Basic Counting. Periklis A. Papakonstantinou. York University

Transcription

1 Basic Coutig Periklis A. Papakostatiou York Uiversity We survey elemetary coutig priciples ad related combiatorial argumets. This documet serves oly as a remider ad by o ways does it go i depth or is it complete. Despite this it is self-cotaied ad over-explaatory (o-dese. The mai use of this documet is to skim over the prerequisites from coutig eeded for COSC Basic coutig priciples Oe way to formally itroduce coutig priciples is through coutable sets, the cardiality of these sets, operatios amog sets, mappigs betwee these sets 2 ad formal power series. We follow a high-level approach (also adopted i most itroductory textbooks i Discrete Mathematics as log it is well uderstood how we ca techically formalize the argumets. Workig from basic priciples ad usig elemetary tools we develop the basic theory i its full geerality. Defiitio 1 (Priciple of Sum. Say that a task ca be performed i ways. Say that a secod task ( urelated to the first oe ca be performed i m ways. The, the combied task of doig somethig either from the first or from the secod task ca be performed i + m ways. Defiitio 2 (Priciple of Product. Say that a task ca be performed i ways. Say that a secod task ( urelated to the first oe ca be performed i m ways. The, the combied task of performig first the first task ad the the secod task ca be performed i m ways. I the above defiitios urelated meas that a realizatio of performig oe task does ot affect the umber of ways the secod task is performed. Example 1. A uiversity cosists oly of two faculties, amely Sciece ad Egieerig. O Tuesday at 7pm there are 10 exams that ca take place from the faculty of Sciece. At the same time there are 5 exams that ca take place from the faculty of Egieerig. The, for this Uiversity there are = 15 exams that ca take place (o Tuesday, 7pm. Example 2. O the same uiversity there are 15 exams that ca be scheduled o Tuesday at 7pm ad 5 (differet from the first exams that ca be scheduled o Tuesday at 8pm. The o Tuesday (at 7 ad 8pm there are 15 5 = 75 differet schedules of exams. 1 Please otify the author for ay typos, errors ad suggestios. papakos@cs.yorku.ca 2 I the last sectio we refer the reader to more advace material.

2 2 Periklis Papakostatiou (Summer 2005 Here are two more, straightforward examples where the questio is to cout the steps of a iterative algorithm. Example 3. How may times is the procedure Do-Somethig called i Nothig-Special- I whe the iput is (, m? Nothig-Special-I[, m] 1 for i 1 to 2 do 3 Do-Somethig[i] 4 for j 1 to m 5 do 6 Do-Somethig[j] By applyig the priciple of sum we have + m calls. Example 4. How may times is the procedure Do-Somethig called i Nothig-Special- II whe the iput is (, m? Nothig-Special-II[, m] 1 for i 1 to 2 do 3 for j 1 to m 4 do 5 Do-Somethig[i,j] By applyig the priciple of product we have m calls. Remark 1. All over this documet we use the terms differet ad distict iterchageably. We also use iterchageably the terms idistiguishable ad idetical. Example 5 (Number of arragemets. We have 10 differet slots (say that we have them umbered. We also have 10 differet balls. To cout the differet arragemets of these balls ito the slots we fix a order amog the slots (i.e. ame this slot to be the first, this slot to be the secod ad so o. I the first slot we ca put oe from the 10 balls. I the secod slot we ca put oe out of the (remaiig 9 balls...cotiuig i the same way we see that i the last slot we oly have oe choice. I total, we ca arrage the balls i = differet ways. What if we could t distiguish amog the slots? That is, suppose that the balls are distict say they are umbered as 1, 2,..., 10, but the slots are idistiguishable. I this case the arragemet 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 would be exactly the same as the arragemet 2, 3, 1, 4, 5, 6, 7, 8, 9, 10. Actually, every arragemet is idistiguishable from every other. Therefore, there is oly oe way to arrage the 10 balls i the 10 idistiguishable slots.

3 Basic Coutig (prerequisites for COSC We would also have oly oe way to arrage them if the balls were idistiguishable but the slots were distict; ad we would also have oe way to arrage them if both the balls ad the slots were idistiguishable. The questio of whether the slots ad the balls are distict is based o the defiitio of the coutig problem we wat to solve. What if istead of balls ad slots we had the followig problem? How may 10-digit decimal umbers (we allow the umbers to start with a zero exist, whe we restrict the umbers to have each digit differet from the others? It should be clear that we ca model our problem i a equivalet way as the balls-to-slots problem. I a iteger the positio where each digit appears is clearly distict ad every digit is distict from every other. These two trivial problems give rise to a geeral trick applied i coutig argumets. The techique is much stroger tha its applicatio i the previous cotext. (actually, the two previous examples - 10 balls distict to distict slots ad distict digit 10-digit umbers- are straightforwardly the same problem. To make the power of the techique well-uderstood, ad well-appreciated i its geerality we refer the reader to sectios 4.2 ad 4.3. Say that A ad B are two sets. A fuctio f is a mappig from A oto B, f : A B; that is, for every x A there exists oe f(x B. If there do ot exist two distict x 1, x 2 A (i.e. x 1 x 2 such that f(x 1 = f(x 2 the the fuctio is called 1 1 (or ijective. If for every elemet y B there exists a x A, such that f(x = y the the fuctio is called oto (or surjective. A fuctio f that it is both 1 1 ad oto is called 1 1 correspodece (or bijective. 1 1 correspodeces play a crucial role i coutig argumets. Cosider two fiite sets A ad B. Say that there exists a f : A B which is a 1 1 correspodece. The A = B. I other words: Suppose that we kow how to cout the elemets of a set A. Say that = A. If we maage to show that there exists a 1 1 correspodece f betwee A ad B, the we have also couted the elemets of B! They simply are. To make it eve more explicit: I the above cotext the set A is the set cotaiig as its elemets arragemets of 10 distict balls to 10 distict slots. You ca realize such a elemet as a sequece; example: 2, 3, 1, 4, 5, 7, 6, 8, 9, 10. The set B cosists of 10-digit (with distict digits umbers. Therefore, a 1 1 correspodece ca be the oe that for example maps 2, 3, 1, 4, 5, 7, 6, 8, 9, 10 to the umber It is easy to show that uder this mappig(fuctio: (i there are o two arragemets (of balls-to-slots that are mapped to the same umber ad (ii for every 10-digit umber there exists oe arragemet (of balls-to-slots that correspods to it. Therefore, if we have already couted the umber of balls-to-slots arragemets ad havig established the 1 1 correspodece, we also kow the umber of 10-digit umbers. Defiitio 3. Let N. If 1, defie! = ( Defie 0! = 1.

4 4 Periklis Papakostatiou (Summer 2005 Stirlig s approximatio Here is a quite useful approximatio of!:! = 2π ( ( Θ( e Perhaps a more coveiet form is the followig: 2π e <! < 2π e Arragemets of objects We deote by A( the umber of arragemets of distict objects (ito distict positios. The, it should be clear that A( = ( =!. Now, cosider the two problems we have just looked. For both of these problems it is clear that there exists a 1 1 correspodece betwee them ad the set of all arragemets of 10 distict objects (whose cardiality is A(10. Oe last remark before cocludig this sectio: Cosider agai the problem of determiig the umber of 10-digit itegers where each digit is distict. We solve a differet problem by ot allowig itegers to start with 0. How may itegers exist uder this modificatio? The aswer is A(10 A(9 = 10! 9! = (10 19! = 9 9!. Why? Oe way to argue is the followig: We apply the priciple of sum. therefore, # itegers startig with zero + # itegers ot startig with zero = # itegers # itegers ot startig with zero = # itegers - # itegers startig with zero Alteratively, we could have argued as follows: for the most sigificat digit we have 9 available choices (we exclude 0. The, for the secod most sigificat digit we have 9 choices (we exclude the oe we used for the first - but ow we ca use zero. For the third most sigificat digit we have 8 choices, etc. Therefore we have = 9 9!. What s the differece betwee the above two differet (valid argumets for the same eumeratio problem? I the first argumet we use the fact that we already kow how to cout arragemets of objects, but sice the problem is ot a mere arragemet problem we have to exclude some cases (which agai we cout usig A(. I the secod argumet we start from scratch. This is global all over mathematics. Oe could have stopped readig this documet here ad still be able to solve every coutig problem. Actually, coutig argumets is a brach of Discrete Mathematics where people ca very quickly get familiar with, havig very little backgroud. Despite this, without the material surveyed i the followig sectios some of the problems might get quite tedious ad difficult to be solved. Readig the forthcomig sectios keep i mid that each time you have to solve a coutig problem you should first sped some time uderstadig it ito detail ad the try to model it appropriately (so that you ca use the material itroduced i this documet.

5 Basic Coutig (prerequisites for COSC Permutatios ad Combiatios without repetitio 2.1 Permutatios Let, k N, k. We deote by P (, k the differet ways ito which we ca arrage k objects by choosig from a set of objects (without reusig objects. Fix a order o k positios. For the first positio we have choices, for the secod we have 1... for the last we have ( k + 1. Therefore, P (, k = ( 1... ( k + 1 =! ( k!. 2.2 Combiatios Basics Lets solve a similar problem, where the order of objects does ot cout. That is, give a set of distict objects we wat to cout how may differet subsets of k objects we ca costruct. We deote by C(, k or ( k the possible distict combiatios of k objects chose from a set of (distict objects. Usig the priciple of product we have: C(, k (# arragemets of k objects = P (, k That is, the umber of permutatios P (, k equals to the umber of ways of first choosig sets of k objects (without carig about their order-arragemet ad the arragig (i.e. itroducig orderig the objects of each set i all possible ways. Therefore, C(, k ( k = P (,k =. A(k k!( k! Oe immediate property of combiatios ( k (read as choose k is that ( k = ( k. Biomial Coefficiets Oe applicatio of combiatios is to determie coefficiets of the polyomial (s + t (actually, the reaso for iitially itroducig combiatios was these coefficiets. Biomial Coefficiets are the costats of the terms of the polyomial (s + t (this is a polyomial o two variables s ad t. Lets look o a simpler problem first. Cosider the polyomial (1 + x say over the field of reals. It is clear that this polyomial is of degree. Just for fu, lets expad (1 + x for some small values of. For = 2 we have: ad for = 3: (1 + x 2 = (1 + x(1 + x = 1 + x + x + x 2 = 1 + 2x + x 2 (1+x 3 = (1+x(1+x 2 = (1+x(1+x+x+x 2 = 1+x+x+x 2 +x+x 2 +x 2 +x 3 = 1+3x+3x 2 +x 3 I geeral we have (1 + x = c 0 + c 1 x + c 2 x c x ad we wish to determie the values of the costats c 0, c 1, c 2,..., c. Oe (o-trivial thig we could try is to guess the value of c i ad the prove it by iductio. We will follow a alterative, combiatorial argumet. We observe that i (1 + x k the term x k appears oly oce. Give a distict ame

6 6 Periklis Papakostatiou (Summer 2005 to each term (1 + x of the product (1 + x 1 (1 + x 2... (1 + x }{{. We observe that by the } (1+x times distributivity of multiplicatio if we write (1 + x = (1 + x k (1 + x k ad we fix what we mea by (1+x k ad by (1+x k, there will be oly oe x k due to the fixed (1+x k. This fixed x k will evetually appear ito the expasio if we successively apply the distributive law. So the problem i determiig c k reduces i coutig how may differet x k we ca have as a result of the expasio of (1 + x. That is, i how may differet ways we ca write (1 + x k. We have cosidered each parethesis (1 + x of the product (1 + x as distict. Therefore, we have ( k ways of choosig sets of differet parethesis (1 + x k. (1 + x = k=0 ( x k k Substitutig x = s t we get the followig: (s + t = k=0 ( s k t k k Properties of Biomial Coefficiets Here are some basic properties of the biomial coefficiets: Let, r N. 1. ( ( r = r 2. ( ( r = 1 ( r + 1 r 1 3. ( ( r = 1 r r 1 4. ( r+k ( k=0 k = r ( m ( m( r = ( r r m r 6. ( r ( s ( k=0 k k = r+s Property 2 yields a represetatio called Pascal s triagle. Cosider a arragemet of biomial coefficiets o a triagle. O oe vertex put ( ( 0 0. The cosider 1 ( 0 ad 1 1 ad put them below ( ( 0 0 ad alig them to the left ad to the right of 0 0. I the same fashio cosider ( ( 2 0, 2 ( 1 ad 2 2 ad alig them o the third row. I this way we get the Pascal s triagle, part of which is depicted below. ( 3 0 ( 0 ( 1 0 ( 1 ( 0 2 ( 2 1 ( 2 0 ( 1 3 ( 3 2 ( By property 2 a biomial coefficiet is the sum of the two biomial coefficiets i the triagle that are just above ad just to the left ad to the right.

7 Basic Coutig (prerequisites for COSC Apart from a proof based o algebraic maipulatios the above properties (1-6 ca be show usig combiatorial argumets (as almost every argumet preseted i this ( documet. For example, cosider property 3. Algebraically the proof is trivial: 1 r r 1 = ( 1! = ( r(r 1!( r! r. Here is a combiatorial proof: I order to choose r objects from, it suffices to choose 1 ad the choose r 1 amog (the remaiig 1. For the choice of the first object we have choices ad for the remaiig we have ( 1 r 1. Therefore, i total we have ( 1 r 1. But i this way we distiguish betwee the first chose object ad the rest (i.e. we pose a orderig. Hece, ( ( 1 r 1 = ( r r. That is, 1 r 1 equals to the umber of ways of choosig (without a order r( objects from ad the choosig oe of these r objects as the special object. Hece, 1 ( r r 1 = r. Exercise 1. It s a good exercise to give a combiatorial proof for property A applicatio of the (Biomial coefficiets of the polyomial (s + t Say that a ad b are costats. We wish to show that (x+a b = Θ(x b. We have determied that (x+a b = b ( b k=0 k x k a b k. Sice a ad b are costats, for every k the factor ( b k a b k is also a costat. Therefore, (x+a b = Θ(x b. (if you were asked to show that (x+a b = Θ(x b, as part of oe of your assigmets or tests you should have give the full proof. 3 Arragemets of objects cotaiig idistiguishable objects Cosider objects ad say that there are r groups each cotaiig idetical (idistiguishable objects. Say that the i-th group cosists of q i idetical elemets. Say that the umber of distict arragemets of the objects are N. We apply the followig combiatorial trick. Sice we kow how to cout arragemets of distict objects we will use this fact: For the momet modify the elemets (for example, by addig a distict subscript withi each group so as to make them distict. Note that N (stillcouts the objects as they iitially are (i.e. with groups of idistiguishable objects. Arrage the objects of the first group. The, arrage the objects of the secod group etc. Therefore, we have Nq 1!q 2!... q r! =!. That is, the umber of arragemets of object cosistig of r groups of idetical objects is! q 1!q 2!... q r! For the special case where there are oly two groups havig q 1 ad q 2 idetical elemets (i.e. = q 1 + q 2, the umber of differet arragemets is:! q 1!q 2! =! q 1!( q 1! = ( ( = q 1 q 2

8 8 Periklis Papakostatiou (Summer Permutatios ad Combiatios with repetitios 4.1 Permutatios with repetitios Suppose that we have distict objects but ulike the previous cases, ow we allow (ifiitely may repetitios of each object. We have objects ad we cout the umber of differet ways we ca arrage k of them whe repetitio is allowed. Fix a order o k slots. For the first slot we have choices. For the secod we also have choices (sice we allow repetitio of objects etc. Therefore, there are k differet permutatios with repetitios. 4.2 Applicatio of permutatios with repetitios Give distict objects we wish to cout all differet subsets (icludig the empty oe we ca costruct out of these objects. That is, we wat to determie the value of ( k=0 k. Oe easy way to do so is to associate (i.e. establish a 1 1 correspodece betwee -digit biary umbers (we allow the biary umbers to start with zero ad subsets of objects. Fix a orderig o the objects. A object ca either be icluded i a subset or ot. For a arbitrary subset, we associate with every chose object a 1 ad with every o-chose object a 0. For example, cosider the 3 objects {a, b, c}. Fix a orderig, say a, b, c. The, the biary umber 011 correspods to the subset {b, c}. Note that for every object there is a uique biary umber that correspods to it. Ad, for every subset there exists a biary umber which correspods to it. That is, the mappig is a 1 1 correspodece. Now, i order to determie ( k=0 k it suffices to determie the umber of distict -digit biary umbers. This is just the umber of permutatios with repetitio of two digits (0 ad 1 ito positios. That is, k=0 ( = 2 k 4.3 Combiatios with repetitios Coutig the umber of combiatios (i.e. the order does ot cout with repetitio is more trickier. I order to cout the differet subsets of size (cardiality k from a set of objects whe repetitio is allowed, this calls for the coutig trick where we establish a 1 1 correspodece betwee a set of thigs we kow how to cout ad the target set of thigs we wat to cout. We give some hits for the geeral proof through a illustrative example. Suppose that we have 4 objects say a, b, c, d ad we wat to cout how may combiatios with repetitio we have resultig sets of 5 objects. Sice we allow repetitio we formally talk about multisets of 5 objects. A multiset is a set where a elemet appears together with its multiplicity (or equivaletly may appear more tha oce. Oe possible choice is the multiset {a, b, b, c, d} ad aother is {a, b, a, d, c}. I order to establish the 1-1 correspodece we itroduce some orderig ito the problem. Note that the way i which we will itroduce this orderig does ot hurt geerality (i.e. the uordered ature of the

9 Basic Coutig (prerequisites for COSC multiset. Fix a orderig o the five objects, say a b c d e. We list the resultig multisets accordig to this order. For example: {a, b, b, c, d} {a, a, b, c, d} {a, a, a, d, d} {b, b, b, b, b} {a, a, a, a, a}. Our goal is to establish a 1 1 correspodece with a set that we kow how to cout. We use 4 1 = 3 vertical bars to somehow desigate the begi ad the ed of each group of idetical objects. For example, a bb c d. The first vertical bar correspods to objects a that are o the left of this bar. Betwee the first ad the secod vertical bar we have objects b. Betwee the secod ad the third vertical bar are objects c. Ad to the right of the third vertical bar we have objects d. Note that the itroduced sematics for the vertical bars allows us to use oly oe symbol to deote objects. That is, if we write x xx x x this uiquely correspods to {a, b, b, c, d}. Hece, we have: {a, b, b, c, d} x xx x x {a, a, b, c, d} xx x x x {a, a, a, d, d} xxx xx {b, b, b, b, b} xxxxx {a, a, a, a, a} xxxxx. It is also easy to show that every arragemet of 5 x s ad 3 vertical bars correspods to a multiset. Therefore, we have established a 1 1 correspodece from the multisets we wat to cout to the arragemets of 5 + (4 1 = 8 objects, where we have oe group of 5 idetical objects ad oe group of 4 1 = 3 idetical objects. As we have already see the umber of these differet arragemets is 8! 5!3!. It is a good exercise to attempt to cout the above umber of combiatios with repetitios startig from basic priciples (i.e. the priciples of sum ad product. The tempted reader will shortly realize that oe could easily get ito trouble. It worths stressig out that: (i we may wish to establish a 1 1 correspodece betwee the elemets of the set we wat to cout (here a elemet is a multiset of size 4 ad the elemets of a set we kow how to cout (here a elemet is a sequece of 5 x s ad 3 vertical bars ad (ii the elemets of the set we already kow how to cout might have quite differet structure tha the elemets of the set we wat to cout. I the above example, a multiset does ot have a order, where the sequeces (arragemets do; the multiset cosisted of 5 elemets where the arragemet of 8; the multiset cosists of 4 differet elemets where the arragemet of 2..

10 10 Periklis Papakostatiou (Summer 2005 Now, it is ot hard to give a geeral proof of the followig fact: The umber of combiatios where we choose with repetitio k objects from distict objects is ( + k 1 k Exercise 2. Show that the umber of combiatios where we choose with repetitio k objects from distict objects is ( +k 1 k. 4.4 Applicatios of combiatios with repetitio Example 6. We wat to cout the umber of calls to Do-Somethig (lie 7. Nothig-Special[] 1 for i 1 to 2 do 3 for j 1 to i 4 do 5 for k 1 to j 6 do 7 Do-Somethig[i,j,k] It seems cumbersome to work from basic priciples. We observe that if we try to cout ordered permutatios correspodig to (i, j, k we may also ru ito the same troubles as if we were workig from basic priciples. The reaso is that (5, 3, 2 is valid, where (5, 10, 2 is ot (j ca ever get a higher value tha i. Therefore, workig i this way it seems that we have to exclude cases with complex iteractios. It appears that a differet explicitly uordered approach yields a straightforward aswer. If we cosider (multisets of three elemets istead of permutatios ad we agree o the covetio that the largest elemet correspods to i, the secod largest to j ad the smallest to k the we have resolved the problem. Sice i, j, k may take the same value the we are talkig about combiatios with repetitios. Note that coutig all the three elemet multisets which are subsets of {1, 2,..., } we have every possible assigmet of values for i, j ad k. Therefore, the umber of calls to Do-Somethig is ( ( = Example 7. The followig problem falls ito a well-kow ad well-studied category of problems kow as balls-to-bis. Suppose that we have distict bis ad r idistiguishable balls. What is the umber of differet placemets of balls ito bis? Note that a bi might be empty. We realize the problem as assigig bis to balls. We assig the bis as follows: (i choose r bis, (ii every bi ca be chose more tha oce ad (iii the order does ot cout sice the balls are idistiguishable. Therefore, the umber of placemets is ( +r 1 r. Before readig further you may wish to covice yourself about (i, (ii ad especially for (iii.

11 Basic Coutig (prerequisites for COSC Example 8. Suppose that we wat to cout the umber of o-egative iteger solutios to the followig equatio: x 1 + x x = r,, r N This problem is equivalet to the previous oe, sice we may cosider the balls to correspod to 1s ad the bis to x i s. Therefore, the umber of iteger solutios is ( +r 1 r. Remark 2. Regardig examples 7 ad 8, if somebody wated to work from more basic priciples the it is possible that at some poit might come up with the same trick used i the proof sketch of sectio 4.3. Up to ow it should be clear that the followig are equivalet. The combiatios with repetitios of r objects from objects. The umber of placemets of r idistiguishable balls ito distict bis. The umber of o-egative iteger solutios of x 1 + x x = r,, r N 5 Balls to bis It is true that may coutig problems ca be modeled as a balls-to-bis problem. Here is a table givig the umber of ways for placig balls ito bis for some basic balls-to-bis problems. Balls-to-Bis Problem distict bis r distict balls distict bis r distict balls it also couts the order i which we put the balls ito bis distict bis r idistiguishable balls Number of placemets whe we allow empty bis r (+r 1! ( 1! ( +r 1 r Exercise 3. It is a good exercise to prove (usig combiatorial argumets each of the above. 6 Priciple of Iclusio-Exclusio 6.1 The priciple Ofte i coutig problems we wat to cout the total umber of distict elemets of two or more sets of objects. Say that we have two fiite sets A ad B. We wat to cout the umber of elemets that are i the uio of the two sets. Straightforwardly this equals to A B = A + B A B (6.1

12 12 Periklis Papakostatiou (Summer 2005 (that is the total umber of distict elemets of A ad B equals to the sum of their cardialities mius the umber of their commo elemets - i.e. the cardiality of their itersectio. Remark 3. Oe may woder why is it simpler to compute the umber of elemets i A B by the above formula. Of course this is ot always the case. But, i some cases we ca model some coutig problems such that we kow (or we kow how to compute A, B ad we also do kow (or we kow how to compute how to cout their commo elemets. As a rule of thumb cosider the sets A ad B to correspod to the bad cases for a coutig problem. Talkig a bit abstractly cosider the followig example: Say that we have a set S ad we have two ( bad properties o the elemets of S. Our goal is to cout the umber of elemets of S that do ot satisfy ay of the two ( bad properties. Suppose that we are give S = ad the two properties such that P 1 elemets of S satisfy the first ad P 2 satisfy the secod ad that P elemets of S satisfy both. The, the umber of elemets that does ot satisfy ay of the two properties is (P 1 + P 2 P. What if we have more tha two sets. This case gives rise to more complicated iteractios. For example A B C = (A B C = A B + C (A B C (it is easy to verify that is distributive for ad vice-versa = A B + C (A C (A B = A + B + A B + C (A C (A B = A + B + A B + C ( (A C + (A B (A C (A B = A + B + C + A B (A C (A B + (A B C We derived the above result by repeatedly applyig equatio 6.1. Therefore, for sets A 1, A 2,..., A it is easy to show by iductio that: A i = A i A i1 A i2 + A i1 A i2 A i3... +( 1 A i i=1 i=1 1 i 1 <i 2 1 i 1 <i 2 <i 3 The iterested reader should try to give a combiatorial proof for the above. Up to ow we were discussig problems where we wated to cout i the exact sese. It is quite ofte the case that a upper or lower boud would be sufficiet for a specific problem. It is also the case that for most problems it is hard to compute the umber of elemets i the itersectios of sets. For example, say that after modelig our coutig problem we have 4 sets A 1, A 2, A 3, A 4. Perhaps due to the ature of the problem it may be the case that it is easy to compute the cardiality of the itersectio of ay two sets ( A i A j, i j; but o the other had it gets quite o-trivial to compute the itersectio of three or four sets. A place where the followig iequalities might appear helpful is: (i a upper or lower boud is sufficiet ad (ii we may wish to icrease m up to the poit we achieve the desired accuracy. i=1

13 Basic Coutig (prerequisites for COSC It is easy to show by iductio that the followig iequalities (kow as (Boole- Boferroi iequalities hold. For m odd: A i i=1 i=1 A i 1 i 1 <i 2 A i1 A i2 + 1 i 1 <i 2 <i 3 A i1 A i2 A i i 1 <...<i m m A ik k=1 For m eve: A i i=1 i=1 A i 1 i 1 <i 2 A i1 A i2 + 1 i 1 <i 2 <i 3 A i1 A i2 A i i 1 <...<i m m A ik k=1 6.2 Applicatios of the Priciple of Iclusio-Exclusio Yet aother balls-to-bis problem We wish to cout the umber of placemets of 5 distict balls to 4 distict bis with the costrait that o bi is empty. Here is how we apply the iclusio-exclusio priciple. We already kow that there are r ways (Sectio 5 to place r balls ito bis, whe we allow empty bis. We wish to exclude the umber of placemets which have (at least oe empty bi. Cosider the set A of all placemets where empty bis are allowed. Choose your favorite way of ecodig a placemet i.e. the elemets of A 3. Say that Āi is the set cotaiig all the placemets where the i-th bi is empty. Therefore, we wat to cout A Ā1 Ā2 Ā3 Ā4. Remark 4. Just for fu you may wish to approach this problem from basic priciples. Whe doig so you will realize that there are complex iteractio betwee the cases. That is, you have to cosider the case where the first bi is empty ad the others ot; ad the case where the last two bis are empty ad the first ot; ad so o. I geeral you have to carefully cosider all these complex iteractios that arise i this coutig problem. The Iclusio-Exclusio priciple is othig more tha a systematic way of revealig all these iteractios. The set (Ā1 Ā2 Ā3 Ā4 is othig else but the set cotaiig all placemets where at least oe of these bis is empty. Recall that the uio betwee sets is defied as a logical OR of cotaiig a elemet either from oe or from the other set. Note that this OR is exactly what we wat. Pay attetio to the fact that a placemet which is a elemet i Ā1 might as well be a elemet i Ā2. We wish to compute Ā1 Ā2 Ā3 Ā4 = ( Ā1 + Ā2 + Ā3 + Ā4 ( Ā 1 Ā2 + Ā1 Ā3 + Ā1 Ā4 + Ā2 Ā3 + Ā2 Ā4 + Ā3 Ā4 + ( Ā 1 Ā2 Ā3 + Ā1 Ā2 Ā4 + Ā2 Ā3 Ā4 + Ā1 Ā3 Ā4 + ( Ā 1 Ā2 Ā3 Ā4 3 Perhaps you wat to represet each placemet as a set {(ball i, bi j place ball i ito bi j}. That is, the set A has as its elemets sets of the previous form, that correspod to placemets of the r balls ito bis.

14 14 Periklis Papakostatiou (Summer 2005 How may placemets exist such that the first bi is empty? These are just the umber of placemets of the 5 balls ito the remaiig 3 bis; i.e. Ā1 = 3 5. Similarly, Āi = 3 5, 1 i 4. How may placemets exist such that both the first ad the secod bi is empty? These are just the umber of placemets of the 5 balls ito the remaiig 2 bis. Therefore, Ā1 Ā2 = 2 5. Obviously, Āi Āj = 2 5 for every 1 i < j 4. Similarly, we have that Āi Āj Āk = 1 5, for every 1 i < j < k 4. Also Ā1 Ā2 Ā3 Ā4 = 0 (sice the 5 balls must be placed somewhere. Here is a observatio regardig the equatio i the Iclusio-Exclusio priciple. How may terms Āi exist? There are ( 4 1 = 4. How may terms Ā i Āj (where i j exist? There are exactly ( 4 2 = 6. How may terms Ā i Āj Āk? There are ( 4 3 = 4. Ad there is ( 4 4 = 1 term Ā 1 Ā2 Ā3 Ā4. Therefore, Ā1 Ā2 Ā3 Ā4 = ( ( ( ( = 780. Hece, the umber of placemets we were asked to compute is A Ā1 Ā2 Ā3 Ā4 = = 240 Exercise 4. Geeralize the above result whe the umber of balls is r ad the umber of bis. Exercise 5. Suppose that we have 20 bis ad 30 balls. Istead of coutig the exact umber of placemets (with the costrait that each bi cotais at least oe ball compute a upper ad lower boud, usig Boole-Boferroi iequalities. Compute these bouds with two differet accuracies. The lower boud for 2 ad 4 terms ad the upper boud for 1 ad 3 terms. Exercise 6. Cout the umber of arragemets of the digits 0, 1,..., 9 exist, uder the costrait that the first digit is greater tha 1 ad the last digit is less tha 8? Euler s φ fuctio We are goig to determie a formula for a fuctio kow as Euler s φ fuctio. This fuctio plays a importat role i the aalysis of the Miller-Rabi primality testig algorithm (we will see this algorithm i oe of our last lectures. First we eed some defiitios. Defiitio 4. Let p, Z, p 0. We say that p divides, p, or that p is a divisor of iff there exists a atural umber k 0 such that = kp. We say that p Z, p > 1 is a prime umber iff the oly divisors of p are 1 ad p. Else we say that p is composite. Let a, b be two itegers ot both zero. We deote by gcd(a, b the greatest iteger amog the commo divisors of a ad b. We say that two itegers a ad b are relatively prime (or coprime iff their greatest commo divisor is the uit, gcd(a, b = 1.

15 Basic Coutig (prerequisites for COSC A fudametal theorem of Number Theory is the Uique Factorizatio ito powers of primes. Theorem 1 (Uique Factorizatio. A composite umber Z + ca be uiquely writte as a product of powers of prime umbers p 1, p 2..., p i, = p e 1 1 p e p e i i, where e i Z +. Defiitio 5. For a give positive iteger, the Euler s φ fuctio deotes the umber of positive itegers m < such that ad m are relatively prime. For example, φ(3 = 2, φ(6 = 2 ad φ(7 = 7 1 = 6. Clearly, whe p is prime we have φ(p = p 1. We wat to show that for Z + φ( = p ( 1 1 p where the product rus over all prime divisors of. As i the previous balls-to-bis example, we wat to determie the bad cases. Why? We wat to express φ( i terms of prime divisors of. It seems quite ituitive that the bad cases are the oes i which (i we have this correlatio ad (ii more importatly we do kow how to compute the umber of elemets i each bad case. By the Uique Factorizatio theorem let = p e 1 1 p e p e i i. It is clear that if m < is ot relatively prime to the there exist ot all zero e 1, e 2... e i where 0 e k e k, k = 1,..., i such that m = p e 1 1 p e p e i i. Say that Āl is the set of all itegers a which are less tha or equal to such that p l a. Therefore, the set i l=1 Āl cotais all itegers that are less tha or equal to ad they are ot relatively prime to. If A = {1, 2,..., }, the we have that φ( = A i l=1 Āl = i l=1 Āl. Thus, the problem reduces to the computatio of i l=1 Āl. By the defiitio of Ā k, we have that Āk = p k. For the same reaso Āk Āj = p k p j, for every 1 k < j. Similarly, A k1 A k1... A kv = p k1 p k2...p kv, for every 1 k 1 < k 2 <... < k v. I the iclusio-exclusio formula we have i terms Āk, ( i 2 terms Āk Āj,..., 1 term Ā1 Ā1... Āi. Now it is rather easy to show (by iductio o i that if we factorize ( i k=1 Āk 1 k 1 <k 2 i A k1 A k2 + 1 k 1 <k 2 <k 3 i A k1 A k2 A k3... +( 1 i i k=1 Āk we get i ( i=1 1 1 p i. Therefore, φ( = p ( 1 1 p. Exercise 7. How may are the positive divisors of 2100? 7 Further readig As already metioed i the itroductio, this documet illustrates the very basic coutig priciples. The treatmet is quite elemetary, o-dese ad it lacks completeess. Also, it

16 16 Periklis Papakostatiou (Summer 2005 obviously lacks of examples. The iterested studets are ecouraged to study the prerequisites for coutig by cosultig the material of previous related courses. Also, a excellet text is the followig: Itroductio to Combiatorial Mathematics by C.L. Liu, McGraw-Hill. Studets with a geeral iterest i Combiatorics are strogly ecouraged to study material related to the followig keywords: geeratig fuctios, recurrece relatios, characteristic polyomials (for recurrece relatios, Polyá s theory.