Homework 3. = k 1. Let S be a set of n elements, and let a, b, c be distinct elements of S. The number of k-subsets of S is

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1 Homewor 3 Chapter 5 pp53: Chapter 6 p85: Use combiatorial reasoig to prove the idetity 3 3 Proof Let S be a set of elemets ad let a b c be distict elemets of S The umber of -subsets of S is ad the umber of -subsets of S {a b c} is 3 The the LHS is the umber of -subsets of S that cotais at least of the elemets of {a b c} Such -subsets ca be divided ito 3 types: the -subsets that cotai the elemet a; the -subsets that do ot cotai a but cotai b; ad 3 the -subsets that do ot cotai a b but cotai c The umbers -subsets of type type type 3 are 3 respectively The sum of these umbers is exactly the RHS Let be a positive iteger Prove that { 0 if odd m m m if m 0 Proof Cosider the expasio of x x x O the oe had x x i0 i x i i0 x The the coefficiet of x i the product [ i0 i x i ] [ i0 i x ] is give by i i which is exactly the LHS O the other had 0 x i0 i i x i There are oly eve terms i the expasio Thus the coefficiet of x is zero if is odd; ad the coefficiet of x is m m m if m is eve

2 3 Prove that for all real umbers α ad all itegers ad α α α Proof For < the LHS is zero because 0 The RHS is also zero because α 0 by defiitio For we divide the situatio ito the followig cases: If the LHS αα α!!!! αα α α α α!! RHS If 0 the both LHS ad RHS are both equal to α because α by defiitio If the both LHS ad RHS are both equal to zero because α 0 by defiitio 4 I a partitio of the power set P S of S { } ito symmetric chais fid a formula for the umber of chais of size size ad size respectively Solutio We claim that the umber of symmetric chais of size larger tha is / Cosider a symmetric chai of size l Sice we have A A A l A A l A l A l A l l Hece A A l This meas that each symmetric chai of legth at leat cotais exactly oe -subset ad exactly oe -subset of S Coversely sice the ay symmetric chai that cotais oe -subset ad oe -subset must cotai at leat subsets We thus coclude that the umber of symmetric chais of size larger tha is / / It is clear that the umber of symmetric chais of size is / /

3 5 Assume the expasio formula z z z < 0 Prove by iductio o the followig expasio formula z Proof For it is obviously true because 0 z 0 z z < For suppose z z < The z z i0 0 z z i i i z z Note that for 0 ad l l l l l l l l l We thus have i Cosider the partially ordered set { } whose partial order is the divisibility a Determie a chai of largest size ad a partitio of { } ito the smallest umber of atichais b Determie a atichai of largest size ad a partitio of { } ito the smallest umber of chais a A atichai partitio with four atichais: {} { } {4 6 0} {8 9 } There is oe chai of legth four: { 4 8} b A chai partitio with six chais: { 4 8} {3 6 } {5 0} {7} {9} {} There are several atichais of largest size For istace { } { } { } 3

4 7 Determie the umber of -combiatios of the multiset {4a 3b 4c 5d} Solutio Let S be the set of permutatios of the multiset M { a b c d} A A A 3 A 4 be the sets of permutatios of M such that the umber of a s are more tha 4 the umber of b s are more tha 3 the umber of c s are more tha 4 ad the umber of d s are more tha 5 respectively The S ; A A A A 4 ; A A A A A A 4 A 3 A A A 3 A A A A A A A A 4 A A 3 A A A 3 A 4 ; 0 A A A 3 A 4 0 By the iclusio-exclusio formula the aswer is give by [ ] [ ; ] Determie the umber of permutatios of { 8} i which o eve iteger is i its atural positio Solutio Let S be the set of all permutatios of { 8} The eve itegers i { 8} are 468 Let A A A 3 A 4 be the sets of permutatios that 468 are fixed respectively The S 8! A A A 3 A 4 7! A i A 6! i < 8 A A A 3 A A A 4 A A 3 A 4 A A 3 A 4 5! A A A 3 A 4 4! Thus by the iclusio-exclusio formulas the aswer is give by 8! 4 7! 6 6! 4 5! 4! 4

5 9 Usig combiatorial reasoig to prove the idetity! 0 D 0 D Proof Let S be the set of all permutatios of { } Let A be the set of all permutatios that itegers are fixed at their positios The S! ad A D The idetity follows from the disoit uio S 0 A 0 What is the umber of ways to place six o-attacig roos o the 6-by-6 boards with forbidde positios as show? a b c Recall that the umber R C of ways to place o-attacig roos o the -by- board C with forbidde positios is give by R C r C! 0 where r C is the umber of ways to place o-attacig roos o the board C I all three cases 6 a Sice r 0 r 6 r 3 r 3 8 r 4 r 5 r 6 0 the R 6 C 6! 6 5! 4! 8 3! b Sice the roo polyomial RC x 4x x 3 8x 0x 6x 3 4x 4 4x x x 54x 0x 3 44x 4 48x 5 8x 6 the r 0 r r 54 r 3 0 r 4 44 r 5 48 ad r 6 8 Thus R 6 C 6! 5! 54 4! 0 3! 44! 48! 8 0! 5

6 c Sice the roo polyomial RC x 5x 6x x 3 3x x 8x x 4x 3 9x 4 x 5 the r 0 r 8 r r 3 4 r 4 9 r 5 r 6 0 Thus R 6 C 6! 8 5! 4! 4 3! 9!! How may circular permutatios are there of the multiset {a 3b 4c 5d} so that the elemets of the same type are ot all cosecutively together? Solutio Let S be the set of all circular permutatios of M {a 3b 4c 5d} The S 3!!3!4!5! Let A A A 3 ad A 4 be the sets of circular permutatios that the type a the type b the type c ad the type d elemets are cosecutively together respectively The A! 3!4!5! A!!4!5! A 3 0!!3!5! A 4 9!!3!4! ; A A 0! 4!5! A A 3 9! 3!5! A A 4 8! 3!4! A A 3 8!!5! A A 4 7!!4! A 3 A 4 6!!3! ; A A A 3 7! 5! A A A 4 6! 4! A A 3 A 4 5! 3! A A 3 A 4 4!! ; Thus the aswer is give by A A A 3 A 4 3! 3!!!3!4!5! 3!4!5!!!4!5! 0!!3!5! 9!!3!4! 0! 4!5! 9! 3!5! 8! 3!4! 8!!5! 7!!4! 6!!3! 7! 5! 6! 4! 5! 3! 4!! 3! 6