Some Basic Counting Techniques

Transcription

1 Some Basic Coutig Techiques Itroductio If A is a oempty subset of a fiite sample space S, the coceptually simplest way to fid the probability of A would be simply to apply the defiitio P (A) = s A p(s); () however, this method has two importat drawbacs I all but the simplest cases, equatio () represets a impractically log calculatio For example, the experimet of hirig te ew worers from a pool of oe hudred applicats has more tha 7 trillio possible outcomes, so calculatig P (A) usig () could ru to trillios of additios A brute-force calculatio lie () obscures the bigger picture with a flood of busywor ad blocs the way to discoverig uderlyig coutig priciples Ad, fortuately, these uderlyig coutig priciples do exist; for example, as you must realize did ot calculate () ot by listig all of the possible samples or by coutig them oe-by-oe This hadout will itroduce you to some coutig formulas; the formula I used to calculate the 7 trillio figure will be itroduced i 32 2 Two Basic Rules I may istaces, oe aswers the questio, How may elemets does this set cotai? by tallyig up the umber of ways i which a typical elemet of the set ca be costructed, ad this latter umber is ofte assessed by breaig the costructio of a set elemet ito a sequece of smaller tass As a cosequece, the followig two rules, although they are basic ad obvious, evertheless tur out to be very useful Suppose that you have two tass i frot of you call them T ad T 2 ad suppose that performig tas T ca be result i differet outcomes ad that performig tas T 2 ca be result i m differet outcomes Tas rule # Performig both T ad T 2 ca result i m differet outcomes 2 Tas rule #2 If the tass do ot overlap, performig either T or T 2 (but ot both) ca result i + m differet outcomes To be precise: there are 7, 30, 309, 456, 440 of them 2 The text (p4) calls this the m rule

2 For example: let T be the tas of flippig a coi (outcome set {H, T }, so = 2); ad let T 2 be the tas of rollig a die (outcome set {, 2, 3, 4, 5, 6}, so m = 6) The performig T ad T 2 will result i oe of 2 6 = 2 differet outcomes, amely {(H, ), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, ), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}; ad performig T or T 2 (but ot both) will result i oe of = 8 differet outcomes, amely {H, T,, 2, 3, 4, 5, 6} 3 Coutig Samples May coutig theorems ca be itroduced by aalyzig ad aswerig the followig questio You are goig to sample members of a -member populatio How may differet -elemet samples might you wid up with? (2) Whe phrased i this way, questio (2) is ambiguous; the exact umber of samples depeds upo the aswers to two additioal questios: Is it permissible to sample the same idividual more tha oce? For example: if you are coutig the possible 3-elemet samples draw from {, 2, 3, 4, 5}, are you icludig such samples as {, 2, 2} i your cout? If the aswer is Yes, the samplig process is called samplig with replacemet; if the aswer is No,, the samplig process is called samplig without replacemet 3 2 Does order matter? Agai, suppose that you are coutig the possible 3-elemet samples draw from {, 2, 3, 4, 5} Are you coutig {, 2, 3} ad {3, 2, } as two differet samples? (Ad if the process is samplig with replacemet are you coutig {, 2, 2} ad {2, 2, } as two differet samples?) If the aswer is Yes, the samplig process is called ordered samplig; if the aswer is No,, the samplig process is called uordered samplig I other words, questio (2) is actually four distict questios, ad we will cosider them each i tur, worig our way from the easiest oe to aswer to the hardest oe to aswer 3 Ordered Samplig It is easier to cout possible samples whe order matters tha whe it does t, because the process of choosig a sample ca be broe dow ito a series of tass: Tas T : Tas T 2 : Tas T : 3 The text itroduces these terms o p78 choose the first member of the sample choose the secod member of the sample choose the th member of the sample 2

3 By the first Tas Rule, the, the umber of possible samples is the umber of samples = i= ( ) the umber of ways of choosig the i th member of the sample (3) Equatio (3) will tell us the umber of samples regardless whether samplig is carried out with or without replacemet, but the umber of samples is ot the same i the two cases, because for 2 i, the umber of ways of choosig the i th member of the sample do ot match 3 Ordered Samplig with Replacemet I this case, for each i, the umber of ways of choosig the i th member of the sample is simply, because ay member of the populatio might be chose Therefore, i this case, formula (3) gives: the umber of samples = = (4) For example, i problem 234 o assigmet #3 (text p40), the sample space is the set of ordered 3- elemet samples with replacemet from the 3-elemet populatio {a, b, c}, so the sample space cotais = 3 3 = 27 sample poits 4 32 Ordered Samplig without Replacemet I this case, each sample member after the first oe is draw ot from the etire populatio but oly from the remaiig members of the populatio those who have ot already bee sampled It follows that the umber of ways of choosig the first member of the sample is ; the umber of ways of choosig the secod member of the sample is ( ); the umber of ways of choosig the third member of the sample is ( 2); ad so o It follows that i this case, formula (3) gives: the umber of samples = i=0 The product i formula (5) is commoly deoted P i= i = ( )( 2) ( ( )) (5) or P (, ) P (, ) := ( )( 2) ( + ) [ = ]! 5 (6) ( )! 4 I listed the complete sample space i my wored solutio to this problem 5! Please ote: while is a compact way to represet P (, ), it is the worst possible way to calculate P (, ) ( )! 3

4 32 The meaig of P(, ) Puttig = i (5) will produce the umber of ordered oreplacemet samples of the etire populatio A momet s reflectio will covice you that the differet samples correspod to the differet orders i which the populatio ca be arraged A ordered arragemet of a set is called a permutatio of that set, so P (, ) =! = the umber of orders i which objects ca be arraged = the umber of permutatios of a -elemet set Exercise What aswers do ( 4) ad ( 5) give if <? Do these values mae sese? Why or why ot? 32 Uordered Samplig As metioed above, these samples are more difficult to cout tha are ordered samples Uordered samples without replacemet are both easier to cout ad more importat tha uordered samples with replacemet, so I will tacle these first 32 Uordered Samplig without Replacemet Before turig to the tas of coutig samples of this id, I wat to discuss the samples themselves Selectig a uordered -sample without replacemet from a -elemet set what is meat by the phrase choosig thigs out of thigs For this reaso, the (yet-to-be-determied) umber of such samples is read choose These umbers are called biomial coefficiets, because they appear i the Biomial Theorem (Theorem 3 below), ad symbolically, choose is deoted ( ) A -sample of this sort draw from a set T precisely the same as a -elemet subset of T Therefore, choose = = the umber of -elemet subsets of a -elemet set (7) For example, a complete list of the two-elemet subsets of {, 2, 3, 4} is therefore, 4 = 6 2 {, 2}, {, 3}, {, 4}, {2, 3}, {2, 4}, {3, 4}; It is worth otig that usig (7) to defie ( ( ) leads to the correct iterpretatio that 0) = for all 0 5 Exercise 2 List all of the three-elemet subsets of {, 2, 3, 4, 5} What is the value of? 3 Exercise 3 Use (7 ) to iterpret the equatio 0 = Why is it correct? 0 4

5 32 A Formula for Fix ad, with, ad cosider the list of all P (, ) -elemet ordered samples without replacemet from the populatio {, 2,, } For example: If = 5 ad = 3, this list cosists of P (5, 3) = 60 elemets, which ca be arraged i the followig table Observe that: (, 2, 3) (, 3, 2) (2,, 3) (2, 3, ) (3,, 2) (3, 2, ) (, 2, 4) (, 4, 2) (2,, 4) (2, 4, ) (4,, 2) (4, 2, ) (, 2, 5) (, 5, 2) (2,, 5) (2, 5, ) (5,, 2) (5, 2, ) (, 3, 4) (, 4, 3) (3,, 4) (3, 4, ) (4,, 3) (4, 3, ) (, 3, 5) (, 5, 3) (3,, 5) (3, 5, ) (5,, 3) (5, 3, ) (, 4, 5) (, 5, 4) (4,, 5) (4, 5, ) (5,, 4) (5, 4, ) (2, 3, 4) (2, 4, 3) (3, 2, 4) (3, 4, 2) (4, 2, 3) (4, 3, 2) (2, 3, 5) (2, 5, 3) (3, 2, 5) (3, 5, 2) (5, 2, 3) (5, 3, 2) (2, 4, 5) (2, 5, 4) (4, 2, 5) (4, 5, 2) (5, 2, 4) (5, 4, 2) (3, 4, 5) (3, 5, 4) (4, 3, 5) (4, 5, 3) (5, 3, 4) (5, 4, 3) Every etry o this list ca be thought of as a three-elemet subset of {, 2, 3, 4, 5} 2 All of the etries o ay give row represet the same three-elemet subset, with the elemets arraged i differet orders 3 Each three-elemet subset ca be arraged i P (3, 3) = 3! = 6 differet orders, so ay give row cotais all of the etries that correspod to a give subset 4 Therefore, the umber of three-elemet subsets of of {, 2, 3, 4, 5} equals the umber of rows i the table Sice there are 60 = 0 rows, we ca coclude: 6 5 = = 0 These cosideratios apply for ay ad such that : i the list of all P (, ) ordered -samples chose without replacemet from the populatio {, 2,, }, each -elemet subset of {, 2,, } appears exactly! times oce i each of the! orders i which the set ca be arraged We therefore have Theorem P (, ) ( ) ( + ) = = =!!! ( )!! (8) Exercise 4 Show that the experimet of hirig te ew worers from a pool of oe hudred applicats has exactly 7, 30, 309, 456, 440 differet possible outcomes 5

6 322 Pascal s Triagle The biomial coeffiets are ofte arraged i a triagle called Pascal s Triagle: ( 0 0) ( ) ( 0 ) ( 2 ) ( 2 ) ( 2 0 2) ( 3 ) ( 3 ) ( ) ( 3 3) = As you o doubt have leared, there is a easy way to geerate the umbers i this array, oe row at at time I order to mae a ew row: put s i the outside positios (because ( 0) = ( ) = for ay ); at ay iside positio, simply add the two umbers i the previous row that are above that positio The reaso this wors is the followig theorem: Theorem 2 For ( ), = + (9) Proof Cosider a list of all ( ) subsets of {, 2,, } Sice some of these subsets cotai, while others do ot, we ow that ( ) ( = umber of subsets that do ot cotai }{{} (A) Let us try to evaluate (A) ad (B) ) ( + umber of subsets that cotai }{{} (B) ) (0) (A): The -elemet subsets of {, 2,, } that do ot cotai are exactly the -elemet subsets of {, 2,, ( )}, ad there are ( ) of these Therefore, (A) = () (B): Every -elemet subset of {, 2,, } that does cotai ca be costructed by performig two tass: choose a ( ) elemet subset of {, 2,, ( )} (ca be doe i ( ) ways); add to the set (ca be doe i oe way) By tas rule #, the, a -elemet subset of {, 2,, } that does cotai ca be costructed i ( ) ( ) ways Coclusio: (B) = (2) Fially, substitutig () ad (2) ito (0) gives (9) 6

7 323 The Biomial Theorem As metioed above, biomial coefficiets get their ame from the rôle they play i the extremely useful Biomial Theorem The theorem ca best be uderstood by first observig a few examples of it Below, I have put the first few rows of Pascal s Triagle ext to the expasios of (x + y) for the first few oegative values of x + y x 2 + 2xy + y 2 x 3 + 3x 2 y + 3xy 2 + y 3 x 4 + 4x 3 y + 6x 2 y 2 + 4xy 3 + y 4 = (x + y) 0 = (x + y) = (x + y) 2 = (x + y) 3 = (x + y) 4 As you ca see: at least for 0 4, the coeffiets i the expasio match the umbers i the triagle The Biomial Theorem is the assertio that, i fact, they will match for every value of Theorem 3 (Biomial Theorem) For ay 0, (x + y) = x y 0 + x y + + x y x 0 y = =0 x y (3) Proof Equatio (3) ca be proved by aalyzig the process of carryig out the multiplicatio below: 7 Cosider: (x + y) = (x + y)(x + y) (x + y) : }{{} pairs of paretheses Every term i the expasio is the product of factors, with oe factor selected from each pair of paretheses 2 The factor the selected from each pair of paretheses must be either a x or a y; hece each term is of the form x y for some 0 3 Before gatherig lie terms, the complete expasio is the sum of all of the 2 possible terms 4 Gatherig lie terms Two terms ca be added together if they have the same umber of x s ad y s After gatherig terms, the, the product loos lie c 0 x y 0 + c x y + + c x y + + c x 0 y, (4) where, for 0, c is the umber of terms with y s ad ( ) y s 5 The umber of terms with y s ad ( ) y s is the umber of ways of selectig of the pairs of paretheses out of which to tae the y s (ad tae x s out of the remaiig pairs of paretheses) 6 The umber of ways of maig this selectio is ( ) ; therefore, i (4), c = ( ), 0 7 Fially: you get (3) by replacig each c with ( ) i (4) 6 The expasios of (x + y) 0 ad (x + y) are obviously correct, you doubtless have the expasio of (x + y) 2 memorized The other two are also correct, as you ca easily chec 7 This is ot the easiest proof, but it is the best oe, because it explais why the theorem is true 7

8 324 Dormitories, Multiomial Coefficiets, ad the Multiomial Theorem The Biomial Theorem has sibligs: the Triomial Theorem that tells you how to expad (x + x 2 + x 3 ) ; the Quatriomial Theorem that tells you how to expad (x + x 2 + x 3 + x 4 ) ; ad so o All of these, icludig the Biomial Theorem, ca be combied ito oe overarchig theorem, called the Multiomial Theorem (Theorem 4 o p9), which tells you how to expad (x + x x ) for ay itegers > 0 ad 0 The coefficiets that appear i the statemet of the theorem are called wait for it!! multiomial coefficiets; they iclude the biomial coefficiets as a special case Multiomial coefficiets ca be very comprehesibly itroduced as umber of ways i which the followig tas ca be carried out Choreopsis Uiversity has studets ad dormitories, poetically amed {D, D 2,, D } For i, dorm D i ca hold i studets, ad, lucily for the Uiversity, =, so that the studets ca exactly fit The tas is to assig the studets to the various dorms The groud rules are: Every studet is assiged to oe ad oly oe dorm 2 Assigig particular rooms is ot icluded i the tas; we are merely selectig a uordered set of i studets for dorm D i, for each i (5) 3 It matters which set of studets is assiged to each dorm, so that, if, say, = 2 = 50, assigig studets through 50 to D ad studets 5 through 00 to D 2 is couted as differet from assigig studets through 50 to D 2 ad studets 5 through 00 to D We eed to ascertai the umber of ways i which this tas ca be carried out It will be helpful to cosider specific cases first Suppose that = 2, that = 85, ad that 2 = 95 (so that = 80) The umber of ways the studets ca be assiged is just ( 80) 85, the umber of ways to choose the studets who will go to D (It is also the umber of ways to choose the studets who will go to D 2, amely ( 80) 95 ; these 80! ) umbers are equal, because they both equal 5!95! The geeral = 2 case Let + 2 = Depedig upo which dorm you process first, you fid that either that there are ( ) ( or ) 2 ways to perform the tas These aswers match, because =!! 2! = 2 8

9 I this cotext, biomial coefficiets are writte slightly differetly: i the case = 2 umber of possible dorm assigmets is :=! 2! 2! Suppose that = 3 ad that = 70, 2 = 80, ad 3 = 90 (so that = 240) I this case, the dorm assigmets ca be made i six differet orders Ay of the six orders ca be used to cout the umber of possible assigmets For example, you ca mae the three dorm assigmets i the order (D, D 2, D 3 ) by performig each of the followig subtass ( (240 ) ) Subtas #: Assig 70 of the studets to D 70 ways ( (70 ) ) Subtas #2: Assig 80 of the remaiig studets to D 2 ways Subtas #3: Assig the remaiig 90 studets to D 3 (oe way) It ow follows from tas rule # that the umber of possible dorm assigmets is ( 240)( 70 ) Exercise 5 [a ]: Show that ( 240)( 70 ) = 240! 70!80!90! [b ]: Calculate the umber of possible dorm assigmets usig each of the other five 240! orders, ad show that each of the five differet-looig aswers is equal to 70!80!90! The geeral = 3 case Let = There are six orders i which the idividual dorm assigmets ca be made Each order gives a differet way of calculatig the umber of possible dorm assigmets, but the six aswers match: the umber of possible dorm assigmets is! := 2 3! 2! 3! The geeral case For the geeral problem (box (5) o p8): There are! differet orders i which the dorms ca be processed, ad all of the orders give the same aswer for the umber of possible dorm assigmets: The umber of ways to carry out (5) is ( 2 80 ) :=!! 2!! (6) The Multiomial Theorem is a straightforward geeralizatio of the Biomial Theorem (p7) Theorem 4 For ay 0 ad, (x + x 2 + x ) = = ( ) x 2 x 2 2 x (7) I (7), there is oe term for each way to write as the sum of oegative itegers I will show you some examples i class 9

10 Outlie of Proof The proof is completely parallel to that of the Biomial Theorem I the expasio of (x + x x ) = (x + x x )(x + x x ) (x + x x ), }{{} pairs of paretheses each term is of the form x x 2 2 x, where = After combiig terms with the same values of, 2,,, we have that (x + x 2 + x ) umber of ways = of obtaiig this x x 2 2 x ; term = ad the umber of ways of gettig x x 2 2 x is the umber of ways of carryig out the dormassigmet tas ((5) o p8), amely ( ) 2 The theorem follows 322 Uordered Samplig with Replacemet There is a lot to say about these, but because the text does ot discuss them, I will iclude oly a little iformatio here Notatio The umber of uordered samples chose with replacemet from a -elemet set is sometimes deoted (( )) 8 2 A example I will list all the uordered 3-samples that ca be chose with replacemet from the populatio {2, 3, 4}; i order guard agaist duplicatio, I will arrage each sample i odecreasig order [2, 2, 2] [2, 2, 3] [2, 2, 4] [2, 3, 3] [2, 3, 4] [2, 4, 4] [3, 3, 3] [3, 3, 4] [3, 4, 4] [4, 4, 4] (8) 3 Samples of this sort are ofte called multisets Multisets differ from sets i that elemets ca be icluded multiple times; two multisets are equal if ad oly if they cotai the same umber of elemets, each oe the same umber of times For example, as sets, but the multisets are all differet {2, 3} = {2, 2, 3} = {2, 3, 3}, [2, 3], [2, 2, 3], ad [2, 3, 3] (9) 8 I ow of o stadard way to read (( )) For wat of a better optio, I read it double-choose 0

11 4 You have ecoutered multisets if you have ever wored with the prime factorizatios of itegers As you probably ow, every iteger 2 ca be expressed as a product of primes i a uique way Stated more carefully: every iteger 2 ca be expressed as a product of a uique multiset of primes: you must use certai prime umbers ad use each oe a certai umber of times For example, the three multisets listed i (9) are the factorizatios, respectively, of 6 = 2 3, 2 = 2 2 3, ad 8 = 2 3 3; differet multisets of primes multiply out to differet umbers 5 Seeig a formula for (( )) : a approach that does NOT wor Recall that we were able to fid the formula for ( ) (formula (8) o p5) by dividig P (, ) by!; this wored because i the list of all -permutatios, 9 each -subset appeared exactly! times This approach will ot wor here: i the list of -sequeces, 0 differet -multisets appear differet umbers of times I class, I will use (8) to illustrate 6 Seeig a formula for (( )) : a approach that DOES wor Let us start with a list of all of the (( )) -multisets draw from a -elemet set of umbers; assume that the elemets of each multiset have bee arraged i odecreasig order (as was doe i (8)) The, let us add the vector [0,,, ( )] to each multiset If we were to do this to the multisets i (8), for example, we would geerate the followig table: LEFT: RIGHT: [2, 2, 2] +(0,, 2) = {2, 3, 4} [2, 2, 3] +(0,, 2) = {2, 3, 5} [2, 2, 4] +(0,, 2) = {2, 3, 6} [2, 3, 3] +(0,, 2) = {2, 4, 5} [2, 3, 4] +(0,, 2) = {2, 4, 6} [2, 4, 4] +(0,, 2) = {2, 5, 6} [3, 3, 3] +(0,, 2) = {3, 4, 5} [3, 3, 4] +(0,, 2) = {3, 4, 6} [3, 4, 4] +(0,, 2) = {3, 5, 6} [4, 4, 4] +(0,, 2) = {4, 5, 6} (20) There are several worthwhile observatios to be made about table (20) (a) Sice the elemets i each multiset o the left are odecreasig, the elemets i each sample o the right are strictly icreasig (b) Therefore, i each sample o the right, there are o repeats that is, each sample o the right is a three-elemet set (c) The smallest umber that appears o the right is = 2, ad the largest umber that appears o the right is = 6; therefore, every three-elemet set o the right is a subset of {2, 3, 4, 5, 6} 9 That is, ordered -samples that were sampled without replacemet 0 That is, ordered -samples that were sampled with replacemet There are more iformative proofs, but this oe is the most direct

12 (d) Every three-elemet subset of {2, 3, 4, 5, 6} appears o the right, ad o subset of {2, 3, 4, 5, 6} appears twice o the right ( (3 ) ) (e) Therefore, 3 = ( 5 3) = 0 Furthermore, all of these observatios ca also be made i the geeral case Assume that we have o the left a list of all of the -elemet multisets draw from {, 2,, }, with the elemets of each multiset arraged i odecreasig order, ad assume that as we did i (20) we have added the vector [0,,, ( )] to each multiset o the left to geerate a list of samples o the right (a) Sice the elemets i each multiset are odecreasig, the elemets i each sample o the right are strictly icreasig (b) Therefore, i each sample o the right, there are o repeats that is, each sample o the right is a -elemet set (c) The smallest umber that appears o the right is + 0 =, ad the largest umber that appears o the right is + ( ); therefore, every -elemet set o the right is a -elemet subset of {, 2,, + } (d) To see that that o subset appears twice o the right, observe that subtractig the vector [0,,, ( )] from ay set o the right will get you bac to the multiset that it came from (so that a repetitio o the right would reveal a repetitio o the left, ad there are oe of those) (e) Seeig that every -elemet subset of {, 2,, ( )} appears o the right is a little more ivolved Observe first that a sequece (a, a 2,, a ) of itegers appears o the left as a multiset if ad oly if the sequece satisfies the iequalities Now let a a 2 a (2) T = {b, b 2,, b } (b < b 2 < b ) be ay -elemet subset of {, 2,, + }, ad let M = (a, a 2,, a ) = (b, b 2,, b ( )) be the sequece you get by subtractig the vector [0,,, ( )] from T Observe that i b < b 2 < < b = a a 2 a ; ii b = b 0 = a ; ad iii b + = a = b ( ) ( + ) ( = ) = These three observatios show that sequece M satisfies (2); hece, M appears o the left as a multiset, ad T appears across from M o the right (f) We ca thus coclude that (( ))=( + ) (22) Exercise 6 Each of the questios below ca be aswered by applyig (22 ) [a]: How may -digit itegers have their digits i odecreasig order? [b]: How may -digit itegers have their digits i odecreasig order? [c]: How may terms are beig added o the right side of the Multiomial Theorem [equatio (7 ) o p9 ]? 2