1.3 Binomial Coefficients


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1 18 CHAPTER 1. COUNTING 1. Biomial Coefficiets I this sectio, we will explore various properties of biomial coefficiets. Pascal s Triagle Table 1 cotais the values of the biomial coefficiets ( ) for 0to 6 ad all relevat values. The table begis with a 1 for 0ad 0,because the empty set, the set with o elemets, has exactly oe 0elemet subset, amely itself. We have ot put ay value ito the table for a value of larger tha, because we have t defied what we mea by the biomial coefficiet ( ) i that case. However, sice there are o subsets of a elemet set that have size larger tha, itisatural to defie ( ) to be zero whe >, ad so we defie ( ) to be zero whe >. Thus wecould could fill i the empty places i the table with zeros. The table is easier to read if we do t fill i the empty spaces, so we just remember that they are zero. Table 1.1: A table of biomial coefficiets \ Exercise 1.1 What geeral properties of biomial coefficiets do you see i Table 1.1 Exercise 1.2 What is the ext row of the table of biomial coefficiets? Several properties of biomial coefficiets are apparet i Table 1.1. Each row begis with a 1, because ( 0) is always 1, as it must be because there is just oe subset of a elemet set with 0 elemets, amely the empty set. Similarly, each row eds with a 1, because a elemet set S has just oe elemet subset, amely S itself. Each row icreases at first, ad the decreases. Further the secod half of each row is the reverse of the first half. The array of umbers called Pascal s Triagle emphasizes that symmetry by rearragig the rows of the table so that they lie up at their ceters. We show this array i Table 2. Whe we write dow Pascal s triagle, we leave out the values of ad. You may ow a method for creatig Pascal s triagle that does ot ivolve computig biomial coefficiets, but rather creates each row from the row above. Each etry i Table 1.2, except for the oes, is the sum of the etry directly above it to the left ad the etry directly above it to the right. We call this the Pascal Relatioship, ad it gives aother way to compute biomial coefficiets without doig the multiplyig ad dividig i equatio 1.5. If we wish to compute may biomial coefficiets, the Pascal relatioship ofte yields a more efficiet way to do so. Oce the coefficiets i a row have bee computed, the coefficiets i the ext row ca be computed usig oly oe additio per etry.
2 1.. BINOMIAL COEFFICIENTS 19 Table 1.2: Pascal s Triagle We ow verify that the two methods for computig Pascal s triagle always yield the same result. I order to do so, we eed a algebraic statemet of the Pascal Relatioship. I Table 1.1, each etry is the sum of the oe above it ad the oe above it ad to the left. I algebraic terms, the, the Pascal Relatioship says , (1.6) wheever >0 ad 0 <<. Notice that It is possible to give a purely algebraic (ad rather dreary) proof of this formula by pluggig i our earlier formula for biomial coefficiets ito all three terms ad verifyig that we get a equality. A guidig priciple of discrete mathematics is that whe we have a formula that relates the umbers of elemets of several sets, we should fid a explaatio that ivolves a relatioship amog the sets. A proof usig sets From Theorem 1.2 ad Equatio 1.5, we ow that the expressio ( ) is the umber of elemet subsets of a elemet set. Each of the three terms i Equatio 1.6 therefore represets the umber of subsets of a particular size chose from a appropriately sized set. I particular, the three sets are the set of elemet subsets of a elemet set, the set of ( 1)elemet subsets of a ( 1)elemet set, ad the set of elemet subsets of a ( 1)elemet set. We should, therefore, be able to explai the relatioship betwee these three quatities usig the sum priciple. This explaatio will provide a proof, just as valid a proof as a algebraic derivatio. Ofte, a proof usig subsets will be less tedious, ad will yield more isight ito the problem at had. Before givig such a proof i Theorem 1. below, we give a example. Suppose 5, 2. Equatio 1.6 says that (1.7) Because the umbers are small, it is simple to verify this by usig the formula for biomial coefficiets, but let us istead cosider subsets of a 5elemet set. Equatio 1.7 says that the umber of 2 elemet subsets of a 5 elemet set is equal to the umber of 1 elemet subsets of a4elemet set plus the umber of 2 elemet subsets of a 4 elemet set. But to apply the sum priciple, we would eed to say somethig stroger. To apply the sum priciple, we should be able to partitio the set of 2 elemet subsets of a 5 elemet set ito 2 disjoit sets, oe of which
3 20 CHAPTER 1. COUNTING has the same size as the umber of 1 elemet subsets of a 4 elemet set ad oe of which has the same size as the umber of 2 elemet subsets of a 4 elemet set. Such a partitio provides a proof of Equatio 1.7. Cosider ow the set S {A, B, C, D, E}. The set of two elemet subsets is S 1 {{A, B}, {AC}, {A, D}, {A, E}, {B,C}, {B,D}, {B,E}, {C, D}, {C, E}, {D, E}}. We ow partitio S 1 ito 2 blocs, S 2 ad S. S 2 cotais all sets i S 1 that do cotai the elemet E, while S cotais all sets i S 1 that do ot cotai the elemet E. Thus, ad S 2 {{AE}, {BE}, {CE}, {DE}} S {{AB}, {AC}, {AD}, {BC}, {BD}, {CD}}. Each set i S 2 must cotai E ad the cotais oe other elemet from S. Sice there are 4 other elemets i S that we ca choose alog with E, S 2 ( 4 1). Each set i S cotais 2 elemets from the set {A, B, C, D}, ad thus there are ( 4 2) ways tochoose such a subset. But S 1 S 2 S ad S 2 ad S are disjoit, ad so, by the sum priciple, Equatio 1.7 must hold. We ow give a proof for geeral ad. Theorem 1. If ad are itegers with >0 ad 0 <<, the Proof: The formula says that the umber of elemet subsets of a elemet set is the sum of two umbers. As i our example, we will apply the sum priciple. To apply it, we eed to represet the set of elemet subsets of a elemet set as a uio of two other disjoit sets. Suppose our elemet set is S {x 1,x 2,...x }. The we wish to tae S 1,say,tobethe ( ) elemet set of all elemet subsets of S ad partitio it ito two disjoit sets of elemet subsets, S 2 ad S, where the sizes of S 2 ad S are ( 1) ( ad 1 ) respectively. We ca do this as follows. Note that ( 1) stads for the umber of elemet subsets of the first 1 elemets x 1,x 2,...,x 1 of S. Thusweca let S be the set of elemet subsets of S that do t cotai x. The the oly possibility for S 2 is the set of elemet subsets of S that do cotai x.how ca we see that the umber of elemets of this set S 2 is ( 1 1)? By observig that removig x from each of the elemets of S 2 gives a ( 1)elemet subset of S {x 1,x 2,...x 1 }.Further each ( 1)elemet subset of S arises i this way from oe ad oly oe elemet subset of S cotaiig x. Thus the umber of elemets of S 2 is the umber of ( 1)elemet subsets of S, which is ( 1 1). Sice S2 ad S are two disjoit sets whose uio is S, the sum priciple shows that the umber of elemets of S is ( ) ( + 1 ). Notice that i our proof, we used a bijectio that we did ot explicitly describe. Namely, there is a bijectio f betwee S (the elemet sets of S that cotai x ) ad the ( 1)elemet subsets of S. For ay subset K i S,Welet f(k) bethe set we obtai by removig x from K. Itisimmediate that this is a bijectio, ad so the bijectio priciple tells us that the size of S is the size of the set of all subsets of S.
4 1.. BINOMIAL COEFFICIENTS 21 The Biomial Theorem Exercise 1. What is (x + y)? What is (x +1) 4? What is (2 + y) 4? What is (x + y) 4? The umber of elemet subsets of a elemet set is called a biomial coefficiet because of the role that these umbers play i the algebraic expasio of a biomial x+y. The Biomial Theorem states that Theorem 1.4 (Biomial Theorem) For ay iteger 0 (x + y) x + 0 x 1 y + 1 x 2 y xy y, (1.8) or i summatio otatio, (x + y) i0 x i y i. i Ufortuately whe most people first see this theorem, they do ot have the tools to see easily why it is true. Armed with our ew methodology of usig subsets to prove algebraic idetities, we ca give a proof of this theorem. Let us begi by cosiderig the example (x + y) which by the biomial theorem is (x + y) x + 0 x 2 y + 1 xy y (1.9) x +x 2 y +xy 2 + x. (1.10) Suppose that we did ot ow the biomial theorem but still wated to compute (x + y). The we would write out (x + y)(x + y)(x + y) ad perform the multiplicatio. Probably we would multiply the first two terms, obtaiig x 2 +2xy + y 2, ad the multiply this expressio by x + y. Notice that by applyig distributive laws you get (x + y)(x + y) (x + y)x +(x + y)y xx + xy + yx + y. (1.11) We could use the commutative law to put this ito the usual form, but let us hold off for a momet so we ca see a patter evolve. To compute (x + y),weca multiply the expressio o the right had side of Equatio 1.11 by x + y usig the distributive laws to get (xx + xy + yx + yy)(x + y) (xx + xy + yx + yy)x +(xx + xy + yx + yy)y (1.12) xxx + xyx + yxx + yxx + xsy + xyy + yxy + yyy (1.1) Each of these 8 terms that we got from the distributive law may be thought of as a product of terms, oe from the first biomial, oe from the secod biomial, ad oe from the third biomial. Multiplicatio is commutative, so may of these products are the same. I fact, we have oe xxx or x product, three products with two x s ad oe y, orx 2 y, three products with oe x ad two y s, or xy 2 ad oe product which becomes y.now loo at Equatio 1.9, which summarizes this process. There are ( ( 0) 1way to choose a product with x s ad 0 y s, 1) way to choose a product with 2 x s ad 1 y, etc. Thus we ca uderstad the biomial theorem
5 22 CHAPTER 1. COUNTING as coutig the subsets of our biomial factors from which we choose a yterm to get a product with y s i multiplyig a strig of biomials. Essetially the same explaatio gives us a proof of the biomial theorem. Note that whe we multiplied out three factors of (x + y) usig the distributive law but ot collectig lie terms, we had a sum of eight products. Each factor of (x+y) doubles the umber of summads. Thus whe we apply the distributive law as may times as possible (without applyig the commutative law ad collectig lie terms) to a product of biomials all equal to (x+y), we get 2 summads. Each summad is a product of a legth list of x s ad y s. I each list, the ith etry comes from the ith biomial factor. A list that becomes x y whe we use the commutative law will have a y i of its places ad a x i the remaiig places. The umber of lists that have a y i places is thus the umber of ways to select biomial factors to cotribute a y to our list. But the umber of ways to select biomial factors from biomial factors is simply ( ), ad so that is the coefficiet of x y. This proves the biomial theorem. Applyig the Biomial Theorem to the remaiig questios i Exercise 1. gives us (x +1) 4 x 4 +4x +6x 2 +4x +1 (2 + y) y +24y 2 +8y + y 4 ad (x + y) 4 x 4 +4x y +6x 2 y 2 +4xy + y 4. Labelig ad triomial coefficiets Exercise 1.4 Suppose that I have labels of oe id ad labels of aother. I how may differet ways may I apply these labels to objects? Exercise 1.5 Show that if we have 1 labels of oe id, 2 labels of a secod id, ad! 1 2 labels of a third id, the there are 1! 2!! ways toapply these labels to objects. Exercise 1.6 What is the coefficiet of x 1 y 2 z i (x + y + z)? Exercise 1.4 ad Exercise 1.5 ca be thought of as immediate applicatios of biomial coefficiets. For Exercise 1.4, there are ( ) ways to choose the objects that get the first label, ad the other objects get the secod label, so the aswer is ( ) (.For Exercise 1.5, there are ) 1 ways to choose the 1 objects that get the first id of label, ad the there are ( 1 ) 2 ways to choose the objects that get the secod id of label. After that, the remaiig 1 2 objects get the third id of label. The total umber of labelligs is thus, by the product priciple, the product of the two biomial coefficiets, which simplifies as follows. 1! ( 1 )! 1!( 1 )! 2!( 1 2 )! 1 2! 1! 2!( 1 2 )!! 1! 2!!. A more elegat approach to Exercise 1.4, Exercise 1.5, ad other related problems appears i the ext sectio.
6 1.. BINOMIAL COEFFICIENTS 2 Exercise 1.6 shows how Exercise 1.5 applies to computig powers of triomials. I expadig (x + y + z),wethi of writig dow copies of the triomial x + y + z side by side, ad applyig the distributive laws util we have a sum of terms each of which is a product of x s, y s ad z s. How may such terms do we have with 1 x s, 2 y s ad z s? Imagie choosig x from some umber 1 of the copies of the triomial, choosig y from some umber 2, ad z from the remaiig copies, multiplyig all the chose terms together, ad addig up over all ways ofpicig the i s ad maig our choices. Choosig x from a copy of the triomial labels that copy with x, ad the same for y ad z, sothe umber of choices that yield x 1 y 2 z is the umber of ways to label objects with 1 labels of oe id, 2 labels of a secod id, ad labels of a third. Notice that this requires that 1 2. By aalogy with our otatio for a biomial coefficiet, we defie the triomial coefficiet ( ) 1, 2, to be! 1! 2!! if ad 0 otherwise. The ( ) 1, 2, is the coefficiet of x 1 y 2 z i (x + y + z). This is sometimes called the triomial theorem. Importat Cocepts, Formulas, ad Theorems 1. Pascal Relatioship. The Pascal Relatioship says that wheever >0 ad 0 << , 2. Pascal s Triagle. Pascal s Triagle is the triagular array of umbers we get by puttig oes i row ad colum 0 ad i row ad colum of a table for every positive iteger ad the fillig the remaider of the table by lettig the umber i row ad colum j be the sum of the umbers i row 1 ad colums j 1 ad j wheever 0 <j<.. Biomial Theorem. The Biomial Theorem states that for ay iteger 0 (x + y) x + x 1 y + 1 x 2 y xy y, or i summatio otatio, (x + y) i0 x i y i. i 4. Labelig. The umber of ways to apply labels of oe id ad labels of aother id to objects is ( ). 5. Triomial coefficiet. We defie the triomial coefficiet ( ) 1, 2, to be! 1! 2!! if ad 0 otherwise. 6. Triomial Theorem. The coefficiet of x i y j z i (x + y + z) is ( i,j,).
7 24 CHAPTER 1. COUNTING Problems 1. Fid ( 12) ( ad 12 ) ( 9. What ca you say i geeral about ) ( ad )? 2. Fid the row of the Pascal triagle that correspods to 8.. Fid the followig a. (x +1) 5 b. (x + y) 5 c. (x +2) 5 d. (x 1) 5 4. Carefully explai the proof of the biomial theorem for (x + y) 4. That is, explai what each of the biomial coefficiets i the theorem stads for ad what powers of x ad y are associated with them i this case. 5. If I have te distict chairs to pait i how may ways may I pait three of them gree, three of them blue, ad four of them red? What does this have to do with labelligs?! 6. Whe 1, 2,... are oegative itegers that add to, the umber 1!, 2!,...,! is called a multiomial coefficiet ad is deoted by ( ) 1, 2,...,. A polyomial of the form x 1 + x x is called a multiomial. Explai the relatioship betwee powers of amultiomial ad multiomial coefficiets. This relatioship is called the Multiomial Theorem. 7. Give a bijectio that proves your statemet about ( ) ( ad ) i Problem 1 of this sectio. 8. I a Cartesia coordiate system, how may paths are there from the origi to the poit with iteger coordiates (m, ) if the paths are built up of exactly m + horizotal ad vertical lie segmets each of legth oe? 9. What is the formula we get for the biomial theorem if, istead of aalyzig the umber of ways to choose distict y s, we aalyze the umber of ways to choose distict x s? 10. Explai the differece betwee choosig four disjoit three elemet sets from a twelve elemet set ad labellig a twelve elemet set with three labels of type 1, three labels of type two, three labels of type, ad three labels of type 4. What is the umber of ways of choosig three disjoit four elemet subsets from a twelve elemet set? What is the umber of ways of choosig four disjoit three elemet subsets from a twelve elemet set? 11. A 20 member club must have a Presidet, Vice Presidet, Secretary ad Treasurer as well as a three perso omiatios committee. If the officers must be differet people, ad if o officer may be o the omiatig committee, i how may ways could the officers ad omiatig committee be chose? Aswer the same questio if officers may be o the omiatig committee. 12. Prove Equatio 1.6 by pluggig i the formula for ( ).
8 1.. BINOMIAL COEFFICIENTS Give two proofs that. 14. Give at least two proofs that j j. j j 15. Give at least two proofs that j ( j ) j. 16. You eed ot compute all of rows 7, 8, ad 9 of Pascal s triagle to use it to compute ( 9 6). Figure out which etries of Pascal s triagle ot give i Table 2 you actually eed, ad compute them to get ( 9 6). 17. Explai why ( 1) i 0 i i0 18. Apply calculus ad the biomial theorem to (1 + x) to show that True or False: ( ) ( 2 ) ( ) ( ).IfTrue, give a proof. If false, give a value of ad that show the statemet is false, fid a aalogous true statemet, ad prove it.
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