Counting III. Today we ll briefly review some facts you dervied in recitation on Friday and then turn to some applications of counting.

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1 6.04/18.06J Mathematics for Computer Sciece April 5, 005 Srii Devadas ad Eric Lehma Lecture Notes Coutig III Today we ll briefly review some facts you dervied i recitatio o Friday ad the tur to some applicatios of coutig. 1 The Booeeper Rule I recitatio you leared that the umber of ways to rearrage the letters i the word BOOKKEEPER is: total letters {}}{ 10! }{{} 1! }{{}! }{{}! }{{} 3! }{{} 1! }{{} 1! B s O s K s E s P s R s This is a special case of a exceptioally useful coutig priciple. Rule 1 (Booeeper Rule). The umber of sequeces with 1 copies of l 1, copies of l,..., ad copies of l is ( )! 1!!...! provided l 1,..., l are distict. Let s review some applicatios ad implicatios of the Booeeper Rule Mile Wals I m plaig a 0 miles wal, which should iclude 5 orthward miles, 5 eastward miles, 5 southward miles, ad 5 westward miles. How may differet wals are possible? There is a bijectio betwee such wals ad sequeces with 5 N s, 5 E s, 5 S s, ad 5 W s. By the Booeeper Rule, the umber of such sequeces is: 0! 5! 4 Page 1 of 15

2 Coutig III 1. Bit Sequeces How may bit sequeces cotai exactly oes? Each such sequece also cotais zeroes, so there are by the Booeeper Rule.!! ( )! 1.3 elemet Subsets of a elemet Set How may elemets subsets of a elemet set are there? This questio arises all the time i various guises: I how may ways ca I select 5 boos from my collectio of 100 to brig o vacatio? How may differet 13 card Bridge hads ca be dealt from a 5 card dec? I how may ways ca I select 5 toppigs for my pizza if there are 14 available? There is a atural bijectio betwee elemet subsets of a elemet set ad bit sequeces with exactly oes. For example, here is a 3 elemet subset of {x 1, x,..., x 8 } ad the associated 8 bit sequece with exactly 3 oes: { x 1, x 4, x 5 } ( 1, 0, 0, 1, 1, 0, 0, 0 ) Therefore, the aswer to this problem is the same as the aswer to the earlier questio about bit sequeces. Rule (Subset Rule). The umber of elemet subsets of a elemet set is:! =! ( )! The factorial expressio i the Subset Rule comes up so ofte that there is a shorthad,. This is read choose sice it deotes the umber of ways to choose items from amog. We ca immediately oc off all three questios above usig the Sum Rule: I ca select 5 boos from 100 i 100 ways. 5 There are 5 differet Bridge hads. 13 Page of 15

3 Coutig III 3 There are 14 differet 5 toppig pizzas, if 14 toppigs are available. 5 The elemet subsets of a elemet set are sometimes called combiatios. There are a great may similar soudig terms: permutatios, r permutatios, permutatios with repetitio, combiatios with repetitio, permutatios with idistiguishable objects, ad so o. For example, the Booeeper Rule is sometimes called the formula for permutatios with idistiguishable objects. Broadly speaig, permutatios cocer sequeces ad combiatios cocer subsets. However, the coutig rules we ve taught you are sufficiet to solve all these sorts of problems without owig this jargo, so we ll sip it. 1.4 Word of Cautio Someday you might refer to the Booeeper Rule i frot of a roomful of colleagues ad discover that they re all starig bac at you blaly. This is ot because they re dumb, but rather because we just made up the ame Booeeper Rule. However, the rule is excellet ad the ame is apt, so we suggest that you play through: You ow? The Booeeper Rule? Do t you guys ow aythig??? Biomial Theorem Coutig gives isight ito oe of the basic theorems of algebra. A biomial is a sum of two terms, such as a + b. Now let s cosider a postive, itegral power of a biomial: (a + b) Suppose we multiply out this expressio completely for, say, = 4: (a + b) 4 = aaaa + aaab + aaba + aabb + abaa + abab + abba + abbb + baaa + baab + baba + babb + bbaa + bbab + bbba + bbbb Notice that there is oe term for every sequece of a s ad b s. Therefore, the umber of terms with copies of b ad copies of a is:! =! ( )! by the Booeeper Rule. Now let s group equivalet terms, such as aaab = aaba = abaa = baaa. The the coefficiet of a b is. So for = 4, this meas: (a + b) 4 = a b 0 + a 3 b 1 + a b + a 1 b 3 + a 0 b I geeral, this reasoig gives the Biomial Theorem: Page 3 of 15

4 4 Coutig III Theorem 1 (Biomial Theorem). For all N ad a, b R: (a + b) = a b The expressio is ofte called a biomial coefficiet i hoor of its appearace here. This reasoig about biomials exteds icely to multiomials, which are sums of two or more terms. For example, suppose we wated the coefficiet of i the expasio of (b + o + + e + p + r) 10. Each term i this expasio is a product of 10 3 variables where each variable is oe of b, o,, e, p, or r. Now, the coefficiet of bo e pr is the umber of those terms with exactly 1 b, o s, s, 3 e s, 1 p, ad 1 r. Ad the umber of such terms is precisely the umber of rearragmets of the word BOOKKEEPER: 10! 10 = 1!!! 3! 1! 1!, 1,, 3, 1, 1 Theorem (Multiomial Theorem). For all N ad z 1,... z m R: (z 1 + z z m ) = z 1 1 z... z m m 1,,..., m =0 3 bo e pr The expressio o the left is a example of a multiomial coefficiet ad the otatio o the right is a shorthad. This reasoig exteds to a geeral theorem: 1,..., m N m = You ll be far better off if your remember the reasoig behid the Multiomial Theorem rather tha this mostrous equatio. 3 Poer Hads There are 5 cards i a dec. Each card has a suit ad a value. There are four suits: Ad there are 13 values: spades hearts clubs diamods jac quee ig ace, 3, 4, 5, 6, 7, 8, 9, J, Q, K, A Thus, for example, 8 is the 8 of hearts ad A is the ace of spades. Values farther to the right i this list are cosidered higher ad values to the left are lower. Page 4 of 15

5 Coutig III 5 Five Card Draw is a card game i which each player is iitially dealt a had, a subset of 5 cards. (The the game gets complicated, but let s ot worry about that.) The umber of differet hads i Five Card Draw is the umber of 5 elemet subsets of a 5 elemet set, which is 5 choose 5: 5 total # of hads = =, 598, Let s get some coutig practice by worig out the umber of hads with various special properties. 3.1 Hads with a Four of a Kid A Four of a Kid is a set of four cards with the same value. How may differet hads cotai a Four of a Kid? Here a couple examples: { 8, 8, Q, 8, 8 } { A,,,, } As usual, the first step is to map this questio to a sequece coutig problem. A had with a Four of a Kid is completely described by a sequece specifyig: 1. The value of the four cards.. The value of the extra card. 3. The suit of the extra card. Thus, there is a bijectio betwee hads with a Four of a Kid ad sequeces cosistig of two distict values followed by a suit. For example, the three hads above are associated with the followig sequeces: (8, Q, ) { 8, 8, 8, 8, Q } (, A, ) {,,,, A } Now we eed oly cout the sequeces. There are 13 ways to choose the first value, 1 ways to choose the secod value, ad 4 ways to choose the suit. Thus, by the Geeralized Product Rule, there are = 64 hads with a Four of a Kid. This meas that oly 1 had i about 4165 has a Four of a Kid; ot surprisigly, this is cosidered a very good poer had! Page 5 of 15

6 6 Coutig III 3. Hads with a Full House A Full House is a had with three cards of oe value ad two cards of aother value. Here are some examples: {,,, J, J } { 5, 5, 5, 7, 7 } Agai, we shift to a problem about sequeces. There is a bijectio betwee Full Houses ad sequeces specifyig: 1. The value of the triple, which ca be chose i 13 ways.. The suits of the triple, which ca be selected i 4 ways The value of the pair, which ca be chose i 1 ways. 4. The suits of the pair, which ca be selected i 4 ways. The example hads correspod to sequeces as show below: (, {,, }, J, {, }) {,,, J, J } (5, {,, }, 7, {, }) { 5, 5, 5, 7, 7 } By the Geeralized Product Rule, the umber of Full Houses is: We re o a roll but we re about to hit a speedbump. 3.3 Hads with Two Pairs How may hads have Two Pairs; that is, two cards of oe value, two cards of aother value, ad oe card of a third value? Here are examples: { 3, 3, Q, Q, A } { 9, 9, 5, 5, K } Each had with Two Pairs is described by a sequece cosistig of: 1. The value of the first pair, which ca be chose i 13 ways.. The suits of the first pair, which ca be selected 4 ways. 3. The value of the secod pair, which ca be chose i 1 ways. 4. The suits of the secod pair, which ca be selected i 4 ways. Page 6 of 15

7 Coutig III 7 5. The value of the extra card, which ca be chose i 11 ways. 6. The suit of the extra card, which ca be selected i 4 = 4 ways. 1 Thus, it might appear that the umber of hads with Two Pairs is: Wrog aswer! The problem is that there is ot a bijectio from such sequeces to hads with Two Pairs. This is actually a to 1 mappig. For example, here are the pairs of sequeces that map to the hads give above: (3, {, }, Q, {, }, A, ) (Q, {, }, 3, {, }, A, ) (9, {, }, 5, {, }, K, ) (5, {, }, 9, {, }, K, ) { 3, 3, Q, Q, A } { 9, 9, 5, 5, K } The problem is that othig distiguishes the first pair from the secod. A pair of 5 s ad a pair of 9 s is the same as a pair of 9 s ad a pair of 5 s. We avoided this difficulty i coutig Full Houses because, for example, a pair of 6 s ad a triple of igs is differet from a pair of igs ad a triple of 6 s. We ra ito precisely this difficulty last time, whe we wet from coutig arragemets of differet pieces o a chessboard to coutig arragemets of two idetical roos. The solutio the was to apply the Divisio Rule, ad we ca do the same here. I this case, the Divisio rule says there are twice as may sequeces ad hads, so the umber of hads with Two Pairs is actually: Aother Approach The precedig example was disturbig! Oe could easily overloo the fact that the mappig was to 1 o a exam, fail the course, ad tur to a life of crime. You ca mae the world a safer place i two ways: 1. Wheever you use a mappig f : A B to traslate oe coutig problem to aother, chec the umber elemets i A that are mapped to each elemet i B. This determies the size of A relative to B. You ca the apply the Divisio Rule with the appropriate correctio factor. Page 7 of 15

8 8 Coutig III. As a extra chec, try solvig the same problem i a differet way. Multiple approaches are ofte available ad all had better give the same aswer! (Sometimes differet approaches give aswers that loo differet, but tur out to be the same after some algebra.) We already used the first method; let s try the secod. There is a bijectio betwee hads with two pairs ad sequeces that specify: The values of the two pairs, which ca be chose i ways.. The suits of the lower value pair, which ca be selected i 4 ways. 3. The suits of the higher value pair, which ca be selected i 4 ways. 4. The value of the extra card, which ca be chose i 11 ways. 5. The suit of the extra card, which ca be selected i 4 = 4 ways. 1 For example, the followig sequeces ad hads correspod: ({3, Q}, {, }, {, }, A, ) ({9, 5}, {, }, {, }, K, ) { { 3, 9, 3, 9, Q, 5, Q, 5, A K } } Thus, the umber of hads with two pairs is: This is the same aswer we got before, though i a slightly differet form. 3.4 Hads with Every Suit How may hads cotai at least oe card from every suit? Here is a example of such a had: { 7, K, 3, A, } Each such had is described by a sequece that specifies: 1. The values of the diamod, the club, the heart, ad the spade, which ca be selected i = 13 4 ways.. The suit of the extra card, which ca be selected i 4 ways. 3. The value of the extra card, which ca be selected i 1 ways. Page 8 of 15

9 Coutig III 9 For example, the had above is described by the sequece: (7, K, A,,, 3) { 7, K, A,, 3 } Are there other sequeces that correspod to the same had? There is oe more! We could equally well regard either the 3 or the 7 as the extra card, so this is actually a to 1 mappig. Here are the two sequeces correspodig to the example had: (7, K, A,,, 3) (3, K, A,,, 7) { 7, K, A,, 3 } Therefore, the umber of hads with every suit is: 4 Magic Tric There is a Magicia ad a Assistat. The Assistat goes ito the audiece with a dec of 5 cards while the Magicia loos away. Five audiece members each select oe card from the dec. The Assistat the gathers up the five cards ad reveals four of them to the Magicia, oe at a time. The Magicia cocetrates for a short time ad the correctly ames the secret, fifth card! 4.1 The Secret The Assistat somehow commuicated the secret card to the Magicia just by amig the other four cards. I particular, the Assistat has two ways to commuicate: 1. He ca aouce the four cards i ay order. The umber of orderigs of four cards is 4! = 4, so this aloe is isufficiet to idetify which of the remaiig 48 cards is the secret oe.. The Assistat ca also choose which four of the five cards to reveal i biom54 = 5 differet ways. Of course, the Magicia ca ot determie which of these five possibilities the Assistat selected, sice he does ot ow the secret card. Nevertheless, these two forms of commuicatio allow the Assistat to covertly reveal the secret card to the Magicia. Our coutig tools give a lot of isight ito the magic tric. Put all the sets of 5 cards i a collectio X o the left. Ad put all the sequeces of 4 distict cards i a collectio Y o the right. Page 9 of 15

10 10 Coutig III Y = all X = all sets of sequeces of 4 distict cards 5 cards (8, K, Q, ) {8, K, Q,, 6 } P ((((((((((((( PPPPP (K, 8, Q, ) hhhhhhhhhhhhh (K, 8, 6, Q ) {8, K, Q, 9, 6 } For example, {8, K, Q,, 6 } is a elemet of X o the left. If the audiece selects this set of 5 cards, the there are may differet 4 card sequeces o the right i set Y that the Assistat could choose to reveal, icludig (8, K, Q, ), (K, 8, Q, ), ad (K, 8, 6, Q ). Let s thi about this problem i terms of graphs. Regard the elemets of X ad Y as the vertices of a bipartite graph. Put a edge betwee a set of 5 cards ad a sequece of 4 if every card i the sequece is also i the set. I other words, if the audiece selects a set of cards, the the Assistat must reveal a sequece of cards that is adjacet i the bipartite graph. Some edges are show i the diagram above. What we eed to perform the tric is a matchig for the X vertices; that is, we eed a subset of edges that joi every vertex o the left to exactly oe, distict vertex o the right. If such a matchig exists, the the Assistat ad Magicia ca agree oe i advace. The, whe the audiece selects a set of 5 cards, the Assistat reveals the correspodig sequece of 4 cards. The Magicia traslates bac to the correspodig set of 5 cards ad ames the oe ot already revealed. For example, suppose the Assistat ad Magicia agree o a matchig cotaiig the two bold edges i the diagram above. If the audiece selects the set {8, K, Q, 9, 6 }, the the Assistat reveals the correspodig sequece (K, 8, 6, Q ). The Magicia ames the oe card i the correspodig set ot already revealed, the 9. Notice that the sets must be matched with distict sequeces; if the Assistat revealed the same sequece whe the audiece piced the set {8, K, Q,, 6 }, the the Magicia would be uable to determie whether the remaiig card was the 9 or! The oly remaiig questio is whether a matchig for the X vertices exists. This is precisely the subject of Hall s Theorem. Regard the X vertices as girls, the Y vertices as boys, ad each edge as a idicatio that a girl lies a boy. The a matchig for the girls exists if ad oly if the marriage coditio is satisfied: Every subset of girls lies at least as large a set of boys. Let s prove that the marriage coditio holds for the magic tric graph. We ll eed a couple prelimiary facts: Page 10 of 15

11 Coutig III 11 Each vertex o the left has degree 5 4! = 10, sice there are five ways to select the card ept secret ad there are 4! permutatios of the remaiig 4 cards. I terms of the marriage metaphor, every girl lie 10 boys. Each vertex o the right has degree 48, sice there are 48 possibilities for the fifth card. Thus, every boy is lied by 48 girls. Now let S be a arbitrary set of vertices o the left, which we re regardig as girls. There are 10 S edges icidet to vertices i this set. Sice each boy is lied by at most 48 girls, this set of girls lies at least 10 S /48 S differet boys. Thus, the marriage coditio is satisfied, a matchig exists by Hall s Theorem, ad the tric ca be doe without magic! 4. The Real Secret You might ot fid the precedig aswer very satisfyig. After all, as a ( practical ) matter, the Assistat ad the Magicia ca ot memorize a matchig cotaiig 5 =, 598, edges! The remaiig challege is to choose a matchig that ca be readily computed o the fly. We ll describe oe approach. As a ruig example, suppose that the audiece selects: Q J The Assistat pics out two cards of the same suit. I the example, the assistat might choose the 3 ad 10. The Assistat locates the values of these two cards o the cycle show below: A K Q 3 J For ay two distict values o this cycle, oe is always betwee 1 ad 6 hops clocwise from the other. For example, the 3 is 6 hops clocwise from the 10. The more couterclocwise of these two cards is revealed first, ad the other becomes the secret card. Thus, i our example, the 10 would be revealed, ad the 3 would be the secret card. Therefore: The suit of the secret card is the same as the suit of the first card revealed. Page 11 of 15

12 1 Coutig III The value of the secret card is betwee 1 ad 6 hops clocwise from the value of the first card revealed. All that remais is to commuicate a umber betwee 1 ad 6. The Magicia ad Assistat agree beforehad o a orderig of all the cards i the dec from smallest to largest such as: A... K A... Q A... Q A... Q The order i which the last three cards are revealed commuicates the umber accordig to the followig scheme: ( small, medium, large ) = 1 ( small, large, medium ) = ( medium, small, large ) = 3 ( medium, large, small ) = 4 ( large, small, medium ) = 5 ( large, medium, small ) = 6 I the example, the Assistat wats to sed 6 ad so reveals the remaiig three cards i large, medium, small order. Here is the complete sequece that the Magicia sees: 10 Q J 9 The Magicia starts with the first card, 10, ad hops 6 values clocwise to reach 3, which is the secret card! 4.3 Same Tric with Four Cards? Suppose that the audiece selects oly four cards ad the Assistat reviews a sequece of three to the Magicia. Ca the Magicia determie the fourth card? Let X be all the sets of four cards that the audiece might select, ad let Y be all the sequeces of three cards that the Assistat might reveal. Now, oe o had, we have 5 X = = 70,75 4 by the Subset Rule. O the other had, we have Y = = 13,600 by the Geeralized Product Rule. Thus, by the Pigeohole Priciple, the Assistat must reveal the same sequece of three cards for some two differet sets of four. This is bad ews for the Magicia: if he hears that sequece of three, the there are at least two possibilities for the fourth card which he caot distiguish! Page 1 of 15

13 Coutig III 13 5 Combiatorial Proof Suppose you have differet T shirts oly wat to eep. You could equally well select the shirts you wat to eep or select the complemetary set of shirts you wat to throw out. Thus, the umber of ways to select shirts from amog must be equal to the umber of ways to select shirts from amog. Therefore: = This is easy to prove algebraically, sice both sides are equal to:!! ( )! But we did t really have to resort to algebra; we just used coutig priciples. Hmm. 5.1 Boxig Isha, famed 6.04 TA, has decided to try out for the US Olympic boxig team. After all, he s watched all of the Rocy movies ad spet hours i frot of a mirror seerig, Yo, you waa piece a me?! Isha figures that people (icludig himself) are competig for spots o the team ad oly will be selected. As part of maeuverig for a spot o the team, he eed to wor out how may differet teams are possible. There are two cases to cosider: Isha is selected for the team, ad his 1 teammates are selected from amog the other 1 competitors. The umber of differet teams that be formed i this way is: 1 1 Isha is ot selected for the team, ad all team members are selected from amog the other 1 competitors. The umber of teams that ca be formed this way is: 1 All teams of the first type cotai Isha, ad o team of the secod type does; therefore, the two sets of teams are disjoit. Thus, by the Sum Rule, the total umber of possible Olympic boxig teams is: Page 13 of 15

14 14 Coutig III Christos, equally famed 6.04 TA, this Isha is t so tough ad so he might as well try out also. He reasos that people (icludig himself) are tryig out for spots. Thus, the umber of ways to select the team is simply: Christos ad Isha each correctly couted the umber of possible boxig teams; thus, their aswers must be equal. So we ow: = 1 This is called Pascal s Idetity. Ad we proved it without ay algebra! Istead, we relied purely o coutig techiques. 5. Combiatorial Proof A combiatorial proof is a argumet that establishes a algebraic fact by relyig o coutig priciples. May such proofs follow the same basic outlie: 1. Defie a set S.. Show that S = by coutig oe way. 3. Show that S = m by coutig aother way. 4. Coclude that = m. I the precedig ( example, ) ( S ) was the set of all possible Olympic boxig teams. ( Isha ) 1 computed S = + 1 by coutig oe way, ad Christos computed S = by 1 coutig aother. Equatig these two expressios gave Pascal s Idetity. More typically, the set S is defied i terms of simple sequeces or sets rather tha a elaborate story. Here is less colorful example of a combiatorial argumet. Theorem 3. 3 = r r r=0 Proof. We give a combiatorial proof. Let S be all card hads that ca be dealt from a dec cotaiig red cards (umbered 1,..., ) ad blac cards (umbered 1,..., ). First, ote that every 3 elemet set has S = 3 Page 14 of 15

15 Coutig III 15 elemet subsets. From aother perspective, the umber of hads with exactly r red cards is r r sice there are ways to choose the r red cards ad r ways to choose the r blac r cards. Sice the umber of red cards ca be aywhere from 0 to, the total umber of card hads is: S = r r Equatig these two expressios for S proves the theorem. r=0 Combiatorial proofs are almost magical. Theorem 3 loos pretty scary, but we proved it without ay algebraic maipulatios at all. The ey to costructig a combiatorial proof is choosig the set S properly, which ca be tricy. Geerally, the simpler side of ( the ) equatio should provide some guidace. For example, the right side of Theorem 3 is 3, which suggests choosig S to be all elemet subsets of some 3 elemet set. Page 15 of 15