Combinatorics II. Combinatorics. Product Rule. Sum Rule II. Theorem (Product Rule) Theorem (Sum Rule)


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1 Combiatorics Combiatorics I Slides by Christopher M. Bourke Istructor: Berthe Y. Choueiry Sprig 26 Computer Sciece & Egieerig 235 to Discrete Mathematics Sectios & of Rose Combiatorics is the study of collectios of objects. Specifically, coutig objects, arragemet, deragemet, etc. of objects alog with their mathematical properties. Coutig objects is importat i order to aalyze algorithms ad compute discrete probabilities. Origially, combiatorics was motivated by gamblig: coutig cofiguratios is essetial to elemetary probability. Combiatorics II Product Rule A simple example: How may arragemets are there of a deck of 52 cards? I additio, combiatorics ca be used as a proof techique. A combiatorial proof is a proof method that uses coutig argumets to prove a statemet. If two evets are ot mutually exclusive (that is, we do them separately), the we apply the product rule. (Product Rule) Suppose a procedure ca be accomplished with two disjoit subtasks. If there are ways of doig the first task ad 2 ways of doig the secod, the there are 2 ways of doig the overall procedure. Sum Rule I Sum Rule II If two evets are mutually exclusive, that is, they caot be doe at the same time, the we must apply the sum rule. (Sum Rule) If a evet e ca be doe i ways ad a evet e 2 ca be doe i 2 ways ad e ad e 2 are mutually exclusive, the the umber of ways of both evets occurrig is + 2 There is a atural geeralizatio to ay sequece of m tasks; amely the umber of ways m mutually exclusive evets ca occur is m + m We ca give aother formulatio i terms of sets. Let A, A 2,..., A m be pairwise disjoit sets. The A A 2 A m A + A A m I fact, this is a special case of the geeral Priciple of IclusioExclusio.
2 Priciple of IclusioExclusio (PIE) I Priciple of IclusioExclusio (PIE) II We caot use the sum rule because we would be over coutig the umber of possible outcomes. Say there are two evets, e ad e 2 for which there are ad 2 possible outcomes respectively. Now, say that oly oe evet ca occur, ot both. I this situatio, we caot apply the sum rule? Why? Istead, we have to cout the umber of possible outcomes of e ad e 2 mius the umber of possible outcomes i commo to both; i.e. the umber of ways to do both tasks. If agai we thik of them as sets, we have A + A 2 A A 2 Priciple of IclusioExclusio (PIE) III More geerally, we have the followig. Lemma Let A, B be subsets of a fiite set U. The. A B A + B A B 2. A B mi{ A, B } 3. A \ B A A B A B 4. A U A 5. A B A B A B A+B 2 A B A\B + B\A 6. A B A B Priciple of IclusioExclusio (PIE) I Let A, A 2,..., A be fiite sets, the A A 2 A i A i A i A j i<j + A i A j A k i<j<k +( ) + A A 2 A Priciple of IclusioExclusio (PIE) II Priciple of IclusioExclusio (PIE) III To illustrate, whe 3, we have Each summatio is over all i, pairs i, j with i < j, triples i, j, k with i < j < k etc. A A 2 A 3 A + A 2 + A 3 [ A A 2 + A A 3 + A 2 A 3 ] + A A 2 A 3
3 Priciple of IclusioExclusio (PIE) IV Priciple of IclusioExclusio (PIE) I I To illustrate, whe 4, we have A A 2 A 3 A 4 A + A 2 + A 3 + A 4 [ A A 2 + A A A A 4 ] A 2 A 3 + A 2 A 4 + A 3 A 4 [ + A A 2 A 3 + A A 2 A 4 + ] A A 3 A 4 + A 2 A 3 A 4 A A 2 A 3 A 4 How may itegers betwee ad 3 (iclusive) are Let. Divisible by at least oe of 3, 5, 7? 2. Divisible by 3 ad by 5 but ot by 7? 3. Divisible by 5 but by either 3 or 7? A { 3 3 } B { 3 5 } C { 3 7 } Priciple of IclusioExclusio (PIE) II I Priciple of IclusioExclusio (PIE) III I How big are each of these sets? We ca easily use the floor fuctio; A 3/3 B 3/5 6 C 3/7 42 For () above, we are asked to fid A B C. By the priciple of iclusioexclusio, we have that A B C A + B + C [ ] A B + A C + B C + A B C It remais to fid the fial 4 cardialities. All three divisors, 3, 5, 7 are relatively prime. Thus, ay iteger that is divisible by both 3 ad 5 must simply be divisible by 5. Priciple of IclusioExclusio (PIE) IV I Priciple of IclusioExclusio (PIE) V I Usig the same reasoig for all pairs (ad the triple) we have Therefore, A B 3/5 2 A C 3/2 4 B C 3/35 8 A B C 3/5 2 For (2) above, it is eough to fid (A B) \ C By the defiitio of setmius, (A B) \ C A B A B C A B C
4 Priciple of IclusioExclusio (PIE) VI I For (3) above, we are asked to fid B \ (A C) B B (A C) By distributig B over the itersectio, we get B (A C) (B A) (B C) B A + B C (B A) (B C) B A + B C B A C So the aswer is B Priciple of IclusioExclusio (PIE) I II The priciple of iclusioexclusio ca be used to cout the umber of oto fuctios. Let A, B be oempty sets of cardiality m, with m. The there are ( ) ( ) ( ) m ( ) m + ( 2) m + ( ) m 2 i.e. i ( )i( i) ( i) m oto fuctios f : A B. See textbook page 46. Priciple of IclusioExclusio (PIE) II II Priciple of IclusioExclusio (PIE) III II How may ways of givig out 6 pieces of cady to 3 childre if each child must receive at least oe piece? This ca be modeled by lettig A represet the set of cadies ad B be the set of childre. To cout how may there are, we apply the theorem ad get (for m 6, 3), 3 6 ( ) 3 ( ) 3 (3 ) (3 2) 6 54 The a fuctio f : A B ca be iterpreted as givig cady a i to child c j. Sice each child must receive at least oe cady, we are cosiderig oly oto fuctios. Deragemets I Deragemets II Cosider the hatcheck problem. A employee checks hats from customers. However, he forgets to tag them. Whe customer s checkout their hats, they are give oe at radom. What is the probability that o oe will get their hat back? This ca be modeled usig deragemets: permutatios of objects such that o elemet is i its origial positio. For example, 2453 is a deragemet of 2345, but 2543 is ot. The umber of deragemets of a set with elemets is [ D!! + 2! 3! + ] ( )! See textbook page 46.
5 Deragemets III The Pigeohole Priciple I Thus, the aswer to the hatcheck problem is Its iterestig to ote that D! e! + 2! 3! + + ( )! So that the probability of the hatcheck problem coverges; D lim e ! The pigeohole priciple states that if there are more pigeos tha there are roosts (pigeoholes), for at least oe pigeohole, more tha two pigeos must be i it. (Pigeohole Priciple) If k + or more objects are placed ito k boxes, the there is at least oe box cotaiig two ore more objects. This is a fudametal tool of elemetary discrete mathematics. It is also kow as the Dirichlet Drawer Priciple. The Pigeohole Priciple II Geeralized Pigeohole Priciple I It is seemigly simple, but very powerful. The difficulty comes i where ad how to apply it. Some simple applicatios i computer sciece: Calculatig the probability of Hash fuctios havig a collisio. Provig that there ca be o lossless compressio algorithm compressig all files to withi a certai ratio. Lemma For two fiite sets A, B there exists a bijectio f : A B if ad oly if A B. If N objects are placed ito k boxes the there is at least oe box cotaiig at least N k I ay group of 367 or more people, at least two of them must have bee bor o the same date. Geeralized Pigeohole Priciple II Geeralized Pigeohole Priciple III A probabilistic geeralizatio states that if objects are radomly put ito m boxes with uiform probability (each object is placed i a give box with probability /m) the at least oe box will hold more tha oe object with probability, m! (m )!m Amog people, what is the probability that two or more will have the same birthday? Here, ad m 365 (igore leapyears). Thus, the probability that two will have the same birthday is 365!.69 (365 )!365 So less tha a 2% probability!
6 Pigeohole Priciple I I Pigeohole Priciple II I Show that i a room of people with certai acquaitaces, some pair must have the same umber of acquaitaces. Note that this is equivalet to showig that ay symmetric, irreflexive relatio o elemets must have two elemets with the same umber of relatios. We ll show by cotradictio usig the pigeohole priciple. Assume to the cotrary that every perso has a differet umber of acquaitaces;,,..., (we caot have here because it is irreflexive). Are we doe? No, sice we oly have people, this is okay (i.e. there are possibilities). We eed to use the fact that acquaitaceship is a symmetric, irreflexive relatio. I particular, some perso kows people while aother kows people. I other words, someoe kows everyoe, but there is also a perso that kows o oe. Thus, we have reached a cotradictio. Pigeohole Priciple I II Show that i ay list of te oegative itegers, A,..., A 9, there is a strig of cosecutive items of the list a l, a l+,... whose sum is divisible by. Cosider the followig umbers. a a + a a + a + a 2. a + a + a a 9 Pigeohole Priciple II II Otherwise, we observe that each of these umbers must be i oe of the cogruece classes mod, 2 mod,..., 9 mod By the pigeohole priciple, at least two of the itegers above must lie i the same cogruece class. Say a, a lie i the cogruece class k mod. The (a a ) k k(mod ) ad so the differece (a a ) is divisible by. If ay oe of them is divisible by the we are doe. Pigeohole Priciple I III Say 3 buses are to trasport 2 Corhusker fas to Colorado. Each bus has 8 seats. Show that. Oe of the buses will have 4 empty seats. 2. Oe of the buses will carry at least 67 passegers. For (), the total umber of seats is seats. Thus there will be empty seats total. Pigeohole Priciple II III By the geeralized pigeohole priciple, with 4 empty seats amog 3 buses, oe bus will have at least empty seats. For (2) above, by the pigeohole priciple, seatig 2 passegers amog 3 buses, oe will have at least passegers.
7 Permutatios I Permutatios II A permutatio of a set of distict objects is a ordered arragemet of these objects. A ordered arragemet of r elemets of a set is called a rpermutatio. The umber of r permutatios of a set with distict elemets is r P (, r) ( i) ( )( 2) ( r + ) i It follows that I particular, P (, r)! ( r)! P (, )! Agai, ote here that order is importat. It is ecessary to distiguish i what cases order is importat ad i which it is ot. Permutatios I Permutatios II How may pairs of dace parters ca be selected from a group of 2 wome ad 2 me? The first woma ca be partered with ay of the 2 me. The secod with ay of the remaiig 9, etc. To parter all 2 wome, we have P (2, 2) I how may ways ca the Eglish letters be arraged so that there are exactly te letters betwee a ad z? The umber of ways of arragig letters betwee a ad z is P (24, ). Sice we ca choose either a or z to come first, there are 2P (24, ) arragemets of this 2letter block. For the remaiig 4 letters, there are P (5, 5) 5! arragemets. I all, there are 2P (24, ) 5! Permutatios III Permutatios III  Cotiued How may permutatios of the letters a, b, c, d, e, f, g cotai either the patter bge or eaf? The umber of total permutatios is P (7, 7) 7!. If we fix the patter bge, the we ca cosider it as a sigle block. Thus, the umber of permutatios with this patter is P (5, 5) 5!. Fixig the patter eaf we have the same umber, 5!. Thus we have Is this correct? 7! 2(5!) No. We have take away too may permutatios: oes cotaiig both eaf ad bge. Here there are two cases, whe eaf comes first ad whe bge comes first.
8 Permutatios III  Cotiued Combiatios I Defiitio eaf caot come before bge, so this is ot a problem. If bge comes first, it must be the case that we have bgeaf as a sigle block ad so we have 3 blocks or 3! arragemets. Altogether we have 7! 2(5!) + 3! 486 Whereas permutatios cosider order, combiatios are used whe order does ot matter. Defiitio A kcombiatio of elemets of a set is a uordered selectio of k elemets from the set. A combiatio is simply a subset of cardiality k. Combiatios II Defiitio Combiatios III Defiitio The umber of kcombiatios of a set with cardiality with k is ( )! C(, k) k ( k)!k! Note: the otatio, ( k) is read, choose k. I T EX use { choose k} (with the forward slash). A useful fact about combiatios is that they are symmetric. etc. ( ) ( ) ( ) ( ) 2 2 Combiatios IV Defiitio Combiatios I I This is formalized i the followig corollary. Corollary Let, k be oegative itegers with k, the ( ) ( ) k k I the Powerball lottery, you pick five umbers betwee ad 55 ad a sigle powerball umber betwee ad 42. How may possible plays are there? Order here does t matter, so the umber of ways of choosig five regular umbers is ( ) 55 5
9 Combiatios II I Combiatios I II We ca choose amog 42 power ball umbers. These evets are ot mutually exclusive, thus we use the product rule. ( ) 55 55! , 7, (55 5)!5! So the odds of wiig are 46, 7, 962 <.6845 I a sequece of coi tosses, how may ways ca 3 heads ad 7 tails come up? The umber of ways of choosig 3 heads out of coi tosses is ( ) 3 Combiatios II II Combiatios I III However, this is the same as choosig 7 tails out of coi tosses; ( ) 3 ( ) 2 7 This is a perfect illustratio of the previous corollary. How may possible committees of five people ca be chose from 2 me ad 2 wome if. if exactly three me must be o each committee? 2. if at least four wome must be o each committee? Combiatios II III Combiatios III III For (), we must choose 3 me from 2 the two wome from 2. These are ot mutually exclusive, thus the product rule applies. ( )( ) For (2), we cosider two cases; the case where four wome are chose ad the case where five wome are chose. These two cases are mutually exclusive so we use the additio rule. For the first case we have ( )( ) 2 2 4
10 Combiatios IV III Biomial Coefficiets I Ad for the secod we have ( )( ) Together we have ( 2 )( 2 4 ) + ( 2 )( 2 5 ), 692 The umber of rcombiatios, ( r) is also called a biomial coefficiet. They are the coefficiets i the expasio of the expressio (multivariate polyomial), (x + y). A biomial is a sum of two terms. Biomial Coefficiets II Biomial Coefficiets III Expadig the summatio, we have (Biomial ) Let x, y be variables ad let be a oegative iteger. The (x + y) j ( ) x j y j j For example, (x + y) ( ) x + ( ) x y + ( 2) x 2 y ( ) xy + ( ) y (x + y) 3 (x + y)(x + y)(x + y) (x + y)(x 2 + 2xy + y 2 ) x 3 + 3x 2 y + 3xy 2 + y 3 Biomial Coefficiets I What is the coefficiet of the term x 8 y 2 i the expasio of (3x + 4y) 2? Biomial Coefficiets I More May useful idetities ad facts come from the Biomial. Corollary By the Biomial, we have 2 ( ) 2 (3x + 4y) j (3x) 2 j (4y) j j So whe j 2, we have ( ) 2 (3x) 8 (4y) 2 2 ( ) k k ( ) ( ) k k ( ) 2 k k k k 2 3 so the coefficiet is 2! 2!8!
11 Biomial Coefficiets II More Biomial Coefficiets III More Most of these ca be prove by either iductio or by a combiatorial argumet. Check textbook for proofs, which are based o: 2 ( + ), (( ) + ), 3 ( + 2). (Vadermode s Idetity) Let m,, r be oegative itegers with r ot exceedig either m or. The ( ) m + r ( )( ) m r r k k k Biomial Coefficiets IV More Biomial Coefficiets V More Takig m r i the Vadermode s idetity. Corollary If is a oegative iteger, the ( ) 2 k ( ) 2 k Let, r be oegative itegers, r. The ( ) + r + jr ( ) j r Corollary Biomial Coefficiets I Pascal s Idetity & Triagle Pascal s Triagle The followig is kow as Pascal s Idetity which gives a useful idetity for efficietly computig biomial coefficiets. (Pascal s Idetity) Let, k Z + with k. The ( ) ( ) + k k + ( ) k Pascal s Idetity forms the basis of a geometric object kow as Pascal s Triagle. ( ) ( ) ( ) ( 2 ) ( 2 ) ( 2 2) ( 3 ) ( 3 ) ( 3 ) ( 3 2 3) ( 4 ) ( 4 ) ( 4 ) ( 4 ) ( ) ( )
12 Pascal s Triagle Pascal s Triagle ( ) ( ) ( ) ( 2 ) ( 2 ) ( 2 2) ( 3 ) ( 3 ) ( 3 ) ( 3 2 3) ( 3 ) ( ) ( 3 4 3) ( 4 ) ( 4 ) ( 4 ) ( 4 ) ( ) ( ) Geeralized Combiatios & Permutatios I Geeralized Combiatios & Permutatios II Sometimes we are cocered with permutatios ad combiatios i which repetitios are allowed. The umber of rpermutatios of a set of objects with repetitio allowed is r. Easily obtaied by the product rule. There are ( ) + r r rcombiatios from a set with elemets whe repetitio of elemets is allowed. Geeralized Combiatios & Permutatios III Geeralized Combiatios & Permutatios IV There are 3 varieties of douts from which we wish to buy a doze. How may possible ways to place your order are there? Here 3 ad we wish to choose r 2. Order does ot matter ad repetitios are possible, so we apply the previous theorem to get that there are ( ) possible orders. 2 The umber of differet permutatios of objects where there are idistiguishable objects of type, 2 of type 2,..., ad k of type k is!! 2! k! A equivalet way of iterpretig this theorem is the umber of ways to distribute distiguishable objects ito k distiguishable boxes so that i objects are placed ito box i for i, 2,..., k.
13 Geeralized Combiatios & Permutatios V Geeratig Permutatios & Combiatios I How may permutatios of the word Mississippi are there? Mississippi cotais 4 distict letters, M, i, s ad p; with, 4, 4, 2 occurreces respectively. Therefore there are permutatios.!!4!4!2! I geeral, it is iefficiet to solve a problem by cosiderig all permutatios or combiatios sice there are a expoetial umber of such arragemets. Nevertheless, for may problems, o better approach is kow. Whe exact solutios are eeded, backtrackig algorithms are used. Geeratig permutatios or combiatios are sometimes the basis of these algorithms. Geeratig Permutatios & Combiatios II Geeratig Permutatios & Combiatios III (Travelig Sales Perso Problem) Cosider a salesma that must visit differet cities. He wishes to visit them i a order such that his overall distace traveled is miimized. This problem is oe of hudreds of NPcomplete problems for which o kow efficiet algorithms exist. Ideed, it is believed that o efficiet algorithms exist. (Actually, Euclidea TSP is ot eve kow to be i NP!) The oly kow way of solvig this problem exactly is to try all! possible routes. We give several algorithms for geeratig these combiatorial objects. Geeratig Combiatios I Recall that combiatios are simply all possible subsets of size r. For our purposes, we will cosider geeratig subsets of The algorithm works as follows. Start with {,..., r} {, 2, 3,..., } Assume that we have a a 2 a r, we wat the ext combiatio. Locate the last elemet a i such that a i r + i. Replace a i with a i +. Replace a j with a i + j i for j i +, i + 2,..., r. Geeratig Combiatios II The followig is pseudocode for this procedure. Algorithm (Next rcombiatio) Iput : A set of elemets ad a rcombiatio, a a r. Output : The ext rcombiatio. i r 2 while a i r + i do 3 i i 4 ed 5 a i a i + 6 for j (i + )... r do 7 a j a i + j i 8 ed
14 Geeratig Combiatios III Geeratig Permutatios Fid the ext 3combiatio of the set {, 2, 3, 4, 5} after {, 4, 5} Here, a, a 2 4, a 3 5, 5, r 3. The last i such that a i i is. Thus, we set a a + 2 a 2 a a 3 a The text gives a algorithm to geerate permutatios i lexicographic order. Essetially the algorithm works as follows. Give a permutatio, Choose the leftmost pair a j, a j+ where a j < a j+. Choose the least item to the right of a j greater tha a j. Swap this item ad a j. Arrage the remaiig (to the right) items i order. So the ext rcombiatio is {2, 3, 4}. Geeratig Permutatios Lexicographic Order Algorithm (Next Permutatio (Lexicographic Order)) Iput : A set of elemets ad a rpermutatio, a a r. Output j 2 while a j > a j+ do 3 j j 4 5 ed : The ext rpermutatio. //j is the largest subscript with a j < a j+ k 6 while a j > a k do 7 k k 8 9 ed //a k is the smallest iteger greater tha a j to the right of a j swap(a j, a k ) r s j + 2 while r > s do 3 swap(a r, a s) ed r r s s + Geeratig Permutatios II Geeratig Permutatios I Ofte there is o reaso to geerate permutatios i lexicographic order. Moreover, eve though geeratig permutatios is iefficiet i itself, lexicographic order iduces eve more work. A alterate method is to fix a elemet, the recursively permute the remaiig elemets. JohsoTrotter algorithm has the followig attractive properties. Not i your textbook, ot o the exam, just for your referece/culture. It is bottomup (orecursive). It iduces a miimalchage betwee each permutatio. Geeratig Permutatios III The algorithm is kow as the JohsoTrotter algorithm. We associate a directio to each elemet, for example: A compoet is mobile if its directio poits to a adjacet compoet that is smaller tha itself. Here 3 ad 4 are mobile ad ad 2 are ot. Algorithm (JohsoTrotter) Iput : A iteger. Output : All possible permutatios of, 2,.... π while There exists a mobile iteger k π do 3 k largest mobile iteger 4 swap k ad the adjacet iteger k poits to 5 reverse directio of all itegers > k 6 Output π 7 ed
15 More s I I As always, the best way to lear ew cocepts is through practice ad examples. How may bit strigs of legth 4 are there such that ever appears as a substrig? We ca represet the set of strig graphically usig a diagram tree. See textbook page 39. I II : Coutig Fuctios I I Let S, T be sets such that S, T m. How may fuctios are there mappig f : S T? How may of these fuctios are oetooe (ijective)? A fuctio simply maps each s i to some t j, thus for each we ca choose to sed it to ay of the elemets i T. Therefore, the umber of such bit strig is 8. : Coutig Fuctios I II : Coutig Fuctios I III Each of these is a idepedet evet, so we apply the multiplicatio rule; m } m {{ m } m times If we wish f to be oetooe (ijective), we must have that m, otherwise we ca easily aswer. Now, each s i must be mapped to a uique elemet i T. For s, we have m choices. However, oce we have made a mappig (say t j ), we caot map subsequet elemets to t j agai. I particular, for the secod elemet, s 2, we ow have m choices. Proceedig i this maer, s 3 will have m 2 choices, etc. Thus we have m (m ) (m 2) (m ( 2)) (m ( )) A alterative way of thikig about this problem is by usig the choose operator: we eed to choose elemets from a set of size m for our mappig; ( ) m m! (m )!!
16 : Coutig Fuctios I IV : Coutig Fuctios II Oce we have chose this set, we ow cosider all permutatios of the mappig, i.e.! differet mappigs for this set. Thus, the umber of such mappigs is m! (m )!!! m! (m )! Recall this questio from the midterm exam: Let S {, 2, 3}, T {a, b}. How may oto fuctios are there mappig S T? How may oetooe (ijective) fuctios are there mappig T S? See, page 46. : Coutig Primes I : Coutig Primes II Give a estimate for how may 7 bit primes there are. Recall that the umber of primes ot more tha is about l See slides o Number Theory, page 33. Usig this fact, the umber of primes ot exceedig 2 7 is 2 7 l 2 7 However, we have over couted we ve couted 69bit, 68bit, etc primes as well. The umber of primes ot exceedig 2 69 is about Thus the differece is 2 7 l l l 269 : More sets I : More sets II How may itegers i the rage k are divisible by 2 or 3? Let A {x x, 2 x} B {y x, 3 y} Clearly, A 5, B 3 33, so is it true that A B ? No; we ve over couted agai ay iteger divisible by 6 will be i both sets. How much did we over cout? The umber of itegers betwee ad divisible by 6 is 6, so the aswer to the origial questio is 6 A B (5 + 33) 6 67
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