Handout 10: Binomial Coefficients

Transcription

1 ENGG 244B: Discete Mathematics fo Enginees Handout 1: Binomial Coefficients Fist Tem Instucto: Anthony Man Cho So Novembe 15, Intoduction Recall that the quantity ( n, which is nown as the binomial coefficient, aises when we count the numbe of diffeent unodeed selections of elements fom an n-element gound set As it tuns out, the binomial coefficients satisfy many useful identities These identities can be poven eithe by invoing the fomula fo ( n o by using counting aguments The fome gives ise to algebaic poofs, while the latte to combinatoial poofs We shall intoduce these concepts shotly 2 Binomial Identities Let us begin with the following simple identity: Poposition 1 Fo any integes n, with n, ( ( n n (1 n Poof As mentioned in the intoduction, one can pove the above identity eithe algebaically o combinatoially This can be done by simply invoing the fomula n!!(n! The dawbac of such poof is that it does not offe any insight into the intepetation of the identity (1 Method II: Combinatoial Poof The idea of a combinatoial poof is that both sides of the identity (1 ae supposed to be two diffeent ways of solving a counting poblem, and ou tas is to explain (i what is the counting poblem and (ii how the two sides of the identity ae solving the counting poblem To poceed, let us define the counting poblem fo (1 It is simply counting the numbe of diffeent unodeed selections of elements fom an n-element gound set This numbe is, by definition, given by the LHS of (1 Now, obseve that to select elements fom an n-element gound set is the same as to select which n elements to be excluded The numbe of ways to pefom the latte is pecisely ( n n It follows that (1 holds Let us now conside some slightly moe involved examples 1

2 Theoem 1 (Pascal s Identity Fo any integes n, with 1 n 1, ( ( ( n n 1 n 1 + (2 1 Poof Using the fomulae fo ( n and n!, the RHS equals (n 1!!(n 1! + (n 1! ( 1!(n! ( (n 1! 1 ( 1!(n 1! + 1 n (n 1! ( 1!(n 1! n (n, as desied Method II: Combinatoial Poof Again, the LHS of (2 suggests that the counting poblem is simply to count the numbe of diffeent unodeed selections of elements fom an n-element gound set To facilitate ou poof, let S be the set of all unodeed selections of elements fom the n-element gound set {1, 2,, n} By definition, we have S What would be anothe way to detemine S? The RHS of the identity (2 involves the sum of two tems, which suggests that the addition pinciple is at wo Following this line of thought, let us patition S into two pats S 1 and S 2, whee S 1 all unodeed selections of elements fom {1, 2,, n} in which 1 does not appea, S 2 all unodeed selections of elements fom {1, 2,, n} in which 1 appeas To undestand the sets S 1 and S 2, let us conside a simple case whee n 4 and 2 Then, S {12, 13, 14, 23, 24, 34}, S 1 {23, 24, 34}, S 2 {12, 13, 14} Let us now etun to the geneal case It is easy to veify that S S 1 S 2 and S 1 S 2 (mae sue you ae comfotable with these statements Hence, S 1, S 2 fom a patition of S By the addition pinciple, we have S S 1 + S 2 Now, obseve that an altenative desciption of S 1 is that it contains all unodeed selections of elements fom {2,, n} (since 1 is not allowed to appea Thus, by definition, 1 S 1 2

3 Fo S 2, since 1 appeas as one of the elements in the selection, we can only select 1 moe elements fom {2, 3,, n} Thee ae ( n 1 1 ways to do this Thus, S 2 Putting eveything togethe yields the identity (2 ( n 1 1 Theoem 2 (Binomial Theoem Let n 1 be an intege Fo any eal numbes x, y, (x + y n x y n (3 The Binomial Theoem genealizes the well-nown expansion (x + y 2 x 2 + 2xy + y 2 As befoe, it can be poven by both algebaic and combinatoial means Poof We poceed by induction on n 1 Fo the base case, the RHS of (3 equals 1 x y 1 y + x x + y, 1 which is the same as the LHS of (3 Fo the inductive step, conside (x + y n+1 (x + y(x + y n (x + y 1 x y n x +1 y n + n+1 x y n x y n +1 x y n +1 (by the change of vaiable + 1 in the fist tem 1 ( + 1 ( n ( n x y n +1 + x y n x n+1 + n + 1 x n+1 + n ( n + 1 y n+1 y n+1 (by applying Pascal s identity (Theoem 1 to the fist tem n x y n +1 3

4 This completes the poof Method II: Combinatoial Poof To gain some insights into the stuctue of the expansion of (x + y n, let us fist conside a simple case whee n 3 The LHS of (3 can be witten as (x + y(x + y(x + y Obseve that the expansion of the above expession can be done by taing one of x o y fom each (x + y tem Fo instance, if we tae the bolded tems below (x + y(x + y(x + y and gathe them, we get the tem xy 2 Of couse, we get the same tem if we gathe the bolded tems below (x + y(x + y(x + y Thus, the tems in the expansion of (x + y(x + y(x + y can be obtained as follows: choice tem (x + y(x + y(x + y x 3 (x + y(x + y(x + y y 3 Moeove, (x + y 3 is simply equal to the sum of the tems in the second column of the above table Motivated by the above obsevation, let us now tacle the geneal case Conside the tem x y n in the expansion of (x + y n, whee n Such a tem can be obtained as follows: Out of the n (x + y tems, we need tae an x fom of them and a y fom the emaining n of them By definition, thee ae ( n ways to pefom this tas Hence, the tem x y n appeas in the expansion of (x + y n By summing the above ove, 1,, n, we obtain the desied identity (3 Inteestingly, the Binomial Theoem can geneate many new identities by substituting vaious values of x and y Fo instance, by setting x y 1, the identity (3 gives 2 n (4 As it tuns out, thee is a combinatoial intepetation of the above identity Conside the diffeent subsets of an n-element gound set S {1,, n} Fo instance, the following ae all the diffeent 4

5 subsets of S {1, 2, 3}: subset numbe of elements in the subset {1} 1 {2} 1 {3} 1 {1, 2} 2 {1, 3} 2 {2, 3} 2 {1, 2, 3} 3 Now, to geneate a subset of S {1,, n} with elements, we need to select elements in S to fom the subset Thus, thee ae ( n diffeent subsets of S with elements Since a subset of S can have, 1,, n elements, by the addition pinciple, the numbe of diffeent subsets of S is given by the RHS of (4 On the othe hand, a subset of S can be geneated via the following odeed sequence of actions: action numbe of choices A 1 : decide whethe 1 belongs to the subset 2 A 2 : decide whethe 2 belongs to the subset 2 A n : decide whethe n belongs to the subset 2 Hence, by the multiplication pinciple, thee ae 2 n diffeent subsets of S, which is the LHS of (4 By genealizing the poof of the Binomial Theoem, we can deive the following esult We leave the poof as an execise to the eade Theoem 3 (Multinomial Theoem Let n 1 be an intege Fo any eal numbes x 1, x 2,, x, (x 1 + x x n n 1,n 2,,n n 1 +n 2 + +n n n! n 1! n 2 n! xn 1 1 xn 2 2 xn As an application of the Multinomial Theoem, we can see that the coefficient of x 2 1 x 2x 3 in the expansion (x 1 + x 2 + x 3 4 is 4! 2!1!1! 12 This, of couse, can be veified by diectly expanding (x 1 + x 2 + x 3 4 5