Counting Functions and Subsets

Transcription

1 CHAPTER 1 Coutig Fuctios ad Subsets This chapte of the otes is based o Chapte 12 of PJE See PJE p144 Hee ad below, the efeeces to the PJEccles book ae give as PJE The goal of this shot chapte is to itoduce the vey impotat umbes, called the biomial coefficiets, ad to pove the Biomial Theoem The biomial coefficiets have may uses i mathematics beyod the biomial theoem i combiatoics, pobability, statistics; they ae eve liked to the Riema hypothesis I this couse they will, fo example, be used i oe of the poofs of Femat s Little Theoem O the way, we itoduce the sets Fu(A, B ad Ij (A, B ad evisit the powe set P(A they ae combiatoial objects impotat i thei ow ight Notatio (PJE defiitios 1213 ad 1215 Give sets X ad Y, deote by Fu (X, Y the set of all fuctios fom X to Y ad by Ij (X, Y the set of all ijectios fom X to Y 11 Popositio (PJE popositios 1212 ad 1214 Let X, Y be fiite sets The ad if m = X Y =, the Fu (X, Y = Y X Ij (X, Y = ( 1 ( m + 1 =! ( m! Poof: See PJE pp 145, 146 But hee I give a o-igoous explaatio why the two fomulas ae tue, i tems of the umbe of choices of fuctios We may assume that X = N m, so that a fuctio f : X Y ca be witte i two-lie otatio: we wite ( 1 2 m a 1 a 2 a m to say that f(1 = a 1, f(2 = a 2 ad so o Hece the set Fu(X, Y has as may elemets as m-tuples (a 1, a 2,, a m of elemets of Y Such m-tuples fo the set Y Y Y = Y m, the Catesia poduct of m copies of Y, of cadiality Y m = Y X 1

2 Coutig Fuctios ad Subsets 2 Ijective fuctios f : X Y coespod, i two-lie otatio, to m-tuples (a 1, a 2,, a m whee all the eties ae distict elemets of Y Note that a 1 ca be chose i ways; to each choice of a 1 thee coespod 1 possible choices of a 2, so that (a 1, a 2 ca be chose i ( 1 ways Similaly, (a 1, a 2, a 3 ca be chose i ( 1( 2 ways Cotiuig i the same way, we coclude that thee ae ideed ( 1 ( m + 1 choices fo (a 1, a 2,, a m ( 1 2 Example Ijective fuctios fom N 2 to {a, b, c} ae, i two-lie otatio,, a b ( ( ( ( ( ,,, ad Oe has Ij (N 2, {a, b, c} = 6 which a c b c b a c a c b 3! agees with the value give by the Popositio The set Fu(N 2, {a, b, c} (3 2! ( ( ( cotais thee o-ijective fuctios,, ad Oe thus has a a b b c c Fu(N 2, {a, b, c} = 9 which is 3 2 as i the Popositio Notatio (PJE defiitio 631 Fo a set A we deote by P (A the powe set of A, which is the collectio of all subsets of A 12 Popositio (PJE theoem 1221 If A is a fiite set, the P (A = 2 A Poof: give C A, costuct a fuctio f : A {0, 1} defied by f(x = 1 if x C, f(x = 0 if x / C The each subset of A coespods to a fuctio i Fu(A, {0, 1}, ad each f Fu(A, {0, 1} coespods to a uique set C = {x A : f(x = 1} We coclude that P(A = Fu(A, {0, 1} which is 2 A by the pevious Popositio Notatio (PJE defiitio 1224 Let P (A deote the collectio of all subsets of A cotaiig exactly elemets P (A = {C A : C = }, o, equivaletly, P (A = {C P (A : C = }

3 Coutig Fuctios ad Subsets 3 Defiitio (PJE defiitio 1224 The biomial umbe o biomial coefficiet, ead as choose, is the cadiality P (A fo ay set A of cadiality, ie = P (A Example (see also PJE example 1225 If A = {a, b, c, d, e} the P 3 (A = { {a, b, c}, {a, b, d}, {a, b, e}, {a, c, d}, {a, c, e}, {a, d, e}, {b, c, d}, {b, c, e}, {b, d, e}, {c, d, e} } ( 5 Also, P 0 (A = { } ad P 5 (A = {A} We obseve that P 3 (A = 10 so = 10 3 Remak: some special cases of the biomial coefficiets If A = the P 0 (A = { } ad P (A = {A}, hece = 1 = 0 fo all 0 Fially, let us ote that if < 0 o >, the set A has o subsets of cadiality, ie, P (A = This implies that = 0 if < 0 o > Questio Does this defiitio of biomial umbe deped o the choice of set A? Aswe No, thakfully ot! The biomial umbe is well-defied, ie, P (A does ot deped o the choice of a set A, as log as A has cadiality See PJE p149 Remak Of couse, witig dow all the ( -elemet subsets of a -elemet set is a extemely iefficiet way to calculate We will peset two useful ways, oe iductive, the othe via a fomula usig factoials 13 Popositio Iductive fomula fo Fo all 1 ad all, = ( 1 + (PJE popositio

4 Coutig Fuctios ad Subsets 4 Poof: Each oe of the subsets C of N = {1,, } with C = belogs to oe of the two disjoit classes: C Equivaletly, C P (N 1 C Equivaletly, C = C {} with C P 1 (N 1 ( ( 1 It follows that = P (N 1 + P 1 (N 1, which is The esult of Popositio 13 is usually epeseted as a uedig tiagle whee each tem, apat fom those at the ed of the ows, ae the sum of the two tems i the lie above Defiitio Pascal s Tiagle ( 0 ( 0 0 ( 1 ( ( 2 ( 2 ( ( 3 ( 3 ( 3 ( ( 1 ( 0 1 ( 1 ( 1 1 ( 1 ( The stat of this is omally witte as We ae eady to pove by iductio the famous 14 Theoem Biomial Theoem (PJE theoem 1231 Let a, b R Fo all 0 we have (a + b = i=0 a i b i i Remak The ight-had side ca also be witte as a + a 1 b + ( 2 a 2 b ( 2 a 2 b 2 + ab 1 + b We use the covetio that x 0 = 1 fo all x

5 Coutig Fuctios ad Subsets 5 Poof Let P ( deote the statemet of the Theoem fo The P (0 eads 1 = 1 (tue Assume k 1 ( P (k 1 : (a + b k 1 k 1 = a k 1 i b i i i=0 to be tue ad wite (a + b k = (a + b (a + b k 1 = a (a + b k 1 + (a + b k 1 b The oly tems a s b t o the ight ae those with s + t = k, ad the tem a k i b i ca occu i two possible ways: ( k 1 as a(a k 1 i b i, whose coefficiet of i a(a + b k 1 is by P (k 1; i( k 1 as (a (k 1 (i 1 b i 1 b, whose coefficiet i (a + b k 1 b is by P (k 1 i 1 ( ( ( k 1 k 1 k Hece the coefficiet of a k i b i i (a + b k is + which equals by i i 1 i Popositio 13 Thus, P (k is tue By iductio i, P ( is tue fo all 0 We ow deduce the well-kow 15 Theoem Factoial fomula fo = (PJE theoem 12210!! (! fo all 0 ad 0, with the covetio that 0! = 1 Poof The theoem is poved i PJE but hee I give a diffeet poof (It would be useful fo the studets to study both the PJE poof ad the poof give hee Fo the exam, it will be sufficiet to kow oe poof Coside the polyomial ( p(x = (1 + x By the Biomial Theoem, p(x = a 0 + a 1 x + + a x whee a = But the coefficiet a ca also be foud as a = p( (0,! whee p ( is the th deivative of p [MATH10201 Calculus ad Vectos A] Sice oe has ( p (x = (1 + x 1, p (x = ( 1(1 + x 2, p ( (x = ( 1 ( + 1(1 + x = = a = 1!! (! (1 + 0 =! (! (1 + x,!! (!

6 Coutig Fuctios ad Subsets 6 Advice fo exam Theoem 14 is the Biomial Theoem, ot Popositio 13 o Theoem Example Fom the sixth lie i Pascal s tiagle we see (a + b 5 = a 5 + 5a 4 b + 10a 3 b a 2 b 3 + 5ab 4 + b 5 The coefficiet of a 7 b 3 i (2a + b 10 is ( = ! = 128 = !7! 3 2