I other words, before the expressios are added, they have the same coefficiets; the oly thig that has chaged are the values of the expoets. But otice that the two multiplicatios effectively shift the rows by oe uit relative to each other, so whe we combie the multiplicatios of the expasio of (L + R) by L ad R, we wid up addig adjacet coefficiets. It s ot too hard to see that this is exactly the same method we used to geerate Pascal s triagle. But oce we re coviced that the biomial theorem works, we ca use it to re-prove some of the thigs we oticed i sectio 3. For example, to show that the umbers i row of Pascal s triagle add to, just cosider the biomial theorem expasio of ( + ). The L ad the R i our otatio will both be, so the parts of the terms that look like L m R are all equal to. Thus ( + ) is the sum of the umbers i row of Pascal s triagle. Similarly, to show that with alteratig sigs the sum is zero, look at the expasio of ( ) 0. 5
5 A Applicatio to Arithmetic A possible itroductio to the previous sectio might be to have the class look at powers of : 0 3 33 6 5 605 6 7756 It s iterestig that up to the fourth power, the digits i the aswer are just the etries i the rows of Pascal s triagle. What is goig o, of course, is that 0 +, ad the aswers are just (0 + ), for various. Everythig works great util the fifth row, where the etries i Pascal s triagle get to be 0 or larger, ad there is a carry ito the ext row. Although Pascal s triagle is hidde, it does appear i the followig sese. Cosider the fial umber, 6 : (0 + ) 6 0 6 000000 + 6 0 5 600000 + 5 0 50000 + 0 0 3 0000 + 5 0 500 + 6 0 60 + 0 0 7756 By shiftig the colums appropriately, the umbers i ay row of Pascal s triagle ca be added to calculate, by usig the umbers i row. Could similar ideas be used to calculate 0 or 00? 6 Combiatorial Aspects of Pascal s Triagle Before goig ito the theory, it s a good idea to look at a few cocrete examples to see how oe could do the coutig without ay theory, ad to otice that the couts we obtai from a certai type of problem (called combiatios ) all happe to be umbers that we ca fid i Pascal s triagle. Let s start with a easy oe: How may ways are there to choose two objects from a set of four? It does t take too log to list them for some particular set, say {A, B, C, D}. After a little searchig, it appears that this is a complete list: AB, AC, AD, BC, BD, CD. The first time studets try to cout them, it s ulikely that they ll come up with them i a logical order as preseted above, but they ll search for a while, fid six, ad after some futile searchig, they ll be coviced that they ve got all of them. The obvious questio is, How do you kow you ve got them all? 6
There are various approaches, but oe might be somethig like, We ll list them i alphabetical order. First fid all that begi with A. The all that begi with B, ad so o. Try a couple of others; say, 3 objects from a set of 5. The set is {A, B, C, D, E} ad here are the 0 possible groups of objects (listed i alphabetical order): ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE Note that the strategy still works, but we have to be careful sice eve while we re workig o the part where we fid all triples that start with A, we still have to fid all the pairs that ca follow. Note that this has, i a sese, bee solved i the previous example, sice if you kow you re begiig with A, there are four items left, ad the previous exercise showed us that there are six ways to do it. Now cout the umber of ways to choose items from a set of five. Use the same set: {A, B, C, D, E}, ad here are the 0 results: AB, AC, AD, AE, BC, BD, BE, CD, CE, DE Is it just luck that there are the same umber of ways of gettig 3 items from a set of 5 ad items from a set of 5? A key isight here is that if I tell you which oes I m ot takig, that tells you which oes I am takig. Thus for each set of items, I ca tell you which they are, or, which 3 they are t! Thus there must be the same umber of ways of choosig from 5 or 3 from 5. Obviously, the same thig will hold for ay similar situatio: there are the same umber of ways to pick thigs out of 7 as there are to pick 6 out of 7, ad so o. This is the sort of thig a mathematicia would call duality. The geeral statemet is this: There are the same umber of ways to choose k thigs from as there are to choose k thigs from, assumig that k. After you ve looked at a few simple situatios, it s easy to get a lot of other examples. The easiest is: How may ways are there to pick item from a set of? The aswer is obviously. Ad from the previous paragraphs, there are also ways to choose items from a set of. A slightly more difficult cocept is this: How may ways are there to choose 0 (zero) items from a set of. The correct aswer is always there is a sigle way to do it: just pick othig. Or aother way to look at it is that there s clearly oly oe way to choose all items from a set of : take all of them. But the duality cocept that we ve just cosidered would imply that there are the same way to choose items from as 0 items from. After lookig at a few of these, we otice that the couts we obtai are the same as the umbers we fid i Pascal s triagle. Not oly that, but, at least for the few situatios we ve looked at, the umber of ways to choose k thigs from a set of seems to be the umber i colum k (startig the colum cout from zero) ad i row (agai, startig the row cout from zero). The oly etry that might seem a little strage is the oe for row zero, colum zero, but eve the, it ought to be, sice there s really oly oe way to choose o items from a empty set: just take othig. With this ecouragemet, we ca try to see why it might be true that combiatios ad the umbers i Pascal s triagle are the same. First, a little otatio. I order to avoid sayig over ad over somethig like, the umber of ways to choose k objects from a set of objects, we will simply say choose k. There are various ways to write it, but ( choose k) works, with the paretheses idicatig a groupig. The most commo form, of course, is that of the biomial coefficiet: ( k), which will tur out to be the same thig. So from our previous work, we ca say that (5 choose ) (5 choose 3) 0, or, alteratively, ( ( 5 ) 5 3) 0. Here s oe way to look at it: We ll examie a special case ad see why it works. The, if we look at the special umbers we ve chose, we ll see that there is othig special at all about them, ad the geeral case is just a 7
particular example. Suppose we eed to fid out how may ways there are to choose thigs from a set of 7, ad let s say that we ve already somehow worked out the couts for all similar problems for sets cotaiig 6 or fewer objects. For cocreteess, let s say that the set of 7 thigs is {A, B, C, D, E, F, G}. If we cosider the sets of four items that we ca make, we ca divide them ito two groups. Some of them will cotai the member A (call this group ) ad some will ot (group ). Every oe of the sets i group has a A plus three other members. Those additioal three members must be chose from the set {B, C, D, E, F, G} which has six elemets. There are (6 choose 3) ways to do this, so there are (6 choose 3) elemets i group. I group, the elemet A does ot appear, so the elemets of group are all the ways that you ca choose items from a set of the remaiig 6 objects. Thus there are (6 choose ) ways to do this. Thus: (7 choose ) (6 choose 3) + (6 choose ) or, usig the biomial coefficiets: ( ) 7 ( ) 6 + 3 ( ) 6. Now there s clearly othig special about 7 ad. To work out the value of ( choose k) we pick oe particular elemet ad divide the sets ito two classes: oe of subsets cotaiig that elemet ad the other of subsets that do ot. There are ( choose k ) ways to choose subsets of the first type ad ( choose k) ways to choose subsets of the secod type. Add them together for the result: ( choose k) ( choose k ) + ( choose k) or: ( ) k ( ) + k ( k ). If we map these back to Pascal s triagle, we ca see that they amout exactly to our method of geeratig ew lies from previous lies. 7 Back to the Biomial Theorem Now, let s go back to the biomial theorem ad see if we ca somehow iterpret it as a method for choosig k items from a set of. Multiplicatio over the real umbers is commutative, i the sese that LR RL we ca reverse the order of a multiplicatio ad the result is the same. If we were to do a multiplicatio of a biomial by itself i a strictly formal way, the steps would look like this: (L + R)(L + R) L(L + R) + R(L + R) LL + LR + RL + RR LL + LR + LR + RR LL + LR + RR. The first step uses the distributive law; the ext uses the distributive law agai, the we use the commutative law of multiplicatio to chage the RL to LR, ad fially, we ca combie the two copies of LR to obtai the product 8
i the usual form well, usual except that we ve writte LL ad RR istead of L ad R for reasos that will become clear later. But suppose for a miute that we caot use the commutative law of multiplicatio (but that we ca rearrage the terms, so that additio is commutative). Usig the distributive law we ca still do all the multiplicatios eeded to geerate (L + R), but we will wid up with a lot of terms that caot be combied. I fact, oe of them ca be combied, ad (L + R) will cotai terms. We ca compute (L + R) + by multiplyig out the expaded form of (L + R) by oe additioal (L + R). The calculatio above shows the result of (L + R) ; we ll use that to geerate (L + R) 3 : (L + R) 3 (L + R)(L + R) (L + R)(LL + LR + RL + RR) L(LL + LR + RL + RR) + R(LL + LR + RL + RR) LLL + LLR + LRL + LRR + RLL + RLR + RRL + RRR. Without goig through the detailed calculatios that we used above, but usig the same method, here is what we would obtai for (L + R) : (L + R) LLLL + LLLR + LLRL + LLRR + LRLL + LRLR + LRRL + LRRR + RLLL + RLLR + RLRL + RLRR + RRLL + RRLR + RRRL + RRRR. Notice that i our expasios i this maer of (L+R), (L+R) 3 ad (L+R), the results are simply all possible arragemets of, 3 or R s ad L s. It s easy to see why. If we multiply out somethig like: (L + R)(L + R)(L + R)(L + R) we are basically makig every possible choice of oe of the two i each set of paretheses, ad sice there are choices per group ad groups, there are 6 possible sets of choices. Now, whe we do have commutativity, we covert terms like RLRL to L R throughout, ad the combie like terms. Let s do that, but i the opposite order: first, we ll combie the terms we kow will result i the same value, as show below. Groups with the same umber of R s ad L s are eclosed i paretheses: (L + R) (LL) + (LR + RL) + (RR) (L + R) 3 (LLL) + (LLR + LRL + RLL) + (LRR + RLR + RRL) + (RRR) (L + R) (LLLL) + (LLLR + LLRL + LRLL + RLLL) + (LLRR + LRLR + LRRL + RLLR + RLRL + RRLL) + (LRRR + RLRR + RRLR + RRRL) + (RRRR). The groups above have sizes: [,, ], the [, 3, 3, ], the [,, 6,, ]. These are the umbers i rows, 3 ad of Pascal s triagle. Stop for a secod ad look closely at these grouped terms to see if there is some way to iterpret them as ( choose k). Here is oe way. Look at the largest group: the six terms with R s ad L s i the expasio of (L + R) : (LLRR + LRLR + LRRL + RLLR + RLRL + RRLL). If we iterpret the four letters as idicatig positios of four items i a set, the a L meas choose the item ad a R meas do ot choose it. Thus LLRR meas to take the first two ad omit the secod two; RLLR meas to take the secod ad third items oly, ad so o. 9
Clearly, sice all the possibilities appear here, the umber of terms (6) is exactly the same as the umber of ways that we ca choose items from a set of. Whe the commutative law of additio is applied to these terms, sice they all have R s ad L s, all will become L R, ad sice there are 6 of them, the middle term of the expasio of (L + R) will be 6L R. Agai, there s othig special about the middle term of the expasio usig the fourth power; the same argumets ca be used to show that every term i every biomial expasio ca be iterpreted i its combiatorial sese. 8 Statistics: The Biomial Distributio I the previous sectio (Sectio 7) we looked at the patters of L ad R that related to raisig a biomial to a power: (L + R). If we are istead lookig at a game that cosists of flippig a coi times, ad are iterested i the patters of heads ad tails that could arise, it will tur out that if we just substitute T for L ad H for R the we will have basically described the situatio. Cosider flippig a fair coi (a coi that has equal chaces of ladig heads or tails which we will deote from ow o as H ad T ) 3 times. If we idicate the result of such a experimet as a three-letter sequece where the first is the result of the first flip, the secod represets the secod, ad so o, the here are all the possibilities: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. () Note that this (except for the + sigs) is exactly what we would get if we do the multiplicatio below without havig the commutative law as we did i the expasios of (L + R) 3 ad (L + R) i Sectio 7: (H + T) 3 HHH + HHT + HTH + HTT + THH + THT + TTH + TTT. There is o reaso to believe, sice the coi is fair, that ay of the patters i is ay more or less likely tha ay other, ad if the oly thig you are iterested i is the umber of heads, the you ca see that HHH ad TTT both occur oce, while results with oe head occurs three times ad similarly for results with two heads. Thus, if you were to repeat the experimet of doig three coi flips, i the log ru, the ratio of times the experimet yielded zero, oe, two or three heads would be roughly i the ratio of : 3 : 3 :. Recall that i Sectio 7 we also iterpreted the expasio of (L + R) as listig selectios of L or R from each term whe they are writte like this: (L + R) (L + R)(L + R)(L + R) (L + R). But selectig a L or R is like tellig whether the coi came up heads or tails i each of the terms. There is othig special about three flips, obviously, so if the experimet is to do flips, the there are possible outcomes, ad if all you care about is the umber of times T occurred, ad ot o the actual order of the T ad H results that geerated it, the there are ( ( 0) ways to obtai zero heads, ) ways to obtai oe head, ad i geeral, ( k) ways to obtai exactly k heads. I probability terms, the probability of obtaiig exactly k heads i flips is: ( ). k What if your experimet is ot with fair cois, but rather a repeated test where the odds are the same for each test? For example, suppose the game is to roll a sigle die times, ad you cosider it a wi if a occurs, but a loss if 0
, 3,..., 6 occurs? Thus you wi, o average, oe time i six, or equivaltetly, with a probability of /6. If you repeat the rollig times, what is the probability of gettig exactly k wis i this situatio? We ca describe ay experimet like this by labelig the probability of success as p ad the probability of failure as q such that p + q (i other words, you either wi or lose there are o other possibilities). For flippig a fair coi, p q /; for the dice experimet described above, p /6 ad q 5/6. The aalysis ca begi as before, where we just list the possible outcomes. Usig W for wi ad L for lose, the results of three repeats are the familiar: WWW, WWL, WLW, WLL, LWW, LWL, LLW, LLL. But the chace of gettig a W is ow differet from the chace of gettig a L. What is the probability of gettig each of the results above. For ay particular set, say LW L, to obtai that, you first lose (with probability q) the you wi (with probability p) ad the you lose agai (with probability q agai). Thus the chace that that particular result occurs is qpq. For the three-repeat experimet, the chaces of 0,, ad 3 wis (P(0), P(), P() ad P(3)) are give by: P(0) qqq q 3 P() pqq + qpq + qqp 3pq P() ppq + pqp + qpp 3p q P(3) ppp p 3 Notice that there s othig special about repeatig the experimet three times. If the experimet is repeated times, the probability of obtaiig exactly k wis is give by the formula: ( ) P(k) p k q k. k Thus if you roll a fair die 7 times, the probability that you will obtai exactly wis is give by: ( ) 7 ( ( 5 ) 5 35 P() 6) 6565 6 79936 79936 0.386 Fially, ote that there s o eed for the same experimet to be repeated. If you take a hadful of te cois ad flip them all at oce, the odds of gettig, say, exactly four heads is the same as the odds of gettig four heads i te idividual flips of the same coi. Just imagie flippig the first, the the secod, ad so o, ad leavig them i order o the table after each flip. 9 Back to Combiatorics OK, ow, i priciple, we ca calculate ay biomial coefficiet simply usig additio over ad over to obtai the etries i the appropriate row of Pascal s triagle. If you eed a umber like (95 choose ), however, this would take a log time, startig from scratch. The goal of this sectio is to show that: ( choose k) ( ) k! k!( k)!.
As before, the best way to begi is with a cocrete example, ad we ll use ( choose 3). Oe approach is to thik about it this way: There are ways to choose the first object, ad after that is chose, 3 ways to choose the secod (sice oe is already picked) ad fially, ways to choose the last oe. So there should be 3 ways to do it. This, of course, coflicts with our previous result that ( choose 3), so what s goig o? Let s agai use the set {A, B, C, D} as the set of four objects. If we make the 3 choices as above, here are the sets we obtai (i alphabetical order, to be certai we ve omitted othig): ABC ABD ACB ACD ADB ADC BAC BAD BCA BCD BDA BDC CAB CAD CBA CBD CDA CDB DAB DAC DBA DBC DCA DCB The problem becomes obvious: we ve icluded lots of groups that are idetical: ABC ACB BAC ad so o. We wat to cout groups where the orderig does t matter ad we ve geerated groups that have a order. Let s regroup the list above so that each row cotais oly simple rearragemets of the same items: ABC ACB BAC BCA CAB CBA ABD ADB BAD BDA DAB DBA ACD ADC CAD CDA DAC DCA BCD BDC CBD CDB DBC DCB Notice that each appears exactly 6 times, so the umber we obtaied has couted each subset 6 times. To fid the true umber of subsets, we have to divide by 6 ad we obtai the correct aswer,. How may rearragemets are there of 3 items? Well, the first ca be ay of 3, the there remai choices for the secod, ad the fial item is determied. The result is 3 3! 6. Similarly, there are 3 rearragemets of items ad so o. As before, there s othig special about this method to calculate ( choose 3). If we wat to fid out how may combiatios there are of k thigs from a set of, we say that the first ca be ay of, the secod ay of, ad so o, for k terms. But whe we do this, we ll obtai every possible rearragemet of those k terms so we will have couted each oe k(k )(k ) 3 k! times. Puttig this together, we obtai a simple method to do the calculatio. Here are a couple of examples: ( ) 7 (7 choose ) 7 6 5 3 35 ( ) 9 (9 choose 3) 9 8 7 3 3 8 ( ) 0 9 8 7 ( choose 5) 5 5 3 6 Notice how easy this is. If you re choosig k thigs from a set of, start multiplyig the umbers,, ad so o for k terms, ad the divide by the k terms of k!. If we cout carefully, we ca see that the geeral formula looks like this: ( ) ( )( ) ( k + ). k k! The form above is a little icoveiet to use mathematically because of the umerator, but otice that we ca covert the umerator to a pure factorial if we multiply it all the rest of the way dow, which is to say, multiply
the umerator by ( k)( k ) 3 ( k)!. So multiply both umerator ad deomiator of the equatio above by ( k)! to obtai the result we wated: ( )! k k!( k)!. (3) If you eed to do a actual calculatio of this sort, use the first form, sice massive cacelig will occur. I the example: ( ) 0 9 8 7 ( choose 5) 5 5 3 we ca cacel the 0 i the umerator by the 5 ad i the deomiator. The i the deomiator cacels the 8 upstairs to a, ad the 3 similarly cacels with the 9 yieldig 3, ad the problem reduces to: ( ) ( choose 5) 3 7 6. 5 The form i equatio 3 is much easier to calculate with algebraically. For example, if we took this as the defiitio of the terms i Pascal s triagle, we could show that each row is obtaied from the previous by addig the two above it if we could show that: ( ) ( ) ( ) +. k k k Just for the algebraic exercise, let s do this calculatio by covertig the terms to the equivalet factorial forms. We eed to show that:! k!( k)! ( )! k!( k )! + ( )! (k )!( k)!. To do so, all we eed to do is to covert the terms o the right so that they have a commo deomiator ad the add them together. The commo deomiator is k!( k)!. ( )! k!( k )! + ( )! (k )!( k)! ( )!( k) k!( k)( k )! + ( )!k k(k )!( k)! ( )!( k) ( )!k + k!( k)! k!( k)! ( k + k)( )! k!( k)!! k!( k)!, which is what we eeded to show. Notice also that the factorial form shows istatly that ( ) ( k k) ; i other words, that choosig which of the k items to iclude gives the same value as choosig the k items to omit. Fially, it s probably a good idea if the studets have t see it, to poit out that these biomial coefficiets ca be used to fid thigs like lottery odds. If you eed to make 6 correct picks from 50 choices to wi the lottery, what are the chaces of wiig? Well, there are ( ) 50 6 5890700 equally likely choices, so you ll wi about oe time i every 6 millio. 3
0 A Urelated Problem Suppose we have a grid of city streets, with m orth-south streets ad east-west streets. Figure 5 illustrates a example with m 9 although there is o eed for the two to be the same. The goal is to fid the umber of paths from oe corer to the opposite corer (A to B i the figure) that are the shortest possible distace, i other words, with o backtrackig. A typical shortest route is show as a bold path o the grid i the figure. We will examie this problem i a couple of differet ways. Oe way to thik of it (usig the example i the figure) is that the etire route has to iclude 8 steps dow (ad 8 to the right, of course). But those 8 dowward steps have to occur distributed amog the 9 streets that go dow. I the example route, steps dow are take o the fourth street, more o the fifth street, ad more o each of the eighth ad ith streets. If you thik about it, simply kowig how may of the dowward steps are take o each of the 9 streets completely determies the route. A Figure 5: Routes through a grid B So the problem is equivalet to the followig: How may ways are there to assig 8 idetical balls (steps dow) ito 9 labeled boxes (the up-dow streets)? This is similar to the choose k type problems, but ot quite the same. But here s a ice way to visualize the idetical balls i o-idetical boxes problem. Imagie that the boxes are placed side-by-side ext to each other, ad that we use a vertical bar to idicate the boudary betwee adjacet boxes. Sice there are 9 boxes i this example, there will be 8 boudary walls. Similarly, let s represet the balls by stars, ad there will be 8 of those. We claim that every listig of vertical bars ad asterisks correspods to exactly oe valid shortest-path through the grid. For example, the path i the example correspods to this: Thus the umber of paths simply correspods to the umber of arragemets of 8 bars ad 8 stars. Well, we are basically writig dow 6 symbols i order, ad if we kow which 8 of them are stars, the others are bars, so there must be (6 choose 8) 870 ways to do this. If there are m streets by streets, there are steps dow to be distributed amog m streets. Thus there will be m vertical bars ad asterisks. The aswer is that there are (m + choose ) routes i a m grid. It s actually probably worth coutig the streets i a few simple grids before pullig out the big gus ad obtaiig a geeral solutio i a classroom settig. Now look at the same problem i a differet way. Suppose we begi at poit A. There is exactly oe way to get there: do othig. Now look at the poits o the horizotal street from A or the vertical street from A. For every oe of those poits, there s oly oe shortest path, the straight lie. What we re goig to do is label all the poits o the grid with the umber of shortest paths there are to get there. From these simple observatios, all the labels o the top ad left edges of the grid will be labeled with the umber. Now for the key observatio: to get to ay poit iside the grid, you either arrived from above or from the left. The total umber of uique routes to that poit will be the total umber of routes to the poit above plus the total umber of routes to the poit o your left. The upper left corer of the resultig grid will thus look somethig like what is illustrated i figure 6. 3 5 6 3 6 0 5 0 0 35 56 5 5 35 70 6 6 56 6 5 Figure 6: Coutig paths i a grid
Notice that this is just Pascal s triagle, tured o its side! I fact, if we start at the apex of Pascal s triagle ad take paths that always go dow, but ca go either to the right or left at each stage, the the umbers i the triagle idicate the umber of paths by which they ca be reached. I the early examples with the biomial theorem, we used R ad L, that we ca ow thik of as right ad left. Whe we reach the positio i Pascal s triagle correspodig to (7 choose 3) what that really amouts to is the umber of paths from the apex that are 7 steps log, ad which cotai 3 moves to the left (ad hece 7 3 moves to the right). Biomial Coefficiet Relatioships Here is a series of idetities satisfied by the biomial coefficiets. Some are easy to prove, ad some are difficult. ( ) ( ) ( ) ( ) ( ) + + + + + (a) 0 3 ( ) ( ) ( ) ( ) ( ) + + + ( ) 0, > 0 (b) 0 3 ( ) + ( ) + ( ) + ( ) + + ( ) + (c) 0 3 3 + + ( ) ( ) ( ) ( ) ( ) 0 + + + 3 + + (d) 0 3 ( ) ( ) ( ) ( ) ( ) ( ) + + + + + (e) 0 3 ( ) ( ) ( ) ( ) ( ) { + + + ( ) 0 : m + 0 3 ( ) m( ) m (f) m : m ( ) ( ) ( ) ( ) ( ) ( ) + + + + + +, > 0 (g) 0 6 8 0 ( ) ( ) ( ) ( ) ( ) ( ) + + + + + +, > 0 (h) 3 5 7 9 Relatios (a) ad (b) were proved earlier i this article (see Sectio 3). Relatios (c) ad (d) ca be proved by stadard methods, but a quick proof is available if we use a trick from calculus. We begi by oticig that the biomial theorem tells us that: ( ) ( ) ( ) ( ) ( + x) + x + x + + x. 0 If we take the derivative of both sides with respect to x, we obtai: ( ) ( ) ( ) ( ) ( + x) + x + 3 x + x. 3 Substitute for x ad we obtai relatio (d). 5
If we start from the same equatio but itegrate istead, we obtai: ( + x) + ( ) x + ( ) x + ( + 0 3 ) x 3 + + + ( ) x + + C, where C is a costat to be determied. Substitutig x 0, we obtai C /( + ), ad the if we substitute x, we obtai relatio (c). If you do t wat to use calculus, here s a differet approach. For relatio (d) ote that: ( ) ( ) k! k k ( k)!k! ( )! ( k)!(k )!. k The ( ) ( 0 + 0 ( ( 0 ) + ) + ( ) + 3 ( ( 3 ) + + ) + + ( Relatio (c) ca be proved similarly, startig from the fact that: ( ) ( ) +. k + k + k + ( ) ) ). We ll leave the proof as a exercise. There are differet ways to prove relatio (e), but my favorite is a combiatorial argumet. Sice ( ) ( k ca rewrite the sum of the squares of the biomial coefficiets as: ( )( ) ( )( ) ( )( ) ( )( ) + + +. 0 0 k) we Now let s cosider a particular way to calculate ( ). This just couts the umber of ways to choose items from a set of. Imagie that we divide the items ito two sets, A ad B, each of which cotais items. If we choose oe of the items i set A, we have to choose all from set B. If we choose item from A we have to choose from B, ad so o. I geeral, for each of the ( ( k) ways we ca choose k items from set A, there are ( k) ways to choose the remaiig items from B, ad this will add )( k k) more ways to choose items from. Sum them up, ad we obtai relatio (e). A more traditioal way to do the calculatio is as follows. First we ote that: ( ) ( ) ( ) ( ) (x + y) x + x y + x y + + y 0 ( ) ( ) ( ) ( ) x + x y + x y + + y 0 If we multiply (x + y) by itself, oce represeted by the first versio above ad oce represeted by the secod, we ca see that the term cotaiig x y will have a coefficiet equal to the desired sum of products of biomial coefficiets. But this is just the same as the coefficiet of x y i the expasio of (x+y) (x+y) (x+y), ad that is just ( ). Fially, a third way to see that the sum is correct is to cosider a special case of coutig routes through a grid that we solved i Sectio 0. 6
A P Cosider a square grid such as the oe displayed i Figure 7. I that figure there are five horizotal ad five vertical paths, but the argumet we will make will ot deped P o that. Suppose we wat to cout the umber of paths from poit A to poit B i that figure. From the work we did i Sectio 0, we kow that the umber is ( 8 P ). 3 But ay path from A to B must pass through exactly oe of the poits P, P,..., P 5. P The umber of paths from A to B passig through poit P i is the umber of paths from A to P i multiplied by the umber of paths from P i to B. P 5 B Agai, usig methods from Sectio 0, there are ( ) 0 paths from A to P ad ( 0) paths Figure 7: Routes i a from P to B. There are ( ) paths from A to P ad ( ) paths from P to B, ad so o. square grid Thus the total umber of paths, which we kow to be ( 8 ), is: ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( ) 8 + + + +. 0 0 3 3 It should be clear that there is othig special about a 5 5 grid, ad that the same argumet ca be used to cout paths through ay square grid, ad this yields the result expressed i relatio (e). The secod method that we used to compute relatio (e) ca be used to prove relatio (f), ad all we have to do is to multiply (x y) by (x + y) usig the followig two expasios: ( ) ( ) ( ) ( ) (x + y) x + x y + x y + + y 0 ( ) ( ) ( ) ( ) (x y) x x y + x y + ( ) y 0 Agai, the desired term will be the coefficiet of x y i the expasio of (x + y) (x y), but this is just (x y ). If is odd, this coefficiet is clearly zero, ad if is eve, say m, the it is ( ) m( m m ). The case where is odd is totally obvious sice the values i Pascal s triagle are symmetric, ad there is o middle term for odd, so each term with a positive sig will be matched by the correspodig symmetric term with a egative sig. Relatios (g) ad (h) are easy to obtai. Cosider the followig two expasios: ( ) ( ) ( ) ( ) (x + y) x + x y + x y + + y 0 ( ) ( ) ( ) ( ) (x y) x x y + x y + ( ) 0 If we add the equatios together, we obtai twice the sum i relatio (g) ad if we subtract them, we obtai twice the sum i (h). The simply set x ad y to, ad we see that the sum or differece is. Sice this is twice the desired result, we see that i both cases, the sum is. y 7
Harmoic Differeces Write dow the fractios i the harmoic series: /, /, /3,..., ad i each row below that, calculate the differeces of the adjacet umbers. You will obtai a table that looks somethig like this: 3 6 3 5 0 30 6 0 30 5 60 60 30 6 If this table is rotated util the fractio / is o top, ad the if the top row is multiplied by, the secod row by, the third by 3 ad so o, we obtai a triagle that is exactly the same as Pascal s triagle except that the umbers are i the deomitors istead of the umerators. This is easy to prove if we simply write dow the fractios i the table above i terms of the biomial coefficiets ad show that they satisfy the differece equatios. Followig is the table above writte i that form. ( ( 0 )) ( ( 0 )) ( ( 3 )) ( ( 3 )) ( ( 3 5 )) ( ( 6 5 5)) ( ( )) ( ( 0 3 )) ( ( 3 )) ( ( 5 )) ( ( 3 6 5 )) ( ( 3 )) ( ( 0 3 )) ( ( 5 )) ( ( 6 5 )) 3 ( ( 3 )) ( ( 0 5 )) ( ( 6 5 )) ( ( 5 )) ( ( 0 6 5 )) ( ( 6 5 )) 0 It is obvious that the first row is just the harmoic series, so all that eeds to be show is that successive etries i the table ca be obtaied by subtractig the two etries above. This is equivalet to showig that for ay ad k: ( + ) ( ) k ( + ) ( + k+ ) ( + ) ( +). k If we write the biomial coefficiets above i their factorial form, this is equivalet to: ( + )k!( k)!! ( + )(k + )!( k)! ( + )! We ca put everythig over a commo deomiator of!: ( + )k!( k)!! so sice the deomiators are the same, we are doe if: (k + )!( k)!! ( + )k!( + k)!. ( + )! k!( + k)!,! ( + )k!( k)! (k + )!( k)! k!( + k)! 8
k!(( + )( k)! (k + )( k)!) k!( + k)! k!( + k )( k)! k!( + k)! k!( k + )( k)! k!( + k)! k!( + k)! k!( + k)! 3 Fidig Formulas for Sequeces Suppose you come across the followig sequece: 5, 7,, 53, 09, 95,37, 8,..., ad you would like to fid a geeral formula f() such that f(0) 5, f() 7, f(), f(3) 53, ad so o. I may, may cases the followig techique works. 5 7 53 09 95 37 8... 3 56 86 6... 8 30 36... 6 6 6 6 6... 0 0 0 0... First list the umbers i a lie as i the table above, ad o each successive lie, write the differece of the pair of umbers above it. This is sort of the opposite of what you do to form Pascal s triagle. Cotiue i this way, ad if you are lucky, you will wid up with a lie that remais costat (the lie of 6 s above), ad the, if you were to cotiue, the ext lie ad every other lie would be completely filled with zeros. (If this does ot happe, the the followig techique will ot work, although if you do fid a patter, other techiques may work.) The example above happeed to degeerate to all zeroes i the fifth lie. This may occur i more or fewer lies. The sequece, 3, 5, 7, 9,... is all zeros o the third lie, ad other sequeces may require more lies before they degeerate to all zeroes. This is t exactly a formula, but it does provide a easy method to compute successive values i the origial list, assumig the patter cotiues. I this case, sice the lie of 6 s cotiues, just add aother 6 to it. That meas that the umber followig the i the lie above must be + 6 8. The same reasoig tells us that the umber followig the 6 must be 6 + 8 ad repeatig the reasoig, the ext umber i our sequece will be 8 + 693. We ca repeat this as ofte as we like to obtai a arbitrary umber of terms. The most importat thig to ote is that the etire table is completely determied by the umbers o the left edge; i this case they are: 5,, ad 6 followed by a ifiite sequece of zeroes. Although the method above will certaily work, it would still be paiful to calculate the millioth term this way. It would be much better to have a explicit formula for the th term. As we already observed, the geeratio of tables like those above is similar to what we do to obtai Pascal s triagle except that we subtract istead of add. Because of this it is ot surprisig that a explicit formula ca be obtaied based o coefficiets i Pascal s triagle. We will simply state the method for obtaiig such a formula ad the we will explore why it works. I fact, our formula will use oly the umbers o the left edge of our table (which, 9
as we observed, geerate the etire table). I the example, they are 5,,, 6, 0, 0,.... Here is the formula: ( ) ( ) ( ) ( ) ( ) ( ) 5 + + + 6 + 0 + 0 + 0 3 5 Obviously oce we arrive at the ifiite sequece of zeros, oe of the remaiig terms will cotribute aythig, so the formula above really oly has four o-zero terms. I its curret state the formula is correct, but it ca be simplified with a bit of algebra: ( ) ( ) ( ) ( ) f() 5 + + + 6 0 3 ( ) ( )( ) f() 5 + + + 6 6 f() 5 + + 6( ) + ( 3 3 + ) f() 3 + 3 + 5. You ca check that this formula is correct by pluggig i differet values for ad verifyig that f(0) 5, f() 7, f(), ad so o. The method seems to work great, but why? Oe way to see this is to apply the method to a diagoal of Pascal s triagle where f(x) ( 3). (There is othig special about this row except that it is the most complex part of the formula we geerated for the example above.) We will make the sesible assumptios that ( ( 0 3) ( 3) 3) 0. I other words there is o way to choose 3 objects from a set cosistig of zero, oe or two objects. Here is a table of successive differeces with ( 3) as the top row: 0 0 0 0 0 35... 0 0 3 6 0 5... 0 3 5...... 0 0 0 0... What jumps out is that this is just a part of Pascal s triagle o its side. Each successive row i the table is aother diagoal i Pascal s triagle. If you look at how the umbers are geerated, it is obvious why this occurs, ad that o matter what diagoal of the triagle you begi with, a similar patter will occur. Note also that the geeratig patter of umbers o the left of the table is, i this case, 0, 0, 0,, 0, 0,.... Thus 0, 0, 0,, 0, 0,... geerates the row ( ( 3). If we had begu with the row ), the correspodig geeratig patter would have bee 0, 0,, 0, 0,..., or for ( 5), it would have bee 0, 0, 0, 0, 0,, 0, 0,... ad so o. If the geeratig patter for some sequece were 0,, 0, 0,, 0, 0,..., it should be clear that the iitial lie will look like: ( ) ( ) ( ) ( ) ( ) 0 + + 0 + 0 +. 0 3 Also ote that if we had multiplied every term i the top row by a costat, the every row of successive differeces would be multiplied by the same costat, ad the fial patter will simply have that costat i the approprite spots i the list of umbers o the left of the table. Thus our techique of multiplied the appropriate coefficiet of Pascal s triagle by the costat i that row ad summig them should yield a formula for the sequece. 0
As a fial example, let s work out a formula for the followig sum: f() 0 3 + 3 + 3 + 3 3 + + 3. We ca work out the first few values by had, ad usig our method, obtai the followig table. (Note the iterestig fact that all the umbers i the first row happe to be perfect squares.) 0 9 36 00 5 78... 8 7 6 5 6 33... 7 9 37 6 9 7... Accordig to our aalysis, the result should be: ( ) f() 0 + 0 8 30 36... If we apply some algebra, we obtai: ( ) ( ) ( ) f() 0 + + 7 + 0 ( ) f() 0 + + 7 f() 0 + + 7 ( ) f() + f() 3 + 6 6 6 6... 0 0 0... ( ) + 7 ( 3 + ). ) + 6 ( ) + ( ) ( )( ) 6 ( ) + 6 3 + 6 ( ). + ( 3 3 + ) + 6 3 + 6 + 83 + 6 + 6 3 + 6 ( ( + ) ( )( )( 3) The result above may be more familiar as: sice + + 3 + ( + )/. 0 3 + 3 + 3 + 3 (0 + + + + )
3 3 6 5 0 0 5 6 5 0 5 6 7 35 35 7 8 8 56 70 56 8 8 9 36 8 6 6 8 36 9 0 5 0 0 5 0 0 5 0 55 65 330 6 6 330 65 55 66 0 95 79 9 79 95 0 66 3 78 86 75 87 76 76 87 75 86 78 3 9 36 00 00 3003 33 3003 00 00 36 9 5 05 55 365 3003 5005 635 635 5005 3003 365 55 05 5