1.2 Binomial Coefficients and Subsets

Transcription

1 1.2. BINOMIAL COEFFICIENTS AND SUBSETS Biomial Coefficiets ad Subsets The loop below is part of a program to determie the umber of triagles formed by poits i the plae. for i =1 to for j = i +1 to for k = j +1 to if poits i, j, ad k are ot colliear add oe to triagle cout How may times does the loop check three poits to see if they are colliear? Coutig Subsets of a Set I Exercise 1.2-1, we have a loop imbedded i a loop which was embedded i aother loop. Because the secod loop bega at the curret i value, ad the third loop bega at the curret j value, our code examied each triple of values i, j, k with i<j<kexactly oce. Thus oe way i which we might have solved Exercise would be to compute the umber of such triples. As with the case of two-elemet subsets earlier, the umber of such triples is the umber of three-elemet subsets of a -elemet set. Thus we could try to use the method we itroduced for computig the umber of two-elemet subsets of a set for solvig this problem. First, how may lists of three distict umbers (betwee 1 ad ) ca we make? There are choices for the first umber, ad 1 choices for the secod, so by the product priciple there are ( 1) choices for the first two elemets of the list. For each choice of the first two elemets, there are 2 ways to choose a third (distict) umber, so agai by the product priciple, there are ( 1)( 2) ways to choose the list of umbers. This does ot tell us the umber of three elemet sets, though, because a give three elemet set ca be listed i a umber of ways. How may? Well, give the three umbers, there are three ways to choose the first umber i the list, give the first there are two ways to choose the secod, ad give the first two there is oly oe way to choose the third elemet of the list. Thus by the product priciple oce agai, there are 3 2 1=6waystomakethelist. Sice there are ( 1)( 2) lists of three distict elemets chose from a -elemet set, ad each three-elemet subset appears i exactly 6 of these lists, there are ( 1)( 2)/6 three-elemet subsets of a -elemet set. If we would like to cout the umber of k-elemet subsets of a elemet set, ad k>0, the we could first compute the umber of lists of k distict elemets chose from a k-elemet set which, by the product priciple, will be ( 1)( 2) ( k + 1), the first k terms of!, ad the divide that umber by k(k 1) 1, the umber of ways to list a set of k elemets. This gives us Theorem For itegers ad k with 0 k, the umber of k elemet subsets of a elemet set is!/k!( k)!

2 14 CHAPTER 1. COUNTING Proof: Essetially give above, except i the case that k is 0; however the oly subset of our -elemet set of size zero is the empty set, so we have exactly oe such subset. This is exactly what the formula gives us as well. The umber of k-elemet subsets of a -elemet set is usually deoted by ( k) or C(, k), both of which are read as choose k. These umbers are called biomial coefficiets for reasos that will become clear later. Notice that it was the secod versio of the product priciple, the versio for coutig lists, that we were usig i computig the umber of k-elemet subsets of a -elemet set. As part of the computatio we saw that the umber of ways to make a list of k distict elemets chose from a -elemet set is ( 1) ( k +1)=!/( k)!, the first k terms of!. This expressio arises frequetly; we use the otatio k,(read to the k fallig) for ( 1) ( k + 1). This otatio is origially due to Kuth Pascal s Triagle I Table 1 we see the values of the biomial coefficiets ( k) for = 0 to 6 ad all relevat k values. Note that the table begis with a 1 for =0adk = 0, which is as it should be because the empty set, the set with o elemets, has exactly oe 0-elemet subset, amely itself. We have ot put ay value ito the table for a value of k larger tha, because we have t defied what we mea by the biomial coefficiet ( k) i that case. Table 1.1: A table of biomial coefficiets \k Do you see some geeral properties of biomial coefficiets i the table of values we created? What do you thik is the ext row of the table of biomial coefficiets? How do you thik you could prove you were right i the last two exercises? There are a umber of properties of biomial coefficiets that are obvious from the table. The 1 at the begiig of each row reflects the fact that ( 0) is always oe, as it must be because there is just oe subset of a -elemet set with 0 elemets, amely the empty set. The fact that each row eds with a 1 reflects the fact that a -elemet set S has just oe -elemet subset,

3 1.2. BINOMIAL COEFFICIENTS AND SUBSETS 15 S itself. Each row seems to icrease at first, ad the decrease. 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 k. Table 1.2: Pascal s Triagle I school we are taught a method for creatig Pascal s triagle that ivolves ot computig biomial coefficiets, but creatig each row from the row above. I fact, otice that each etry i the table (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. If we eed to compute a umber of biomial coefficiets, this makes a much easier way to do so tha the multiplyig ad dividig formula give above. But do the two methods of computig Pascal s triagle always yield the same results? To verify this, it is hady to have a algebraic statemet of the Pascal Relatioship. To figure out this algebraic statemet of the relatioship, it is useful to observe how it plays out i Table 1, our origial table of biomial coefficiets. You ca see that i Table 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 = k 1 + k 1 1 k wheever >0 ad 0 <k<. 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. We give such a explaatio i the proof that follows. (1.1) Theorem If ad k are itegers with >0 ad 0 <k<,the 1 1 = +. k k 1 k Proof: The formula says that the umber of k-elemet subsets of a -elemet set is the sum of two umbers. Sice we ve used the sum priciple to explai other computatios ivolvig additio, it is atural to see if it applies here. To apply it, we eed to represet the set of k-elemet subsets as a uio of two other disjoit sets. Suppose our -elemet set is S = {x 1,x 2,...x }. The we wish to take S 1,say,tobethe ( k) -elemet set of all k-elemet subsets ad partitio

4 16 CHAPTER 1. COUNTING it ito two disjoit sets of k-elemet subsets, S 2 ad S 3, where the sizes of S 2 ad S 3 are ( 1) k 1 ad ( 1) ( k respectively. We ca do this as follows. Note that 1 ) k stads for the umber of k elemet subsets of the first 1elemetsx 1,x 2,...,x 1 of S. Thus we ca let S 3 be the set of k-elemet subsets of S that do t cotai x. The the oly possibility for S 2 is the set of k-elemet subsets of S that do cotai x. How ca we see that the umber of elemets of this set S 2 is ( 1 k 1)? By observig that removig x from each of the elemets of S 2 gives a (k 1)-elemet subset of S = {x 1,x 2,...x 1 }. Further each (k 1)-elemet subset of S arises i this way from oe ad oly oe k-elemet subset of S cotaiig x. Thus the umber of elemets of S 2 is the umber of (k 1)-elemet subsets of S,whichis ( 1 k 1).SiceS2 ad S 3 are two disjoit sets whose uio is S, this shows that the umber of elemet of S is ( 1) ( k ) k The Biomial Theorem What is (x +1) 4?Whatis(2+y) 4?Whatis(x + y) 4? The umber of k-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. Wearetaught i school that or i summatio otatio, (x + y) = x + (x + y) = x 1 y + 1 i=0 x 2 y x i y i. i Ufortuately whe we first see this theorem may of us do t really have the tools to see why it is true. Sice we kow that ( k) couts the umber of k-elemet subsets of a -elemet set, to really uderstad the theorem it makes sese to look for a way to associate a k-elemet set with each term of the expasio. Sice there is a y k associated with the term ( k), it is atural to look for a set that correspods to k distict ys. Whe we multiply together copies of the biomial x + y, wechoosey from some of the terms, ad x from the remaider, multiply the choices together ad add up all the products we get from all the ways of choosig the ys. The summads that give us x k y k are those that ivolvig choosig y from k of the biomials. The umber of ways to choose the k biomials givig y from the biomials is the umber of sets of k biomials we may choose out of, so the coefficiet of x k y k is ( k). Do you see how this proves the biomial theorem? If I have k labels of oe kid ad k labels of aother, i how may ways may I apply these labels to objects? Show that if we have k 1 labels of oe kid, k 2 labels of a secod kid, ad k 3 =! k 1 k 2 labels of a third kid, the there are k 1!k 2!k 3! ways to apply these labels to objects What is the coefficiet of x k 1 1 xk 2 2 xk 3 3 i (x + y + z)?

5 1.2. BINOMIAL COEFFICIENTS AND SUBSETS Ca you come up with more tha oe way to solve Exercise ad Exercise 1.2-7? Exercise ad Exercise have straightforward solutios For Exercise 1.2-6, there are ( ) k ways to choose the k objects that get the first label, ad the other objects get the secod label, so the aswer is ( ( k). For Exercise 1.2-7, there are ) k 1 ways to choose the k1 objects that get the first label, ad the there are ( k 1 ) k 2 ways to choose the objects that get the secod labels. After that, the remaiig k 3 = k 1 k 2 objects get the third labels. The total umber of labeligs is thus, by the product priciple, the product of the two biomial coefficiets, which simplifies to the ice expressio show i the exercise. Of course, this solutio begs a obvious questio; amely why did we get that ice formula i the secod exercise? A more elegat approach to Exercise ad Exercise appears below. Exercise shows why Exercise is importat. I expadig (x + y + z), we thik of writig dow copies of the triomial x + y + z side by side, ad imagie choosig x from some umber k 1 of them, choosig y from some umber k 2,adz from some umber k 3, multiplyig all the chose terms together, ad addig up over all ways of pickig the k i s ad makig our choices. Choosig x from a copy of the triomial labels that copy with x, ad the same for y ad z, so the umber of choices that yield x k 1 y k 2 z k 3 is the umber of ways to label objects with k 1 labels of oe kid, k 2 labels of a secod kid, ad k 3 labels of a third. Exercises E1.2-1 I the local ice cream shop, you may get a three-way sudae with three of the te flavors of ice cream, ay oe of three flavors of toppig, ad ay (or all or oe) of whipped cream, uts ad a cherry. How may differet sudaes are possible? Assume you have to choose three distict flavors of ice cream, ad that the order that you choose them i does t matter.! E1.2-2 Whe 1, 2,... k are oegative itegers that add to, theumber 1! 2! k! is called a multiomial coefficiet ad is deoted by ( ) 1, 2,..., k. A polyomial of the form x 1 +x 2 + +x k is called a multiomial. If you expad (x 1 +x 2 + +x k ),what will the coefficiet of x k 1 1 xk 2 2 xk k k be, assumig each k i is a o-egative iteger ad i k i =? You might try a example. E.g. if k =3,ad = 4, what is the coefficiet of x 2 1 x 2x 3. Why is the coditio i k i = importat. E1.2-3 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? E1.2-4 What is the formula we get for the biomial theorem if, istead of aalyzig the umber of ways to chose k distict ys, we aalyze the umber of ways to choose k distict xs? E1.2-5 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 3, ad three labels of type 4. What is the umber of ways of choosig three disjoit four elemet subsets form a four elemet set? (Clarificatio: whe a set is labelled, we assume each elemet gets exactly oe label.)

6 18 CHAPTER 1. COUNTING E1.2-6 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. (Clarificatio: the Presidet, Vice Presidet, Secretary ad Treasurer are the officers.) E1.2-7 Give at least two proofs that k k j = j. j k j