Chapter 1 : Combinatorial Analysis

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1 STAT/MATH 394 A - PROBABILITY I UW Autum Quarter 205 Néhémy Lim Chapter : Combiatorial Aalysis A major brach of combiatorial aalysis called eumerative combiatorics cosists of studyig methods for coutig the umber of ways that certai patters ca be formed from fiite sets. We will see how it applies to probability theory. Basic coutig priciples Example. Birthday : 50 studets, 2 moths. At least 2 studets were bor i the same moth.. Pigeohole priciple Theorem. (Pigeohole priciple. Dirichlet, 834. If items are to be put ito m cotaiers, with > m, the at least oe cotaier must cotai more tha oe item. Proof. Let us assume that objects ad boxes are labeled respectively o,..., o ad b,..., b m. Without loss of geerality (w.l.o.g., we ca put o ito b,..., o m ito b m. Therefore, there remai m > 0 objects, amely o m+,..., o that eed to be assiged. So at least oe cotaier will cotai more tha oe item. The pigeohole priciple relates to the cocept of ijectio o fiite sets. Theorem.2 (Ijectios o fiite sets. Let E ad F be fiite sets ad f : E F be a fuctio from E to F. If E > F the f is ot ijective. I other words, if E > F, there is o ijectio from E to F. Lemma.3. If items are to be put ito m cotaiers, with > m, the at least oe cotaier must cotai at least /m items. Remiders : Floor fuctio : x R, x sup{ Z x} Ceilig fuctio : x R, x if{ Z x} Before provig the above lemma, let us first show the followig result : Property.. x R, x < x + (

2 Proof. Let us use a Reductio ad Absurdum argumet. Let us assume there is some x 0 R such that x 0 x 0 +. I that case, x 0 is a iteger smaller tha x 0 satisfyig x 0 x 0, which cotradicts the defiitio of x 0 We are ow ready to prove the lemma. Proof. All items are put ito the m cotaiers. We will use a Reductio ad Absurdum argumet. Let us assume that all cotaiers cotai at most /m items. The, the maximum umber of objects i the boxes is m ( /m < m(/m which cotradicts the fact that the items are all i the boxes. Example bis. Birthday : 50 studets, 2 moths. At least 50/2 5 studets were bor i the same moth..2 Rule of sum Propositio. (Rule of sum. Let us cosider r evets. If there are possible outcomes for the first evet,..., r possible outcomes for the rth evet ad if ay two distict evets caot both occur (evets are mutually exclusive, the there are r i i total possible outcomes for the evets. The rule of sum relates to the followig set theory property. Propositio.2. If S,..., S are pairwise disjoit sets the S i S i i i Example 2. You pla a trip. You hesitate betwee differet destiatios : Europe (3 cities / Asia (4 cities / South America (2 cities. But you ca oly choose oe of them due to a restricted budget. How may differet destiatios are possible? Rule of product Propositio.3 (Rule of product. If r experimets are to be performed sequetially ad the first experimet ca be performed i ways,..., the rth experimet i r ways, the there are r i i ways to perform the r experimets. The rule of product relates to the cocept of cartesia product. 2

3 Propositio.4. Let S,..., S be sets the S... S S i Example 3. You pla a tour of Wester Europe ad wat to visit Lodo, Paris ad Rome. There are 2 recommeded roads from Lodo to Paris ad 3 from Paris to Rome Example 4. Let us cosider 3 sogs cosistig of 20 lies. You ca compose , 486, 784, 40 differet sogs of 20 lies where lie comes from lie of ay of the 3 sogs,..., lie 20 comes from lie 20 of ay of the 3 sogs. It taes aroud 6,634 years to liste to all the sogs at the pace of sog per miute. Example 5. You go to a restaurat. There you ca either choose oe starter ad oe course or oe course ad oe dessert, but you caot tae a starter, a course ad a dessert. That day, the restaurat proposes 4 starters, 4 courses ad 3 desserts. How may differet meus are possible? Permutatios Example 6. I wat to visit 0 people, each of them livig i differet cities. How may differet orders are possible to visit them? , 628, 800 i Defiitio 2. (Permutatio. A ordered raig of N distict elemets is called a permutatio Propositio 2.. There are (... permutatios of N distict elemets. Proof. Rule of product Defiitio 2.2 (Factorial. N, By covetio 0! i i Example 6 bis. I wat to visit 0 people, each of them livig i differet cities. Amog these 0 people, there are 6 relatives of mie. How may differet orders are possible if I wat to visit my family first? 6!4! , 280 Example 7. I have DVDs that I wat to put o my shelf. Of these, 4 are actio movies, 2 are sciece-fictio movies, 3 are fatasy movies, ad 2 are comedy movies. I wat to arrage my DVDs so that all the DVDs dealig with the same subject are together o the shelf. How may differet arragemets are possible? 4!4!2!3!2! 3,824 3

4 3 Partial permutatios Arragemets Defiitio 3. (Partial Permutatio. A partial permutatio or a (combiatorial arragemet is a ordered raig of p items amog N elemets (p Propositio 3.. The umber of arragemets of p items amog N elemets (p is deoted A p ad is equal to Proof. Rule of product A p (... ( p + ( p! Example 8. Formula racig. There are 22 drivers. Oly the first 0 drivers crossig the fiish lie ear champioship poits. A ! (22 0! 4 Combiatios We are ofte iterested i determiig the umber of differet groups of r objects that could be formed from a total of objects Defiitio 4. (Combiatio. A combiatio is a uordered collectio of p items amog N elemets (p Propositio 4.. The umber of combiatios of p items amog N elemets (p is deoted ( p (read choose p ad is equal to ( p p!( p! Proof. If we perform all the permutatios of each combiatio, we obtai all the arragemets of p items amog : A p p! ( p. Example 9. Teis. ATP World Tour Fials. There are 00 players. Oly the first 8 i the ATP raig at the ed of the year get access to the fials. ( 00 8 Property 4.., p N, ( + p + ( ( +, p (2 p p + Proof. OK for a aalytic proof. Let us thi of a combiatorial argumet. Cosider a set S {e,..., e + } of + distict elemets ad focus o a particular elemet, say e w.l.o.g. Combiatios of p + elemets amog + ca be divided as follows : combiatios cotaiig e, there are ( ( p such combiatios ad there are p+ combiatios without e. 4

5 Aother useful combiatorial idetity is Property 4.2., p N, Proof. Thi of a combiatorial argumet ( (, p (3 p p Theorem 4. (Biomial theorem. For all x, y R ad N, (x + y 0 ( x y (4 Proof. By iductio. Let x, y be 2 real umbers. For a give N, deote by P the propositio : (x + y ( 0 x y. First, let us prove P. We have : ( ( (x + y x + y x 0 y 0 + x y 0 Assumig P is true for some N, we have : (x + y + (x + y(x + y ( (x + y x y 0 ( x + y ( x i y + i + i i x + + x + + i ( x y + 0 ( x i y + i + i i {( + i ( + x + y + ( ( + x i y + i i0 i ( x y + ( x y + + y + ( } x i y + i + y + i ( + i i x i y + i + ( + 0 x 0 y + 0 5

6 5 Multiomial Coefficiets We shall ow determie the umber of ways to divide items ito r distict groups of respective sizes,..., r such that r i i. W.l.o.g., there are ( choices for the first group; for each choice of the first group, there are ( 2 for the secod group; ad so o. From the rule of product there are : ( ( (... r possible divisios. 2 r r i r! Defiitio 5.. Let,,..., r N such that r i i, we defie the multiomial coefficiet by (,..., r r i r! Thus, (,..., r represets the umber of possible divisios of distict objects ito r distict groups of respective sizes,..., r. It also relates to the problem of fidig the umber of permutatios of a set of objects whe certai of the objects are idistiguishable from each other. (5 6