6.4 Binomial Coefficients

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1 64 Bioial Coefficiets Pascal s Forula Pascal s forula, aed after the seveteeth-cetury Frech atheaticia ad philosopher Blaise Pascal, is oe of the ost faous ad useful i cobiatorics (which is the foral ter for the study of coutig ad listig probles Theore 1 (Pascal s Forula Let ad r be positive itegers ad suppose r The ( ( ( 1 r r 1 r Proof (algebraic versio Let ad r be positive itegers ad suppose r By the forula for C(, r, ( (! r 1 r (r 1!( r 1!! r!( r!! r r(r 1!( r 1!!( r 1 r!( r!( r 1! r!! r! r!( r 1! r!( r 1!! r!! r! r!( r 1!!! r!( r 1!!( 1 r!( r 1! ( 1! r!( r 1! ( 1 r Proof (cobiatorial versio Let ad r be positive itegers ad suppose r Suppose S is a set with 1 eleets The uber of subsets of S of size r ca be calculated by thiig of S as cosistig of two pieces: oe with eleets {x 1, x 2,, x } ad the other with oe eleet {x 1 } Ay subset of S with r eleets either cotais x 1 or it does ot If it cotais x 1, the it cotais r 1 eleets fro the set {x 1, x 2,, x } If it does ot cotai x 1, the it cotais r eleets fro the set {x 1, x 2,, x } Subsets of size r of {x 1, x 2,, x 1 }: subsets of size r subsets of size r that cosist etirely that cotai x 1 of eleets fro ad r 1 eleets {x 1, x 2,, x } fro {x 1, x 2,, x } There ( are There are ( r of these r 1 of these By the additio rule, uber of subsets of uber of subsets of {x 1, x 2,, x 1 } {x 1, x 2,, x } of size r of size r 1 uber of subsets of {x 1, x 2,, x } of size r

2 By cobiatio forula, the set {x 1, x 2,, x 1 } has ( 1 ( r subsets of size r, the set {x1, x 2,, x } has r 1 subsets of size r 1, ad the set {x1, x 2,, x } has ( r subsets of size r Thus ( ( ( 1 r r 1 r The Bioial Theore I algebra a su of two ters, such as a b, is called a bioial The bioial theore gives a expressio for the powers of a bioial (a b, for each positive iteger ad all real ubers a ad b Cosider what happes whe you calculate the first few powers of a b Accordig to the distributive law of algebra, you tae the su of the products of all cobiatios of idividual ters: (a b 2 (a b(a b aa ab ba bb (a b 3 (a b(a b(a b aaa aab aba abb baa bab bba bbb (a b 4 (a b(a b(a b(a b aaaa aaab aaba aabb abaa abab abba abbb baaa baab baba babb bbaa bbab bbba bbbb Now focus o the expasio of (a b 4 (It is cocrete, ad yet it has all the features of the geeral case A typical ter of this expasio is obtaied by ultiplyig oe of the two ters fro the first factor ties oe of the two ters fro the secod factor ties oe of the two ters fro the third factor ties oe of the two ters fro the fourth factor Sice there are two possible values - a or b - for each ter selected fro oe of the four factors, there are ters i the expasio of (a b 4 Now soe ters i the expasio are lie ters ad ca be cobied Cosider all possible orderigs of three a s ad oe b, for exaple By the techiques of the Sectio 63, there are ( of the Ad each of the four occurs as a ter i the expasio of (a b 4 : aaab aaba abaa baaa By the coutative ad associative laws of algebra, each such ter equals a 3 b, so all four are lie ters Whe the lie ters are cobied, therefore, the coefficiet of a 3 b equals ( 4 1 Siilarly, the expasio of (a b 4 cotais the ( differet orderigs of two a s ad two b s, aabb abab abba baab baba bbaa, all of which equal a 2 b 2, so the coefficiet of a 2 b 2 equals ( 4 2 By a siilar aalysis, the coefficiet of ab 3 equals ( 4 3 Also, sice there is oly oe way to order four a s, the coefficiet of a 4 is 1 (which equals ( 4 0, ad sice there is oly oe way to order four b s, the coefficiet of b 4 is 1 (which equals ( 4 4 Thus, whe all of the lie ters are cobied, ( ( ( ( ( (a b 4 a 4 a 3 b a 2 b 2 ab 3 b a 4 4a 3 b 6a 2 b 2 4ab 3 b 4 The bioial theore geeralizes this forula to a arbitrary oegative iteger Theore 2 (The Bioial Theore Give ay real ubers a ad b ad ay oegative iteger, (a b 0 ( a b a ( a 1 b 1 ( ( a 2 b 2 a 1 b 1 b 2 1

3 Note that the secod expressio equals the first because ( 0 1 ad ( 1, for all oegative itegers, provided that b 0 1 ad a 1 It is istructive to see two proofs of the bioial theore: a algebraic proof ad a cobiatorial proof Both require a precise defiitio of iteger power Defiitio 1 For ay real uber a ad ay oegative iteger, the oegative iteger powers of a are defied as follows: { a 1 if 0 a a 1 if > 0 I soe atheatical cotexts, 0 0 is left udefied Defiig it to be 1, as is doe here, aes it possible to write geeral forulas such as i0 xi 1 1 x without havig to exclude values of the variables that result i the expressio 0 0 The algebraic versio of the bioial theore uses atheatical iductio ad calls upo Pascal s forula at a crucial poit However, algebraic versio of the proof of the bioial theore is rather too log to fit here Please refer to the ed of this lecture otes to see the algebraic proof Proof of the Bioial Theore (cobiatorial versio [The cobiatorial arguet used here to prove the bioial theore wors oly for 1 If we were givig oly this cobiatorial proof, we would have to prove the case 0 separately Sice we have already give a coplete algebraic proof that icludes the case 0, we do ot prove it agai here] Let a ad b be real ubers ad a iteger that is at least 1 The expressio (a b ca be expaded ito products of letters, where each letter is either a or b For each 0, 1,,, the product a b a } a {{ a a } factors b b b }{{} factors occurs as a ter i the su the sae uber of ties as there are orderigs of ( a s ad b s But this uber is (, the uber of ways to choose positios ito which to place the b s [The other positios ( will be filled by a s] Hece, whe lie ters are cobied, the coefficiet of a b i the su is Thus ( (a b a b This is what was to be proved Corollary Let be a oegative iteger The 0 0 ( 2 Proof Usig the bioial theore with a 1 ad b 1, we see that This is the desired result 2 (1 1 0 ( ( Theore 3 (Vaderode s Idetity Let,, ad r be oegative itegers with r ot exceedig either or The r ( r r 0

4 Proof Suppose that there are ites i oe set ad ites i a secod set The the total uber of ways to pic r eleets fro the uio of these sets is ( r Aother way to pic r eleets fro the uio is to pic eleets fro the secod set ad the r eleets fro the first set, where is a iteger with 0 r Because there are ( ways to choose eleets fro the secod set ad ( r ways to choose r eleets fro the first set, the product rule tells us that this ca be doe i ( r ( ways Hece, the total uber of ways to pic r eleets fro the uio also equals r ( 0 r ( We have foud two expressios for the uber of ways to pic r eleets fro the uio of a set with ites ad a set with ites Equatig the gives us Vaderode s idetity Proof (of the bioial theore, algebraic versio Suppose a ad b are real ubers We use atheatical iductio ad let the property P ( be the equatio (a b 0 BASIS STEP Whe 0, the bioial theore states that (a b ( a b ( 0 a 0 b But the left-had side is (a b 0 1 (by defiitio of power ad the right-had side is 0 0 ( 0 a 0 b ( 0 a 0 0 b 0 0 0! 0! (0 0! , sice 0! 1, a 0 1, ad b 0 1 Hece P (0 is true INDUCTIVE STEP We wat to show that for all itegers 0, if P ( is true, the P ( 1 is also true Let 0 be a iteger, ad suppose P ( is true That is, suppose We eed to show that P ( 1 is true: (a b (a b Now, by defiitio of the ( 1st power, 0 0 a b ( 1 a ( b (a b (a b (a b,

5 so by substitutio fro the iductive hypothesis, (a b (a b a b 0 a a b b a b 0 0 a b a b 1 0 We trasfor the secod suatio o the right-had side by aig the chage of variable j 1 Whe 0, the j 1 Whe, the j 1 Ad sice, the geeral ter is ( ( ( a b 1 a (j 1 b j a j b j 0 Hece the secod suatio o the right-had side above is a j b j j1 j1 But the j i this suatio is a duy variable; it ca be replaced by the letter, as log as the replaceet is ade everywhere the j occurs: ( a j b j a b 1 Substitutig bac, we get (a b 0 1 a b 1 a b 1 [The reaso for the above aeuvers was to ae the powers of a ad b agree so that we ca add the suatios together ter by ter, except for the first ad the last ters, which we ust write separately] Thus (a b ( 0 ( 0 a a b a 0 b 0 1 [( a b 1 1 [( ( ] 1 ( 1 ( a b 1 ] a b b ( 1 1 But by Pascal s forula, [( ( ] ( 1 1 Hece 1 (a b a a b b 1 1 a b, 0 which is what we eeded to show Note that ( ( 0 1 a ( b

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