Basic Combinatorics. Math 40210, Section 01 Spring Homework 7 due Monday, March 26

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1 Basic Combiatorics Math 40210, Sectio 01 Sprig 2012 Homewor 7 due Moday, March 26 Geeral iformatio: I ecourage you to tal with your colleagues about homewor problems, but your fial write-up must be your ow wor You should preset your fial homewor solutios clearly ad eatly Keep i mid that whe you write a homewor solutio, you are tryig to commuicate the solutio to someoe other tha yourself, so icomplete seteces ad persoal shorthad is ot helpful! Due to mapower issues, I will oly grade selected homewor problems, but I pla to quicly post solutios to all the problems soo after I ve collected them up Readig: Itroductio to Chapter 2 Sectio 21 Sectio 22 Sectio 23 Problems: Sectio 21: 1, 3, 6, 11 Sectio 22: 2, 3, 4, 7(d,e, 8 (just for fallig powers Sectio 23: 2, 7, 9(d,e, 10 (For problem 2, it should be clarified that the steps must always be tae i a positive directio: you ca go from (x, y, z to ay of (x + 1, y, z, (x, y + 1, z or (x, y, z + 1, but ot for example to (x 1, y, z 1

2 Solutios: 21 1a: 53 choice for iitial character, 63 for all the rest, so (53(63(63(63(63 i total 21 1b: 53 with oe character, with two characters, so (53+(53(63+(53(63(63+ (53(63(63(63 + (53(63(63(63(63 i total 21 1c: 53 with oe character, 53 with two characters (first character determies secod, with three characters (first character determies third, with four characters (first two characters determies third ad fourth, with five characters (first two characters determie fourth ad fifth, so (53 + (53 + (53(63 + (53(63 + (53(63(63 i total 21 3a: 30! 21 3b: (14(13( c: ( ( d: I the wester divisio there are i total 45 ceters from which three must be chose, 60 guards from which four must be chose, ad 75 forwards from which five must be chose, leadig to a total of ( 45 3 ( 60 4 ( : I thi that this questio has some ambiguities! Each multiple choice questio has possible aswers (you ca choose a arbitrary subset of {a, b, c, d} There are 10 such questios, so possibilities for the multiple choice part Each T/F questio has 2 possible aswers (oe is required to give a aswer, so 2 8 possibilities for the true/false part I the defiitio sectio, EITHER for each of seve terms, there is a set of 10 defiitios from which oe must be chose, leadig to a total 10 7 possibilities for this part, OR there is a set of seve terms ad 10 defiitios, ad you have to match the terms ito the defiitios, i which case there are 10 7 for this part (assumig that o defiitio wors for multiple terms I do t uderstad the last istructio EITHER it meas that you do the multiple choice, ad the do either the T/F or the defiitios, i which case there are ( (or ( possibilities i total, OR it meas that you do oe of the multiple choice, the T/F or the defiitios, i which case there are (or possibilities i total 21 11: A which is a positive iteger divisor of N has prime factorizatio p α 1 1 p α 2 2 p αm m, where each α i satisfies 0 α i i So there are i + 1 choices for each α i, leadig to ( 1 + 1( ( m + 1 distict positive iteger divisors of N 22 2: A algebraic proof is easy: (!!(! ( 1! ( 1!(! 2 ( 1! ( 1!(( 1 ( 1! ( 1 1

3 For a combiatorial proof, otice that multiplyig through by oe gets ( ( 1 1 This is exactly the committee-chair idetity from Quiz 4 (the olie solutio for which gives a coutig proof 22 3: We have ( ( m!!!(! m!( m!! 1 (! m!( m!! 1 m! ( m!(!! ( m! m!( m! ( m!(! ( ( m m m 22 4: Either he first selects the paitigs to display from the, ad the selects the m paitigs from the to display promietly, leadig to a cout of ( (, m or he first chooses the m paitigs from the to display promietly, the chooses the remaiig m from the remaiig m to also display (but less promietly, leadig to a cout of ( ( m m m Notice that this is a geeralizatio of the committee-chair idetity from Quiz 4 (which is the case m d: This is a tricy oe! Without usig ( (, I ow of o way to approach this (o algebraic or iductive proof, for example Usig ( (, we have ( 2 ( ( The right had-side is coutig the umber of ways of selectig a set of size from a set {a 1,, a, b 1,, b } of size 2, by first decidig how may of the comes from the a i s ( of them, leadig to a cout of (, forcig the remaider to comes from the bi s ( of them, leadig to a cout of ( ( But by a direct cout, we get that this is just 2 (Notice 3

4 that this is a example of the vadermode covolutio from page 142, with m l i the displayed equatio above (211 I summary: ( 2 ( e: If 0 ad m is egative, the the sum is 0 (it is empty If 0 ad m 0, the the sum is 1 That deals with 0; so from ow o we assume 1 For 1, if m < 0, the the sum is 0 (it is empty For m 0, there s just oe term, ad the sum is 1 For m, the sum is the same as if we stopped at, so it s 0, as we proved i class So the remaiig (ad most iterestig cases are 2 ad 1 m 1 A little experimetatio with Pascal s triagle suggests that i this rage: ( ( 1 ( 1 ( 1 m m m For each fixed 2, we prove this by iductio o m, with the case m 0 trivial For m > 0 we have ( ( ( 1 + ( ( 1 m m m 1 ( ( 1 ( 1 m + ( 1 m 1 m m 1 (( ( 1 ( 1 m m m 1 ( 1 ( 1 m m the secod equality usig the iductio hypothesis ad the last equality usig Pascal s idetity 22 8: I do t ow of a easy way to do this algebraically or by iductio If we allow ourselves to oly prove the result for x, y positive itegers, the there are two easy approaches: First, we ca write ( x y!!(! x y y (! x! (! ( ( x y! So, by vadermode s covolutio, we have ( x y ( ( x y!! 4 ( ( x y! ( x + y

5 But also So we have the idetity ( x + y (x + y! Here s aother, more combiatorial way: the left had side directly couts the umber of ways of taig elemets from a set of size x+y, say {a 1,, a x, b 1,, b y }, ad arragig the elemets i order Aother way to do this is to select elemets ( ruig from 0 to from {a 1,, a x } ad arrage them i order (x ways to do this, tae elemets from {b 1,, b y } ad arrage them i order (y ( ways to do this, ad the merge the two ordered sets to get a ordered list of elemets from the full set of a s ad b s ( ( ways to do this - just choosig the slots ito which the a s go So the right had side also couts the umber of ways of taig elemets from a set of size x+y ad arragig them i order 23 2: I order to reach (a, b, c from (0, 0, 0 taig steps parallel to (ad i the same direct as (1, 0, 0, (0, 1, 0 ad (0, 0, 1, we eed to tae exactly a + b + c steps a of these steps must be steps of the form (1, 0, 0, b of them must be of the form (0, 1, 0, ad c of them must be of the form (0, 0, 1 So we completely determie a path by partitioig the set {1,, a + b + c} ito three classes, class 1 of size a, class 2 of size b ad class 3 of size c, with i fallig ito class 1 idicatig that the ith step is of the form (1, 0, 0, etc There are exactly ( a+b+c a,b,c such partitios 23 7: The left had side couts the umber of ways of partitioig a set of size m +, say {a 1,, a m } {b 1,, b } ito three classes, the first of size a, the secod of size b ad the third of size c The cout is direct Aother (idirect way to cout the same thig is to first decide how may of the a i s go ito class 1 (say α of them, how may of the a i s go ito class 2 (say β of them, ad how may of the a i s go ito class 3 (say γ of them, the cout the umber of partitios that actually achieve this split (the summad of the right had side couts exactly this: if α of the a i s go ito class 1, the a α of the b i s must, etc, the sum this quatity over all possible choices of α, β ad γ (for which the oly costrait is α + β + γ m, sice all of the a i s must go ito some class This is exactly the right-had side NB: I m ot vouchig for the 100% accuracy of the umbers from here o - please let me ow if you spot errors! 23 9d: We have 6 A s, 2 K s, 2 L s, 2 S s, 1 N, ad 1 U r 3: Total choice for word type xxx, each with 1 aagram; 20 choices for word type xxy, each with 3 aagrams; 20 choices for word type xyz, each with 6 aagrams r 4: Total choice for word type xxxx, each with 1 aagram; 5 choices for word type xxxy, each with 4 aagrams; 5

6 6 choices for word type xxyy, each with 6 aagrams; 40 choices for word type xxyz, each with 12 aagrams; 15 choices for word type xyzw, each with 24 aagrams r 14: ( 14 6, 2, 2, 2, 1, e: We have 5 A s, 4 L s, ad 3 W s r 4: Total 80 14! 6!2!2!2! 2 choices for word type xxxx, each with 1 aagram; 6 choices for word type xxxy, each with 4 aagrams; 3 choices for word type xxyy, each with 6 aagrams; 3 choices for word type xxyz, each with 12 aagrams; 0 choices for word type xyzw, each with 24 aagrams r 5: Total choice for word type xxxxx, each with 1 aagram; 4 choices for word type xxxxy, each with 5 aagrams; 3 choices for word type xxxyy, each with 10 aagrams; 3 choices for word type xxxyz, each with 20 aagrams; 3 choices for word type xxyyz, each with 30 aagrams; 0 choices for all other word types r 12: ( 12 12! 5, 4, 3 5!4!3! 23 10: The extra words that ca be formed from the letters of Bobo, Mississippi are exactly the words with either oe or two B s First, the cout for words with 2 B s: 0 choices for word type xxxx; 0 choices for word type xxxy; 4 choices for word type xxyy, each with 6 aagrams; 10 choices for word type xxyz, each with 12 aagrams; 0 choices for word type xyzw This gives a cout of 144 Next, the cout for words with 1 B: 0 choices for word type xxxx; 2 choices for word type xxxy, each with 4 aagrams; 0 choices for word type xxyy, each with 6 aagrams; 16 choices for word type xxyz, each with 12 aagrams; 10 choices for word type xyzw, each with 24 aagrams This gives a cout of 440 The grad total, 584, is greater tha 267 6