MA3210 Lecture 8: The Pigeonhole Principle I 1

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1 MA30 Lecture 8: The Pigeohole Priciple I Erik E. Westlud The Pigeohole Priciple (simple: If +, or more objects are placed ito boxes, the at least oe box will cotai two or more objects. Proof. If ot, the maximum umber of objects is a cotradictio. {}}{ < +, Note: we do t kow which box cotais more tha elemets, oly that oe exists. (No-costructive existece proof. Also called Dirichlet drawer priciple or Shoebox priciple. Ex. Amog ay group of 367 people, there are at least two with the same birthday. Ex. At ay give time, there are at least two people i Califoria who were bor o the same secod of the year. There are (365(4(60(60 3,536,000 secods ad at least 36,000,000 people i Califoria. Ex. At ay give time, there are at least two people i Lodo that have the exact same umber of hairs o their heads. Roughly 5,000 hairs o average head, so i o case more tha,000,000 hairs. Lodo has more tha,000,000 people. Ex. You eed oly select + people out a set of married couples to guaratee a married couple has bee selected. There exists o ijectio ϕ : A B o fiite sets such that the codomai is smaller tha the domai. If A B ad ϕ is ijective, the ϕ is surjective, i.e., If objects are put ito boxes ad o box gets more tha oe object, the each box has a object i it. If A B ad ϕ is surjective, the ϕ is ijective, i.e., If objects are put ito boxes ad o box is empty, the each box has a object i it. Updated: May 6, 009.

2 Ex. Does there exist a perfect cover of a chessboard prued by cuttig out two diagoally opposite corers? Solutio. There are a total of 6 squares, w.l.o.g., 3 black ad 30 white. Let A be the black ad B be the white squares. Each domio covers both a black ad white square. The domioes iduce a - correspodece betwee white ad black squares. But Pigeohole forbids a ijectio ϕ : A B. Ex. Durig a 30-day moth, a baseball team plays at least game a day, but o more tha 45 games. Prove that there exists a successio of cosecutive days durig which the team will have played exactly 4 games. Proof. Let a j # games played o or prior to j th day. a, a,..., a 30 is a icreasig sequece of distict positive itegers ad a j 45 for j 30. Also, a + 4, a + 4,..., a is a icreasig sequece of distict positive itegers, where 5 a j The elemets of the 60-set {a,..., a 30, a + 4,..., a } are all less tha or equal to 59. By Pigeohole Priciple, i, j with i, j 30 such that a i a j + 4, i.e., exact 4 games where played from day j + to day i. Ex. Prove that amog ay + positive itegers ot exceedig, there must be a iteger that divides oe of the others. Proof. Let {a, a,..., a + } Z, where a i for i +. Factor out as may s as possible, i.e., a j k j q j, where k j Z + ad q j is odd. Clearly, {q,..., q + } are odd positive itegers ad q j <. As there are oly odd positive itegers less tha, by the Pigeohole Priciple, q i q j q for some i, j, so that a i k i q ad a j k j q. If k i < k j a i a j ad if k j < k i, the a j a i. (Note: this is optimal i the sese that havig chose oly 0 itegers ot exceedig 0 0, e.g., {,, 3, 4, 5, 6, 7, 8, 9, 0}, the o iteger divides oe of the others. With itegers, we are guarateed oe that divides the other.

3 Ex. The Chiese Remaider Theorem (simple: if m ad are relatively prime positive itegers, the x Z + such that x a (mod m ad x b (mod. (I fact, x is uique modulo m. E.g., If m 4, 9, the gcd(4, 9. Pick a 3 ad b, the a x that satisfies x 3 (mod 4 ad x (mod 9 is x. Proof. Cosider S {a, m + a, m + a,..., ( m + a}. If for some 0 i < j <, we have im + a jm + a r (mod im + a q i + r ad jm + a q j + r, so that (j im (q i q j (j im (j i, as gcd(, m. But 0 < j i, so (j i, a cotradictio, hece all elemets i S must have differet remaiders whe divided by. Hece, by Pigeohole Priciple, each of 0,,..., occurs as a remaider, ad so b is a remaider. Let x q k m + a be the umber that has b as its remaider. Hece, x p + b, so that x a (mod m ad x b (mod. Ex. Let {(x i, y i, z i : i 9, ad x i, y i, z i Z} be a set of ie poits i xyz-space. Prove that the midpoit of at least oe pair of poits also has iteger coordiates. Proof. Let (x i, y i, z i ad (x j, y j, z j be two poits where i < j 9. The midpoit has coordiates (xi + x j, y i + y j, z i + z j. These will be iteger coordiates if ad oly if x i, x j have same parity, y i, y j have same parity, ad z i, z j have same parity. There are 3 possible triples for parity, e.g. (Eve, Odd, Odd or (Eve, Odd, Eve, etc. By Pigeohole, at least two poits must have the same parity triple. Ex. I a room with 0 people, whose ages (i whole umbers are betwee ad 60 iclusive, prove that we ca always fid two disjoit groups of people whose ages sum to the same umber. Proof. There are 0 03 ways to select a o-empty subset of 0 people. The largest the age sum ca be is , thus by Pigeohole, at least two groups must have the same age. If they itersect, remove the commo people, to get disjoit sets with the same age sum. 3

4 MA30 Lecture 9: The Pigeohole Priciple II Erik E. Westlud Pigeohole Priciple (Strog. For some positive itegers, q, q,..., q, if q + q + + q + objects are placed ito boxes, the either the first box cotais at least q objects,..., or the th box cotais at least q objects. Proof. If ot, the we have o more tha (q i objects total, a cotradictio. i If q i for all i, we get the Pigeohole Priciple (Simple. If q i r for all i, we get the Geeralized Pigeohole Priciple: if (r + objects are placed ito boxes, the at least oe box will cotai at least r objects. Alteratively, if D objects are placed ito boxes, the at least oe box will cotai at least D/ objects. Proof. If ot, the the umber of objects is at most ( (( D D < + D (We ca get the Geeralized Pigeohole Priciple by settig D (r +. Ex. Amog 00 people, there are at least 00/ 9 that were bor o the same moth. Ex. What is the least umber of area-codes eeded to guaratee that the 5 millio phoes i a state have distict umbers (NXX-NXX-XXXX where N {,..., 9}? There are (8(0 6 8, 000, 000 umbers of the form NXX-XXXX. Therefore, at least 5, 000, 000/8, 000, of them are idetical, so we eed at least 4 area codes. Updated: May 6, 009.

5 Ex. Suppose I have 7 blue ad black socks i my drawer, it s completely dark, ad I eed to pack for a coferece. What is the miimum umber of socks I eed to take out the drawer to guaratee I have ( at least oe matchig pair? ( six socks of oe color? (3 six black socks? (4 three matched pairs (possibly icludig pairs of each color? ( I eed to pick at least three. ( I eed to pick D (r + objects so some box gets at least r 6, thus D (5 + socks. (3 Pigeohole does t apply, so we take worst-case sceario, i.e. all 7 blue socks are chose. Thus, we eed 3 socks draw. (4 Three socks gets oe matched pair, aother three gets aother matched pair. The two leftovers may ot be matched, so we eed a seveth sock. Ramsey Theory. I every group of 6 people, there are always at least three people who all kow each other, or all do t kow each other. Proof. Let P A, B, C, D, E, F be the people. Out of {B, C, D, E, F } either three or more kow A or three or more do t kow A. (Puttig D 5 objects ito boxes, at least oe box cotais 5/ 3 objects. If, say B, C, D kow A ad ay pair kow each other, the we re doe. Otherwise, B, C, D are three o-aquaitaces. If a, a,..., a N is a sequece of legth, the a i, a i,..., a ik is a subsequece, provided that i < i < < i k N, ad is icreasig if a i a i a ik or decreasig if a i a i a ik. Ex. Every sequece of + real umbers cotais either a icreasig subsequece or a decreasig subsequece of + umbers. E.g. 8,, 9,, 4, 6,, 0, 5, 7 is a sequece of real umbers., 4, 6, is a icreasig subsequece,, 9, 6, 5 is a decreasig subsequece. Proof. Let a, a,..., a + be a sequece ad let m k be the legth of the logest icreasig subsequece that starts with a k : a k a i a i a imk,

6 where k +. If m k +, for some k, we re doe. Otherwise, suppose that k +, we have m k. By Pigeohole Priciple, at least + of the umbers m, m..., m + are equal, call them m k m k m k+, where k < k < < k + +. If for some i, we have a ki < a ki+, the m ki > m ki+, a cotradictio. Hece, a ki a ki+, so that a k a k a k+ is a decreasig subsequece of legth +. 3

7 MA30 Lecture 6: The Multiomial Theorem ad Newto s Biomial Theorem Erik E. Westlud Goal: establish a geeralizatio of the Biomial Theorem to multiomials. Recall: The Biomial Theorem, for N, ( ( (x + y x i y i x i y i i i i0 But, what about expadig (x + y + z? (x + x + x k? i0 i0 ( x i y i i Revisit permutatios of multisets: if S { s, s } is a multiset of two differet types, the for +, the # of permutatios of S is ( (!!!!!(!. I.e., ( is the # of -subsets of a -set, ad the # of permutatios of S. If,,..., k are o-egative itegers, ad k, the the multiomial coefficiet is: (! k!! k!. ( ( ( ( E.g., for, k N, +. k k k k k k k Therefore, the Biomial Theorem may be rewritte as (x + y ( x y + Pascal s Formula for Multiomial Coefficiets:, k,,,..., k N, where + + k, we have ( k ( k i i i+ k i Updated: Jue 9, 009.

8 Proof. k ( i i i+ k i k i k i (!! i!( i! i+! k! i (!! i! k! (!!! k! k i (!!! k!!!! k! i Ex. Expad (x + x + x 3 out ad combie like terms. (x + x + x 3 (x + x + x 3 x + x + x 3 + x x + x x 3 + x x 3 The Multiomial Theorem. Let, k Z,, k. The for all x, x,..., x k, ( (x + x + + x k x x x k k, k where i N for all i k. + + k Iductive Proof. Too complicated. Double iductio o k ad, or iductio ad use of multi-idex otatio. For a fixed k, say k 3, we ca use repeated applicatio of The Biomial Theorem. Here s the idea: (x + x + x 3 ((x + x + x 3 ( (x + x x 3 + Applyig the Biomial Theorem agai to (x + x, (x + x + x 3 ( [ ( k k + k +k Pullig everythig ito the iterior sum, ( ( x k k k xk x 3 + k +k k +k + ( x k xk k k ] x 3 x k xk x 3

9 Combiatorial Proof. Write (x + + x k as the product of factors: (x + + x k (x + + x k (x + + x k (x + + x k }{{} terms Every moomial is formed by selectig oe of the x i s from each of the terms. Hece, we form k moomial terms, ad each term ca be arraged i the form x x x k k, where i N, ad + + k. A arbitrary moomial term is obtaied by pickig x i of the factors, x i of the remaiig factors,..., ad x k i k of the remaiig k k factors. Thus, the # of ways to do this is: ( ( ( k k!!! k!. Ex. Fid the expasio of (x + x + x 3 3 usig the Multiomial Theorem. (x + x + x ( 3 3 x x x 3 3 The total # of terms is The # of distict moomials x x x 3 3 is equivalet to the umber of o-egative itegral solutios to , for which there are ( ( ( x 3 + 3x x + 3x x 3 + 3x x + 6x x x 3 + 3x x 3 + x 3 + 3x x 3 + 3x x 3 + x

10 Ex. What is the coefficiet of x x x 5 x 0 6 i the expasio of (x + x + x 3 + x 4 + x 5 + x 6 + x 7 5? Now, x x x 5 x 0 6 x x x 0 3x 0 4x 5x 0 6 x 0 7, so the coefficiet is ( 5 5! 4, 46, 857, !!!0! There are a total of 7 5, 34, 068, 69, 663, 964, 900, 807 terms. Ex. What is the coefficiet of x x x 3x 5 i the expasio of Note: (x 3x + 5x 3 + x 4 x 5 7? (x 3x +5x 3 +x 4 x 5 7 ( Hece, the coefficiet is: ( 7 ( , If r R, ad k Z, the defie r(r (r k+ ( r k! if k if k 0 k 0 if k (x ( 3x (5x 3 3 x 4 4 ( x 5 5 The expressio r(r (r k + is a fallig factorial, deoted (r k. Pascal s formula holds for geeral ( r k as well, e.g., ( 5 (5 (5 (5 ( ( ! 3 Right aroud the time the Black Plague was ragig i Lodo, Isaac Newto ( geeralized the Biomial Theorem as follows: Geeralized Biomial Theorem. If α, x, y R, where x < y, the ( α (x + y α x k y α k k k0 4

11 Equivaletly, if z x/y, the (x + y α y α (z + α, ad so for ay z R, with z <, we have ( α ( + z α z k k Proof Sketch. Form the Taylor series cetered at z 0 for f(z ( + z α (as f(z has derivatives at all orders at z 0 usig k0 f (k (0 z k + αz + k! k0 α(α z +! α(α (α z 3 + 3! Usig the Ratio Test, it ca be show f has radius of covergece R, i.e. the Taylor series for f coverges to some fuctio o (,. Use Taylor s Theorem to show it actually coverges to f o this iterval. Hece, as z <, we are iside the radius of covergece, ad the result holds. Ex.. ( + 0. /, ad as z 0. <, we have ( 0.5. (0. k + 0. k (0. + (0.3 ( k0 If α, for some Z +, the for all z < ( + z ( z k k k0 ( ( ( k + z k k! k0 ( k (( + ( + k z k k! k0 ( + k ( k z k k ( + z k0 A few useful corollaries, for z < ( z ( z ( + k z k k k0 + Updated: Jue 9,

12 + z ( k z k z + z z 3 + z 4 k0 z z k + z + z + z 3 + z 4 + k0 The last two are ifiite geometric series, ad recall for a, r 0, ad r <, we have ar i a + ar + ar + ar 3 + a r. i0 6

13 MA30 Lecture 7: The Priciple of Iclusio-Exclusio Erik E. Westlud How may elemets are i A B? If A B, the A B A + B. What if A B? Clearly, A + B will cout the elemets i A B two times, hece, A B A + B A B Ex. A discrete mathematics class cosists of 5 math majors, 3 c.s. majors, ad 8 joit math/c.s. majors. If every studet i the class is either a math major, c.s. major, or joit major, how may studets are i the class? Let A : { math majors} ad B : { c.s. majors}. We seek A B. A B A + B A B Ex. How may positive itegers ot exceedig 000 are divisible by 7 or? Let A : {x Z + : x 000, 7 x} ad B : {x Z + : x 000, x}. We seek A B. Now, A , B , ad if 7 x ad x, the lcm(7, x 77 x, i.e. A B Therefore, A B Ex. Suppose there are,807 freshme at your school. Of these, 453 are takig a course i C.S., 567 are takig a course i Math, ad 99 are takig courses i both C.S. ad Math. How may freshme are ot takig a course i either Math or C.S.? Let A : { freshme i a Math course}, B : { freshme i a C.S. course}, so that A 567, B 453, ad A B 99. We seek A B A B, the umber of freshme either i A or B. By Subtractio Priciple, A B S A B S A B + A B If A ad A are subsets of objects of S havig properties P ad P, respectively, the A A is the subset of objects havig both properties, ad A A is the subset of objects havig either property. A A A A S A A + A A Updated: Jue 9, 009.

14 What about A A A 3? What about A A A k? The sum A + A + A 3 couts objects i exactly oe of the three sets oce, i exactly two of the three sets twice, i all three sets three times. The sum A + A + A 3 A A A A 3 A A 3 couts objects i exactly oe of the three sets oce, i exactly two of the three sets oce, i all three sets zero times. Fially, we add back i the elemets i all three sets. A A A 3 A + A + A 3 A A A A 3 A A 3 + A A A 3. Ex. 3 studets have take a course i Spaish (S, 879 have take a course i Frech (F, ad 4 have take a course i Russia (R. Further, 03 have take S ad F, 3 have take S ad R, ad 4 have take F ad R. If 09 studets have take at least oe course i S, F, or R, how may studets have take a course i all three? Note, S 3, F 879, ad R 4, ad S F 03, S R 3, ad F R 4. Also, S F R 09, so S F R S F R 7. Ex. Fid the umber of itegers betwee ad 0,000 that are either perfect squares or perfect cubes. Let A {x Z : x 0000, x y y Z}, the set of perfect squares. Let A {x Z : x 0000, x y 3 y Z}, the set of perfect cubes. As , the squares of the umbers {,,..., 00} are less tha 0,000, so A 00. Likewise, as 3 96, ad 3 0, 648, the cubes of the umbers {,,..., } are all less tha 0,000, so A. Now, if x A A, the x y z 3, for some y, z Z, ad so x t 6, ad so x is a perfect 6 th power. As , ad 5 6 5, 65, the 6 th powers of the umbers {,, 3, 4} are all less tha 0,000, i.e., A A 4. Hece, there are itegers of the required form.

15 The Priciple of Iclusio-Exclusio. Let A, A,..., A m be fiite sets. The, A A A m A i A i A j + A i A j A k i m i<j m i<j<k m + ( m+ A A A m Proof. Let a A A A m ad suppose that a is a member of exactly r of the sets A i, where r m. Now, i evaluatig the summatios o the RHS, we see a is couted ( r times i i m A i, a is couted ( r times i i<j m A i A j, a is couted ( r 3 times i i<j<k m A i A j A k, ad i geeral, for t r, a is couted ( r t times i the sum ivolvig the summatios of itersectios of t of the sets A i. Therefore, the elemet a is couted exactly (r ( ( ( r r r + + ( r+ 3 r times by the RHS. But, we have ( ( ( ( r r r r + + ( r 0, 0 r so ( ( ( ( r r r r + ( r. 0 r Hece, a is couted exactly oce by the RHS. Ex. Give a formula for the umber of elemets i the uio of four sets. A A A 3 A 4 A + A + A 3 + A 4 A A A A 3 A A 4 A A 3 A A 4 A 3 A 4 + A A A 3 + A A A 4 + A A 3 A 4 + A A 3 A 4 A A A 3 A 4 Furthermore, if there is a uiversal set S, the A A A m A A A m S A A A m. 3

16 Ex. How may of the permutatios of the 6 letters of the Eglish alphabet do ot cotai ay of the strigs FISH, RAT, or BIRD? Let S { permutatios of the alphabet } A { permutatios that cotai FISH } A { permutatios that cotai RAT } A 3 { permutatios that cotai BIRD } We seek A A A 3. Clearly, S 6!. Elemets i A are permutatios of the symbol FISH, ad the other letters, i.e., A 3!. Similarly, A 4!, ad A 3 3!. Elemets i A A are permutatios of the symbols, F ISH, RAT, ad the 9 other letters, i.e. A A!. Next, A A 3 A A 3 0, for FISH ad BIRD share a I, ad RAT ad BIRD share a R. For the same reaso, A A A 3 0. Hece, there are 6! (3!+4!+3!+(! , 69, 359, 78, 336, 797, 900, 800, 000 permutatios. 4

17 MA30 Lecture 8: Applicatios of the Priciple of Iclusio-Exclusio Erik E. Westlud Goal: preset ad discuss some iterestig applicatios of The Priciple of Iclusio-Exclusio. Applicatio : The Euler-Phi Fuctio The followig fuctio ϕ( is ubiquitous to umber theory ad algebra: ϕ : N N defied by ϕ( {m N : m ad gcd(, m } (The umber of positive itegers less tha that are relatively prime to. Let p a pa pa k k be the prime factorizatio of a positive iteger. For i k, let A i : {m N : m ad p i m}. Clearly, x A A A k gcd(, x. Thus, Hece, k A i i ϕ( A A A k. A i p i p i A i A j p i p j p i p j A i A j A l p i p j p l p i p j p l A A A k k i p i + k i i<j k p i + k ( pi i. i<j k Updated: Jue 4, p p p k p p p k p i p j i<j<l k + + ( k p i p j p l p p p k + + ( k p i p j p i p j p l p p p k i<j<l k

18 The last equality follows from the fact that for all x i R, we have ( x i i x i + x i x j x i x j x k + + ( x x x. i i<j i<j<k (You ca quickly prove this usig iductio! Applicatio : Coutig Prime Numbers Cout the umber of primes betwee ad a fixed positive iteger. Note: if m is composite, the m ab with < a b a m a m there exists a prime p m ad p m. Hece, every composite iteger m has a prime factor p m, so excludig these umbers will remove all composite umbers from {,,..., }. Ex. Let 0, the every composite iteger < m 0 has a prime factor p , i.e. p {p, p 3, p 3 5, p 4 7}. Let A i {x Z : x ad p i x}. Hece, A A A 3 A 4 0 ( ( ( However, we have t couted the primes,, 3, 5, ad 7 ad we have couted the oprime. So, the umber is Applicatio 3: The Number of Surjective Fuctios We kow that for X ad Y, where X m ad Y.. The umber of fuctios f : X Y is m.. The umber of ijective fuctios f : X Y is { 0 if m > ( ( m + if m 3. The umber of bijective fuctios f : X Y is { 0 if m! if m Questio: What is the umber of surjective fuctios f : X Y?

19 Let Y {y, y,..., y }, ad let A i be the set of all fuctios f : X Y such that y i is ot i the rage of f. Ay fuctio g / i A i is surjective, ad vice versa. Now A i ( m, A i A j ( m, A i A j A k ( 3 m. I geeral, for t hece A i m i A i A i A it ( t m, ( ( m + ( ( ( m + ( m The umber of surjective fuctios f : X Y is 0 if m < ( m + ( i ( i m if m i i Applicatio 4: Combiatios with Repetitio Let S { s,,..., k s k } be a multiset with k differet types. ( if i for all i k r The # of r-combiatios of S ( r + k if i r for all i k r What if < i < r? I.e., what are the # of r-combiatios for geeral repetitio umbers? Ex. Determie the umber of -combiatios of the multiset S {4 s, 3 s, 4 s 3, 5 s 4 }. Cosider the multiset S { s, s, s 3, s 4 }. For i 4, let A i be subsets of -combiatios of S where elemets i A have more tha 4 s s, elemets i A have more tha 3 s s, elemets i A 3 have more tha 4 s 3 s, ad elemets i A 4 have more tha 5 s 4 s. We seek A A A 3 A 4 T A A A 3 A 4, where T is the set of all ( combiatios of S. Note A is the umber of -combiatios that have at least 5 s s, so A ( Similarly, A ( , A3 ( , ad A4 (

20 Now, A A is the umber of -combiatios of S that have at least 5 s( s ad at least 4 s s, i.e. we ca complete this -combiatio i exactly ways. Similarly, A A 3 ( +4 0, A A 4 ( +4 4, ad A A 3 ( , ad A A 4 ( +4 0, ad A3 A 4 ( Furthermore, A A A 3 is the set of -combiatios that have at least 5 s s, at least 4 s s, ad at least 5 s 3 s, a cotradictio, thus A A A 3 0. Similarly, A A A 4 0, A A 3 A 4 0, ad A A 3 A 4 0. Fially, A A A 3 A 4 0. Hece, 455 ( ( ( Usig the above techique, we may ow determie the umber of itegral solutios to x + x + + x k r, where for all i k, we have 0 x i i for o-egative itegers i. Ex. Fid the umber of itegral solutios to x + x + x 3 7 that satisfy x 5, 0 x 4, ad 5 x 3. By lettig y x +, y x, ad y 3 x 3 5, this is the same as the umber of itegral solutios to y + y + y that satisfy 0 y 7, 0 y 4, ad 0 y 3 6. Let S be the set of all o-egative solutios to the above, so ( ( S 0. 4 Let A { o-egative solutios to y + y + y 3 4 where y 8}. Let A { o-egative solutios to y + y + y 3 4 where y 5}. Let A 3 { o-egative solutios to y + y + y 3 4 where y 3 7}. Note, A is the umber of o-egative solutios to z + z + z , 4

21 by lettig z y 8, z y, ad z 3 y 3. Hece, ( ( A 8. 6 Similarly, ( ( ( A 55 ad A Now, A A is the umber of o-egative solutios to u + u + u ( by lettig u y 8, u y 5, ad u 3 y 3. Hece, ( + 3 A A 3. Similarly, ( + 3 A A 3 0 ad A A 3 6 Fially, A A A 3 0, so by the Priciple of Iclusio-Exclusio, we have the umber of itegral solutios to our origial liear equatio is A A A 3 0 ( (

22 MA30 Lecture 9: Deragemets Erik E. Westlud Goal: discuss permutatios that have o fixed poits. Suppose a lazy professor gives his 44 studets a quiz, collects them, mixes them up, ad the radomly passes the quizzes back out to be graded by the studets. What is the probability that o studet grades their ow quiz? (D 44 /44! I recotres, a old Frech card game, the 5 cards of a deck are laid out, ad the cards of a secod deck are laid out with oe o top of each of the previous deck. The score is determied by the umber of matched cards oe from each deck. What is the probability of gettig a score of zero? (D 5 /5! How may aagrams of a word leave o letters fixed? Problems of this kid deal with permutatios that are completely out of order. A deragemet of {,,..., } is a permutatio i i i such that i j j for all j. More geerally, a deragemet of a set S is a bijectio ϕ : S S such that ϕ(x x for all x S. Ex. The permutatio 453 is a deragemet of {,, 3, 4, 5}, but 354 is ot, because 3 is i its atural positio. Ex. List all deragemets of ABCD. BADC, BCDA, BDAC, CADB, CDAB, CDBA, DABC, DCAB, DCBA. Let D be the umber of deragemets of {,,..., }. D is also called the th subfactorial, deoted!. Theorem. For, D! (! +! 3! + + (! Proof. Let X be the set of all! permutatios of {,,..., }. For all j, let A j : {σ i i i X : i j j}. The D A A A. Updated: Jue 7, 009.

23 It is easily see that, for all k, A t A t A tk ( k!, where {t, t,, t k } is ay k-subset of {,,..., }. By the Priciple of Iclusio-Exclusio, we have ( ( ( D! (! + (! + ( 0! D!!! +!! + (!! Ex. The umber of deragemets of ABCD is ( D 4 4!! +! 3! + 4! Ex. Determie the umber of permutatios of {,,..., 8} i which exactly four itegers are i their atural positios. This is the # of permutatios i which exactly four itegers are ot i their atural positios. There are ( 8 4 ways to pick the itegers, ad D4 ways to derage them, so ( 8 4 D4 630 permutatios. From calculus, we have e! +! 3! + 4! Recall the Lagrage Error Boud: if f ad all of its derivatives are cotiuous, ad P (x is the th Taylor polyomial of f about x a, the f(x P (x max f (+ (z x a +, where z is i betwee x ad a. ( +! Note: if P (x is the th MacLauri polyomial of f(x e x, the D!P (, ad e x P (x ( +! x + e D! ( +! Note: e ad D is the closest iteger to!e.

24 D e D /! /( +! So, for 6, e ad D /! agree to at least three decimal places. Therefore, if E D, the Prob(E D! e, for 6. Ex. A ew employee checks the hats of people at a restaurat, forgettig to put claim check umbers o the hats. What is the probability that upo leavig, o customer receives the correct hat? Usig the aforemetioed, formula, we see that the probability is D /! 0.36, ad this probability is essetially the same for 0 customers as it is for billio customers! Theorem. The deragemet umbers satisfy the liear recurrece relatio: D ( (D + D for 3. Let X be the set of deragemets of {,..., }, ad let σ σ σ σ X. Let d i # of deragemets where σ i. Clearly, σ i σ j for all i, j, ad let d be the commo value. Hece, D ( d. Partitio the d deragemets σ σ σ 3 σ accordig to whether σ or σ. I.e., let A {σ σ 3 σ : σ j j, σ } ad B {σ σ 3 σ : σ j j, σ } Thus, d A + B, ad clearly, A is the # of deragemets of {3, 4,..., }, ad B is the # of deragemets of {, 3, 4,..., }. Thus, D ( ( A + B ( (D + D. 3

25 D D ( [D ( D ] ( [D ( D 3 ] ( 3 [D 3 ( 3D 4 ] ( [D D ] ( ( Hece, D D + ( for all. Ex. Prove that D is eve if ad oly if is odd. Proof. If D is eve, the D ± D is odd, hece is odd. If is odd, the D is odd, for otherwise, is odd, a cotradictio. Hece, D D + ( is odd, so that D is eve. 4

26 MA30 Lecture : Ordiary Geeratig Fuctios I Erik E. Westlud Goal: discuss the importace ad basic operatios of geeratig fuctios, ad how they are applied to coutig problems. Aalogy: A geeratig fuctio for a sequece ca be thought of as a clotheslie for displayig a sequece. We wat a way to cocisely represet sequeces ad maipulate them. The formal power series of a sequece {a } is A(x a x a 0 + a x + a x + a 3 x a x + The coefficiets of A(x are the terms i the sequece (a, ad A(x is called the ordiary geeratig fuctio, or GF, of (a. Geeratig fuctios trasform problems about sequeces to problems about algebraic fuctios. Commets: Ulike calculus, we usually do t care about issues of covergece! A(x is ot a fuctio of x i the atural sese of domai ad codomai. A(x may be regarded as a purely algebraic object that ca be maipulated as such. A(x may be expressed i closed form as a fuctio of a formal argumet x, usig the otio of a fuctio ad its power series expasio. x is essetially a placeholder for a. GF s for fiite sequeces, i.e., sequeces with oly fiitely may ozero terms, (a a 0, a,..., a, 0, 0,..., as just polyomials. Ex. The GF s for the sequeces (a, with a 4, a +, ad a, are 4x, ( + x, ad x, respectively. Ex. The GF of (a,,,,...,,..., is A(x + x + x + x x + x Updated: Jue, 009. if x <.

27 Ex. The GF of (a, a, a, a 3,..., a,... is A(x +ax+a x + +a x + +(ax+(ax + +(ax + ax if ax < x < / a. Properties: If A(x a x ad B(x b x, are GF s for (a ad (b, respectively, the A(x + B(x (a + b x (a 0 + b 0 + (a + b x + (a + b x + is the GF for (a + b a 0 + b 0, a + b, a + b,..., a + b,... x k A(x a x +k a 0 x k + a x k+ + a x k+ +, called the right traslatio, is the GF for 0, 0,..., 0, a }{{} 0, a,..., a,... k ca(x ca x ca 0 +ca x+ca x +, is the GF for the scaled sequece, (ca ca 0, ca,..., ca,... A (x d dx [A(x] a x a + a x + 3a 3 x + 4a 4 x 3 +, called the formal derivative, is the GF for A(xB(x a, a, 3a 3,... a }{{} st term,... ( a k b k x, called the discrete covolutio. k0 Ex. Fid the sequece whose GF is C(x ( x. ( C(x ( x x ( x x x k0 (+x Hece, (c,, 3, 4,..., the atural umber sequece startig at, is the correspodig sequece.

28 Alteratively, we could use differetiatio: ( x d [ ] dx x k0 x Ex. The GF for (a k ( ( 0, (, (,..., k,..., for, k Z + is: ( ( ( ( ( A(x x k + x + x + x k ( + x k 0 k Notatio: If A(x is the GF for the sequece (a a 0, a, a,..., a,..., we will write: A(x a 0, a, a,..., a,.... Ex. Fid the GF for the sequece of perfect squares (a 0,, 4, 9,...,,.... Cosider: A (x xa (x d dx [xa (x] x d dx [xa (x] A(x x x x x x x k0,,,...,,..., ( x,, 3,..., +,..., x ( x 0,,, 3,...,,... + x ( x 3, 4, 9,..., ( +,... x( + x ( x 3 0,, 4, 9,...,,... 3

29 MA30 Lecture : Ordiary Geeratig Fuctios II Erik E. Westlud Ex. Fid the GF for (f, the Fiboacci umber sequece. We eed to fid F (x f 0 + f x + f x + + f x +, a GF for the sequece (f 0,, f + f 0, f + f, f 3 + f,... 0,, 0, 0, 0,... x 0, f 0, f, f, f 3,... xf (x + 0, 0, f 0, f, f,... x F (x 0, + f 0, f + f 0, f + f, f 3 + f,... x + xf (x + x F (x Hece, F (x x+xf (x+x F (x F (x x x x x+x +x 3 +3x 4 +5x 5 + I.e., the coefficiets i the power series expasio of F (x are exactly the Fiboacci umbers! Questio: How ca we use geeratig fuctios to cout? Give a set A, let A(x a 0 + a x + a x +, where the coefficiet of x is the # of ways to select objects from A. More geerally, give a task that depeds o a parameter, let the coefficiet of x be the # of ways to perform the task for. Covolutio Rule: Let A(x be the GF for selectig objects from a set A, ad let B(x be the GF for selectig objects from a set B. If A B, the the GF for selectig objects from A B is A(xB(x. Note: to cout the # of ways to select objects from A B, we ca select k objects from A ad k objects from B, where 0 k. As A B, the total is a 0 b + a b + a b + + a b 0 This is the coefficiet of x i A(xB(x. Updated: Jue, 009. Selected examples from Course Notes writte by Prof. Albert R. Meyer ad Prof. Roitt Rubifeld, MIT 005.

30 Ex. Fid the geeratig fuctio for the sequece (a, where a is the umber ways to choose objects from a multiset with k types, each available i ulimited supply. Let S { s, s,..., s k }. The GF for selectig the s s is: x + x + x + x 3 + +, for there is oe way to select zero s s, oe way to select s, oe way to select two s s, etc, ad we may replace s with ay of the s i s ad get the same GF. Hece, the GF for selectig objects from S is: ( x k ( ( ( x x x ( ( ( x x x ( + + k k x ( + k x The coefficiet of x is the umber of oegative itegral solutios to x + x + + x k. We could also derive the GF by usig Newto s Biomial Theorem. Ex. Fid the umber of itegral solutios of y + y + y 3 7 satisfyig y 5, 3 y 6, ad 4 y 3 7. Let S { s, s, s 3 } ad let y i be the # of s i s we pick. The GF for selectig the s s is S (x x + x 3 + x 4 + x 5, because there is zero ways to pick y s s for 0 y, ad 6 y, ad oe way to pick y s s for each y 5. Similarly, the GFs for pickig the s s ad s 3 s are, S (x x 3 + x 4 + x 5 + x 6, ad S 3 (x x 4 + x 5 + x 6 + x 7. Hece, the umber of ways to pick 7 objects from S, subject to the costraits, is the coefficet of x 7 i the expasio of S(x S (xs (xs 3 (x (x +x 3 +x 4 +x 5 (x 3 +x 4 +x 5 +x 6 (x 4 +x 5 +x 6 +x 7 S(x x 9 + 3x 0 + 6x + 0x + x 3 + x 4 + 0x 5 + 6x 6 + 3x 7 + x 8,

31 so there are 3 solutios. Furthermore, the coefficiet of x i S(x is exactly the # of iteger solutios to y + y + y 3 that satisfy the costraits. Ex. Determie the geeratig fuctio for the umber of bags of fruit that cotai a eve umber of apples, at most four orages, at most oe pear, ad the umber of baaas is a multiple of 5, ad o other fruits. We use the Covolutio Rule. The GF for selectig apples is A(x + x + x 4 + x 6 + x The GF for selectig orages is O(x + x + x + x 3 + x 5 x5 x The GF for selectig pears is The GF for selectig baaas is P (x + x B(x + x 5 + x 0 + x 5 + x 5 Hece, the GF for selectig objects amog the four fruits, subject to the costraits is: G(x A(xO(xP (xb(x ( x ( x 5 x ( (+x x 5 ( x. I.e., G(x + x + 3x + 4x 3 +. Hece, there are h + bags of fruits that meet the requiremets. (Note: the above coutig problem was aswered purely through algebraic maipulatio! Ex. I how may ways ca you make chage for a dollar, usig peies, ickels, dimes, ad quarters? This is equivalet to the umber of o-egative itegral solutios to x + 5x + 0x 3 + 5x 4 00 If P (x, N(x, D(x, ad Q(x deote the geeratig fuctios to choosig peies, ickels, dimes, ad quarters, the P (x + x + x + x 3

32 N(x + x 5 + x 0 + x 5 N(x + x 0 + x 0 + x 0 N(x + x 5 + x 50 + x 5 So, the # of solutios to the problem is the coefficiet of x 00 i G(x x x 5 x 0 x 5 Usig Mathematica, we see G(x + x + x + x 3 + x 4 + x 5 + x 6 + x 7 + x 8 +x 9 +4x 0 +4x +4x +4x 3 +4x 4 +6x 5 +6x 6 +6x 7 +6x 8 +6x 9 + 9x 0 + 9x + 9x + 9x 3 + 9x 4 + 3x 5 + 3x 6 + 3x 7 + 3x 8 + 3x 9 + 8x 30 +8x 3 +8x 3 +8x 33 +8x 34 +4x 35 +4x 36 +4x 37 +4x 38 +4x x 40 +3x 4 +3x 4 +3x 43 +3x x x x x x x x 5 +49x 5 +49x x x x x x x x x 6 +73x 6 +73x x x x x x x x x x x x 74 + x 75 + x 76 + x 77 + x 78 + x x x 8 + 4x 8 + 4x x x x x x x x x x x x x x x x x x Agai, G(x aswers more tha we asked for. The coefficiet of x i G(x is the # of ways to make cets out of peies, ickels, dimes, ad quarters. Theorem. Let A(x sequece. The x, be the geeratig fuctio for the geometric A(x e A(x e A(x e r T x is the geeratig fuctio where T is the # of o-egative itegral solutios to e x + e x + + e r x r. Note: there is o closed, short formula i geeral. 4

33 We ca use geeratig fuctios to solve liear recurrece relatios. A sequece (a is said to have a liear recurrece relatio of order k if a β a + β a + + β k a k + γ, where k, β k 0, ad each of the β i ad γ may deped o. Solvig a liear recurrece relatio meas fidig a closed-form expressio for a, i.e., a formula that may deped o, but ot o a i for ay i <. Ex. Solve the recurrece relatio a 3a, for, where a 0. Let A(x a x be the GF for (a. Note that xa(x a x + a x So, A(x 3xA(x a x 3 a x a 0 + Thus, ( 3xA(x A(x a 3 for 0. (a 3a x. 3x A(x ( 3 x, so that Ex. Derive a closed-form for the th Fiboacci umber f, where. x Recall F (x x x is the GF for (f. Usig Partial Fractio Decompositio, we have x F (x x x A ϕx + B ϕx, where ϕ + 5 ad ϕ 5. It s easy to fid that Hece, A ϕ ϕ 5 ad B ϕ ϕ 5. F (x ( 5 ϕx, ϕx F (x ( + ϕx + (ϕx + ( + ϕx + (ϕx + 5 5

34 Hece, F (x 5 ((ϕ ϕx + (ϕ ϕ x + + (ϕ ϕ x +, i.e., f ϕ ϕ 5 5 (( + ( 5 5. Ex. Fid a recurrece relatio for C, the umber of ways to evaluate a matrix product A, A,..., A +, for 0, by addig various paretheses ad determie a closed formula for C. E.g. C 0 C, C, they are: A (A A 3, (A A A 3. C 3 5, they are: ((A A A 3 A 4, (A (A A 3 A 4, (A A (A 3 A 4, A ((A A 3 A 4, A (A (A 3 A 4 I evaluatig C +, the matrix product A A A + eds with the fial operator betwee A k+ ad A k+, for some 0 k. Thus, Let C(x (A A }{{ k+ (A } k+ A }{{ + } C k C + C k C k C k, for. k0 C x be the GF of (C. Now ( ( C(x C x C x So, xc(x So C x ad C(x + ( C k C k x k0 C x ad so xc(x C(x xc(x C(x + 0 C(x 4x x Now, by Newto s Biomial Theorem, ( / ( 4x / ( 4x 6 C + x (!!( +! x+

35 Hece, C(x (!!( +! x ( x + Therefore, the sequece (C is give by C +(, for 0. This is called the Catala sequece, ad is a importat coutig sequece i combiatorics. 7