1 Onto functions and bijections Applications to Counting

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1 1 Oto fuctos ad bectos Applcatos to Coutg Now we move o to a ew topc. Defto 1.1 (Surecto. A fucto f : A B s sad to be surectve or oto f for each b B there s some a A so that f(a B. What are examples of a fucto that s surectve. That s ot surectve? Defto 1.2 (Becto. A fucto f : A B s called a becto f f s both oe-to-oe ad surectve. Bectos are of partcular terest to us. We have bee talg formally about the sze of certa sets. For stace, we have ased how may ways there are to lst elemets from a set of (ether wth or wthout repetto. Ths problem s the same as asg for the sze of the set of such lsts. Bectos ca be thought of as a way of parg elemets of two sets together. Each member of each set must be pared wth exactly oe member of the other set. Thus, tutvely, we wat two sets to have the same sze ff we ca form a becto betwee them. Ths leads to the followg more formal defto of sze: Wth the cocept of becto, we ca formally defe the sze of a set: Defto 1.3 (Sze. A set A s fte f there s a becto from A to [] {1, 2,..., } for some N. I ths case we say that A has sze. Fact 1.4. A fte set must have exactly oe sze. A m, the m. Proof. Actually, the Exercse. Uses ducto. That s, f A ad Defto 1.5 (Composto. Let f : A B ad G : B C. We say that g f : A C s the composto of g wth f, ad s defed by g f(a g(f(a. Fact 1.6. If g ad f are as defed above, the g f s deed a fucto ad furthermore: 1. If g ad f are both oto, the so s g f 2. If g ad f are both oe-to-oe, the so s g f 3. If g ad g are both bectos, the so s g f Proof. To see that g f s a fucto, let a A be gve. We must prove that there s a uque c C so that (a, c g f. (We do t have the rght to use the g f(a c otato utl we prove that g f s a fucto. But we do ow that sce f(a s a uque value of b, that there s a uque c so that g(f(a c. Therefore, by defto, there s exactly oe c so that (a, c g f. Now, I wll prove oly tem (1. I leave tems (2 ad (3 as a exercse. So let g ad f be surectos. Let c C be gve. The sce g s oto there s a b B so that g(b c. Ad sce f s oto there s a a A so that f(a b. Therefore g(f(a g(b c, so g f s oto. 1

2 Fact 1.7. Let f be a becto from A B. The the verse relato of f, defed by f 1 {(y, x (x, y f} s a fucto, ad furthermore s a becto. Proof. Frst we show that f 1 s a fucto from B to A. Suppose that b B. The sce f s a becto, there s a uque a A so that f(a b. Therefore, there s a uque a A so that (a, b f, so there s a uque a A so that (b, a f 1. Therefore, f 1 s a fucto so that f f(a b the f 1 (b a. Now, we show that f 1 s a becto. Here I wll oly show that f s oe-tooe. I leave as a exercse the proof that f s oto. So let f 1 (b 1 f 1 (b 2 a for some b 1, b 2 B ad a A. We eed to show that b 1 b 2. To see ths, otce that sce f s a fucto, ad we ow that f 1 (b a ff f(a b, that f(a b 1 ad f(a b 2. But ths meas that b 1 b 2, because f s a fucto. Theorem 1.8. Two sets A ad B have the same sze ff there s a becto f : A B. Proof. Suppose that A ad B both have sze. The there s a becto g : [] A ad a becto h : [] B. The g 1 s a becto from A to [], so h g 1 s a becto from A to B. Now, suppose there s a becto f : A B, ad suppose that A has sze. The there s a becto g : [] A. The fucto f g s a becto from [] to B, so B must also have sze. 2 Subsets Frst, we wll attempt to cout the umber of subsets of a set. Defto 2.1 (Subset. We say that A B, or A s a subset of B, f each elemet of A s also a elemet of B. Note that ths meas that the set wth o elemets, called, or the empty set s a subset of every set. Oe atural questo s to as, gve a set A wth elemets, how may subsets of A are there? Fact 2.2. The umber of subsets of a elemet set S s 2 Proof. Label the elemets of the set a 1,..., a. To determe for a subset A of S, for each we ust decde whether a s the subset. We thus have a lst of choces, each choce beg yes or o. By the product prcple, there are therefore 2 such lsts of choces. We ca th of the above procedure terms of fuctos. For each subset A of [], for stace, we defe the fucto f A by f A (x 1 f x A, ad f A (x 0 f x / A. Thus, 1 correspods to a yes our lst of choces ad 0 to a o. 2

3 Example Suppose that we have a class of 7 people ad we wat a (o empty set of these 7 people to preset a paper. How may possble groups are there? There are 2 7 possble subsets, but we are ot allowg the empty set here, so the aswer s or 127. Now, suppose that there are o voluteers, ad stead we have to choose a group of three people to preset. How may ways ca we do ths? There are 7!/(7 3! such lsts. But each group of three people ca be lsted 7! 3! ways. So there are a total of (7 3!3! 35 dfferet groups. 3 Bomal Coeffcets The above example suggests a more geeral questo, ad oe that has wdespread applcatos. Suppose we have a set of people. We wat to choose a set of of them for some tas. How may ways ca we choose them? Theorem 3.1. If 0, the the umber of -elemet subsets of a elemet set s gve by the formula: (! (!! Proof. Let s cosder the set of all lsts of elemets choose from a set of elemets. The umber of ways to do ths s!/(!. Furthermore, we ca geerate all such lsts by choosg people frst, ad the lstg them the! proper ways. Thus there are (! ways of lstg people from a group of. Therefore, (!! (!, so (! (!! Fact 3.2. ( ( Proof. Exercse. 4 Usg Pascal s Tragle See the bottom of p.31 Bogart, 3rd edto, for a pctoral vew of Pascal s tragle. Pascal s tragle s cotructed by placg a 1 at the top pot, ad a par of 1s, oe below ad to the left ad oe beloe ad to the rght of the top 1. The further etres are computed by addg the oes above t. It turs out that f we cout the rows startg at row 0, ad the etres at each row from zero (So, the frst row s ow the 0th row, ad so forth, the t turs out that ( s gve by pcg the th elemet of the th row. Ths fact s precsely gve by the followg theorem: Fact 4.1. ( ( 1 ( + 1 wheever 0 < <. 1 Proof. We cosder the ways of pcg elemets out of a set of elemets. Let S be a set of elemets. Let s S be sgled out. To pc elemets of S, we may ether do so by choosg s, or ot choosg s. If we elect to ot choose 3

4 s, the we choose of the remag 1 members, whch ca be doe ( 1 waus. Ad f we elect to choose s, we choose oly 1 of the remag 1 members, whch ca be doe ( 1 1 ways. So we have two dsot collectos of -elemet subsets of S, whch cover all of the possble -elemet subsets of S. By the sum prcple, we therefore get ( ( 1 ( Why do we care about Pascal s tragle? Th computatoally. Suppose we wat to compute ( If we use the formula as stated, we would have 20! 10!10!. 20! ad 10! are very large umbers. But whe we evaluate the fracto, we get somethg more maageable: about 185,000. So ths suggests that there may be a better way of calcluatg ( oe where the umbers do ot get very large. Ad ( ths s Pascal s tragle. It turs out as we sad above, we ca easly get by loog at Pascal s tragle. Ad the etres Pascal s tragle do ot grow as qucly as factorals do, so the computatoal process s much easer. So far, we have oly defed ( for values where t maes sese to choose obects out of. But we ca also defe ( for other values. We do so by otcg that f 0, but < 0 or >, the there s o way to choose obects out of. So these cases we defe ( ( 0. Later o, we wll defe whe s egatve. Ths method of usg Pascal s Tragle to compute ( s a example of recurso. Recurso s whe we try ad compute a value for a fucto from values wth smaller puts. I other words, If we had a fucto f of oe varable, ad we had a way of computg f( from f( 1. 5 Usg Bomal Coeffcets Theorem 5.1 (Bomal theorem. For ay teger 0: (x + y 0 ( x y o Proof. (x + y s a product of factors. Each term the factor s a result of pcg ether x or y from each of the (x + y occurrg the term. x y appears the expaso oly f +. Furthermore t appears as may ways as we ca choose exactly x s out of the factors. So the coeffcet of x y s (. The result follows. See examples 4.1 ad 4.2 Bogart. Here s a more terestg example of ths fact: Ths s called Vadermode s formula Theorem 5.2. Vadermode s Formula s: ( m + ( ( m 0 4

5 Proof. (x + 1 m (x + 1 m ( m x 0 m ( m 0 0 m ( m 0 0 m+ 0 ( 0 0 ( x x ( ( x x x ( ( m x Sce the coeffcet of x s also ( ( m, we get the desred formula. We ve gve a purely algebrac proof of the formula. Let s ow gve a more tutve, combatoral oe: Proof. The left had sde of the formula s ( m+, choosg elemets out of a set of m +. We ca choose these elemets by loog separately at the frst m ad the last elemets. For each, we ca get elemets of the large set of m + elemets by choosg of the frst m ad of the last. By the product prcple, there are ( ( m ways of dog ths. Sce each choce of gves us a completely dfferet collecto of elemets, we get, from the sum prcple: ( m + 0 ( ( m 5