Mathematical Thinking Problem Solving and Proofs Solution Manual Part 2

31 Pages·0.422 MB·English

by  John-P.-D'Angelo,-Douglas-B.-West

63 PartIISolutions Chapter5: CombinatorialReasoning 64
SOLUTIONS FOR PART II choose this, there are n 1 ways to choose the image of a . In general,
(cid:0) 2
for each way to choose the images of a ;...;a , there are n i ways to
1 i (cid:0)
choosetheimageofa . Bytheproductrule,thenumberofwaystoform
i 1
5. COMBINATORIAL REASONING abijectionis in(cid:0)01.n(cid:0)Ci/.
D
5.7. There arQe 12 47 1 48 ways to pick two cards from a standard 52›
(cid:1) C (cid:1)
5.1. Whenrollingndice,theprobabilityis1=2thatthesumofthenumbers card deck such that the (cid:2)rst card is a spade and the second card is not an
obtained is even. There are 6n equally likely outcomes; we show that in Ace. There are 13 ways to start with a spade. If the spade is not the Ace,
half of them the sum is even. For each of the 6n(cid:0)1 ways to roll the (cid:2)rst thenthereare47waystopickthesecondcard,sinceonenon›Acehasbeen
n 1dice,therearesixwaystorollthelastdie,andexactlythreeofthem used. If the spade is the Ace, then there are 48 ways to pick the second
(cid:0)
produceaneventotal. Thusthereare6n=2waystorollanevensum. card. Combiningthetwocasesyieldstheanswer12 47 48
(cid:1) C
Alternatively, one can name a spade for the (cid:2)rst card and a non›Ace
5.2. Probabilitiesforthesumoftworolleddice.
for the second card, eliminating the cases where the same card is named
k 2 3 4 5 6 7 8 9 10 11 12 twice. Thisyields13 48 12,whichequalsthevalueabove.
probability 1 2 3 4 5 6 5 4 3 2 1 (cid:1) (cid:0)
36 36 36 36 36 36 36 36 36 36 36 5.8. Thecoef(cid:2)cientofx4y5intheexpansionof.x y/9is 9 ,bytheBinomial
C 4
5.3. Thenumbersx and14 x areequallylikelytobethesumofthenum› Theorem. Thevalueis 94(cid:1)83(cid:1)72(cid:1)61,whichequals126. (cid:0) (cid:1)
(cid:0) (cid:1) (cid:1) (cid:1)
bers facing up on two dice. Wheneveri; j are two dice rolls that sum to x, 5.9. Probabilities in a 5›card hand. We divide the number of hands with
the numbers 7 i and 7 j are two dice rolls that sum to 14 x, since thedesiredpropertyby 52 ,thetotalnumberofpossiblehands.
1 i 6implie(cid:0)sthat1 (cid:0)7 i 6. Furthermore,thetransforma(cid:0)tionisits 5
(cid:20) (cid:20) (cid:20) (cid:0) (cid:20) a)Handshavingatleastthreecardswiththesamerank. Ifwesimply
owninverse. Thisestablishesabijectionbetweenthesetof(orderedpair) (cid:0) (cid:1)
pickthreecardsofthesamerankandthenpicktwoothercards,wemight
dicerollssummingto x andthesetof(orderedpair)dicerollssummingto
get four of the same rank; such hands would be counted four times. Thus
14 x,sothetwosetsareequallylikelywhentheindividualorderedpairs
(cid:0) we count the two cases separately. By picking a rank and an extra card,
areequallylikelyrolls.
there are 13 48 hands with four of a single rank. For the other case, we
(cid:1)
5.4. There are ml words of length l from an alphabet of size m, and in pickarank,leaveoutoneofthatrank,andpicktwocardsofotherranks,
inliDth1.elwCo1rd(cid:0), ai/leotfttehremmuesatcbhelcehttoesreins.uWsehdenatrmepoesttitoinonces.aFroeraellaocwhepdo,stihteiorne in13b(cid:1))4H(cid:1)an428dswhaayvsi.ngThautsletahsettawnoswcaerrdissw13it(cid:1)h4t8h.e1sCam2e(cid:1)4ra7/n=k.552To. thenumer›
(cid:0) (cid:1) (cid:0) (cid:1)
aQrem choicesateachofthel positions,regardlessofearlierchoices. When ator inpart (a), we could add on the hands having two but not three from
repetitions are forbidden, the number of ways to (cid:2)ll the ith position is asinglerank. Notethatwemustavoiddouble›countingthehandshaving
l 1 i, regardless of how the earlier positions were (cid:2)lled. In each case, apairfromeachoftworanks.
C (cid:0)
multiplyingthesefactorscountsthearrangements,bytheproductrule. Alternatively,wecansubtractfromthetotalthehandshavingnopair
ofcardswiththesamerank. Sincewepick(cid:2)vedifferentranksandonecard
5.5. Given n married couples, there are n.n 1/ ways to form pairs con› fromeach,thereare 13 45 ofthese,andtheansweris1 13 45= 52 . Note
sisting of one man and one woman who are n(cid:0)ot married to each other. We that about half of the5hands have no repeated ranks, s(cid:0)inc5e 13 455= 52
must choose one person of each type. Whichever type we choose (cid:2)rst, we (cid:0) (cid:1) (cid:0) (cid:1) 5 (cid:0) (cid:1) 5 D
can choose such a person in n ways. Whichever person we choose, there 5522(cid:1)4581(cid:1)4540(cid:1)4409(cid:1)3468 D:50708. (cid:0) (cid:1) (cid:0) (cid:1)
(cid:1) (cid:1) (cid:1) (cid:1)
are n (cid:0)1 person of the opposite sex other than that person’s spouse, and 5.10. In2ncoin(cid:3)ips,theprobabilityofobtainingexactlynheadsis 2n =22n.
wechooseoneofthose. n
Thereare22n listsofheadsandtails,allequallylikely. Alistwithn heads
(cid:0) (cid:1)
5.6. There are n! bijections from an n›element set A to an n›element set B. is determined by choosing locations for the n heads in the list. Thus 2n
n
Listtheelementsof A insomeorder, as a ;...;a . Thebijectionassigns outcomeshaven heads.
f 1 ng (cid:0) (cid:1)
anelementof B toeachelementof A. Furthermore,theassignedelements Whenn 10,thevalueis.19 17 13 11/=218 aftercancellation. This
D (cid:1) (cid:1) (cid:1)
are distinct. The image of a can be chosen in n ways. For each way to equalsapproximately:176197.
1
65 PartIISolutions Chapter5: CombinatorialReasoning 66
5.11. Themostcommondifferencebetweentherollsontwodiceis1. Note 5.16. The rule of product, from the rule of sum. In the set T, elements
thatspeci(cid:2)edunorderedpairsofdistinctnumbersappearintwowaysout are formed in k steps. Each element of T can be expressed as a k›tuple in
of 36, while speci(cid:2)ed pairs of equal numbers appear in only one way. In› which the ith coordinate lists the way in which the ith step is performed.
dexingtherowsbytherollofthe(cid:2)rstdieandthecolumnsbytherollonthe We are given that the ith step can be performed in r ways, no matter
i
second die yield the following table of outcomes for the difference. Each how the earlier steps are performed. We use induction on k to prove that
position in the table has probability 1=36 of occurring. Collecting those T k r .
j jD i 1 i
with a particular difference into a single event shows that the differences BasiDsstep: k 1. TheelementsofT arethewaystoperformthestep,
0,1,2,3,4,5occurwithprobabilities 6 , 10, 8 , 6 , 4 , 2 . so T Qr . D
36 36 36 36 36 36 j jD 1
5.12. The probability that three rolls of a die sum to 11 is 1=8. When the Inductionstep: k >1. Wepartition T intosets,dependingonhowthe
(cid:2)rst roll is 1;2;3;4;5, or 6, the numbers of ways to throw the other two (cid:2)rst k 1 steps are performed. The given condition implies that each set
(cid:0)
toreachatotalof11are3;4;5;6;5,or4,respectively. Sinceeachordered hassizerk. Theinductionhypothesisimpliesthatthereare ik(cid:0)11ri setsin
tripleoccurswithprobability1=63,theansweristhus27=216. thepartition. Sinceeachhassizerk,andthesetsarepairwisedDisjoint,the
ruleofsumimpliesthat T k r . Q
5.13. Theprobabilitiesforthenumberof6sinfourrolls. Whenweobtain j jD iD1 i
Wa6eepxiackctltyhek tpiomseitsi,otnhserfoeratrhee(cid:2)v6escihnoic4kesweaaycsh,faonrdthfeorreemacahintihnegr4e(cid:0)arker5ol4l(cid:0)sk. k5.172..TShuepopnolsyestohlauttino!nomfn!!CkmQ!;!bDysky!minmpeotsriyt,ivweeinmteagyearsssiusmneDtmhaDtn1anmd.
waystocompletethelist. Thusthep(cid:0)ro(cid:1)babilityis 4k 54(cid:0)k=64. SiDnce m! > 0, we haveCk! >Dn! (cid:21) m!. Using the de(cid:2)nition of factorial(cid:21), we
dividetheequationbyn!toobtain1 m!=n! k.k 1/ .n 1/. Since1
k 0 1 2 (cid:0)3(cid:1) 4 C D (cid:0) (cid:1)(cid:1)(cid:1) C
andk.k 1/ .n 1/areintegers,m!=n!mustbeaninteger. Sincem n,
instances 625 500 150 20 1 (cid:0) (cid:1)(cid:1)(cid:1) C (cid:20)
this requires m n. Now we have 2 k.k 1/ .n 1/. This requires
probability .4823 .3858 .1157 .0154 .0008 D D (cid:0) (cid:1)(cid:1)(cid:1) C
n 1 2 and k n 1, leaving only the possibility .n;m;k/ .1;1;2/.
C (cid:20) D C D
5.14. Probabilityofsumninthreeselectionsfrom[n]is.n 1/.n 2/=.2n3/. Thispossibilityisindeedasolution.
(cid:0) (cid:0)
Therearen3equallylikelyoutcomes. Thenumberofoutcomesthatsumto
5.18. Sets of six cards with at least one card in every suit. The distribu›
n isthenumberofsolutionsto x x x n inpositiveintegers.
1C 2C 3 D tions over suits can be 3111 or 2211. In the (cid:2)rst case, we pick the suit
Proof 1 (selections with repetition). The number of solutions is the
contributingthreecards,pickthethreecards,andpickonecardfromeach
number of solutions to y y y n 3 in nonnegative integers. This
1C 2C 3 D (cid:0) of the others. In the second case, we pick the two suits contributing two
equals the number of selections of n 3 elements from three types with
repetition allowed, which is n(cid:0)1 , by(cid:0)Theorem 5.31. Thus the probability cards, pick two cards from each, and pick one card each from the remain›
2 ingtwosuits. Ineachcase,theproductruleappliestothesechoices. Thus
is.n 1/.n 2/=.2n3/.
P(cid:0)roof2(cid:0)(summations). W(cid:0) he(cid:1)nx1 Di,therearen(cid:0)i(cid:0)1waystoassign theansweris 41 133 133C 42 123 2132.
wthheic(cid:2)hndaelttewromvinaelusexs3,.sTinhcuesxt2hecannutmhbeenrtoafkseolauntyiovnasliuse frino(cid:0)m12.n1(cid:0)toin(cid:0)(cid:0)1/i.(cid:0)Th1e, 5th.1e9n.uCmobuenrtoinf(cid:0)gdi(cid:1)6s(cid:0)t›din(cid:1)icgtitdniguim(cid:0)ts(cid:1).b(cid:0)eTrhs(cid:1)ebryetahreen9usmucbhernaotfudriasltinnucmt dbiegristsw.iLthetkk b1e
summandsaretheintegersfrom1ton 2(inreverseordDer),sothesumis D
(cid:0) P (sixcopiesofthesamedigit).
.n 1/.n 2/=2.
(cid:0) (cid:0) When k 2, we pick the two digits and choose a sequence with these
5.15. Thesizeoftheunionofk pairwisedisjoint(cid:2)nitesets A1;...;Ak isthe two digits, exDcluding the sequences with all of one type. Thus the answer
sum of their sizes. We use induction on k. Basis step: k 1. The one set is 10 .26 2/ 45 62 2790.
D 2 (cid:0) D (cid:1) D
A1 isalsotheunion,anditssizeisitssize. When k 6, we are arranging six elements from a set of size 10, so
Induction step: k > 1. Let B be the union of A1;...;Ak 1. By the the(cid:0) co(cid:1)untis1D0 9 8 7 6 5 151200.
iAn1d;u.c.t.i;oAnk(cid:0)h1y,paoltshoeAsiks,\jBBjDD?P. NikD(cid:0)o11wAiC.orSoilnlacreyA4k.4i1ssdtiastjeosintthafrto(cid:0)jmAke[acBhj oDf arranWgheefnoukrDof5(cid:1)t,hwe(cid:1)erep(cid:1)micak(cid:1)inth(cid:1)inegoDndeigritespeinattehdedriegmita,ipniicnkgitpsotswitoiopnoss.iTtihonesc,oaunndt
A B . Together,theseyield k A k A . is10 6 9 8 7 6 453600.
j kjCj j i 1 i D i 1j ij 2 (cid:1) (cid:1) (cid:1) (cid:1) D
D D
(cid:12) (cid:12)
(cid:12)S (cid:12) P (cid:0) (cid:1)
(cid:12) (cid:12)
67 PartIISolutions Chapter5: CombinatorialReasoning 68
Whenk 4,wemightusethreeofonedigitortwoeachoftwodigits. Thereductiontothenextinductionisalwaysdoneinthesameway. We
In the (cid:2)rst cDase, counting as in the case k 5 yields 10 6 9 8 7 can combine all the reductions into a single inductive proof (by induction
D 3 (cid:1) (cid:1) (cid:1) D
100800. In the second case, we pick the two repeats, pick two positions on k l) of the more general statement that the product of k consecutive
(cid:0) (cid:1) C
for each, and arranging two other digits in the remaining positions to get naturalnumbersstartingwithl isdivisiblebyk!(seeExercise7.21.)
10 6 4 8 7 226800. Altogetherwehave327600numberswithk 4.
2 2 2 (cid:1) (cid:1) D D m n
When k 3, we can pick three numbers and form 6›tuples from that 5.21. There are rectangles of all sizes formed using segments in a
3(cid:0)›s(cid:1)e(cid:0)t,(cid:1)(cid:0)su(cid:1)btracDtingthe6›tuplesthatdon’tuseallthenumbers. Foragiven grid with m horizo2nt2al lines and n vertical lines. Each such rectangle is
(cid:0) (cid:1)(cid:0) (cid:1)
three,thereare3 3.26 2/bad6›tuples. Thustheansweris 10 [36 3 determined, uniquely, by choosing two vertical lines and two horizontal
26 3] 64800. C (cid:0) 3 (cid:0) (cid:1) linesasboundaries.
C D (cid:0) (cid:1)
Asacheck,9C2790C64800C327600C453600C151200D999999. 5.22. Everyconvexn›gonhas n pairsofcrossingdiagonals.
4
5.20. .n5 5n3 4n/=120 is an integer whenever n is a positive integer. Proof 1 (bijection). The d(cid:0)ir(cid:1)ect proof is that every crossing pair of di›
(cid:0) C
Thenumeratorofthisfractionfactorsas.n 2/.n 1/n.n 1/.n 2/. For agonalsisdeterminedbythefourendpointsofthetwodiagonals,andthis
C C (cid:0) (cid:0)
n 1;2 , the value is 0. For n > 2, the numerator is the product of (cid:2)ve establishesabijectionfromthesetofcrossingpairsofdiagonalstotheset
2 f g
consecutivenaturalnumbers,soitsuf(cid:2)cestoshowthattheproductof(cid:2)ve of 4›tuples of vertices, since each 4›set of vertices can be matched up in
consecutivenaturalnumbersisdivisibleby120. exactlyonewaytoproduceacrossingpairofdiagonals.
Proof1(combinatorialproof). Thenumber.l 4/.l 3/.l 2/.l 1/l=5! Proof 2 (summations). We count the crossings involving diagonals
is the number of ways to choose 5 items from a seCt oflC 4 dCistinctCitems. fromonevertex. Lettheverticesbev1;...;vn inorder. Thediagonalfrom
Themoregeneralstatementthattheproductofanyk coCnsecutivepositive vn to vk is crossed by .k 1/.n k 1/ diagonals not involving vn. Thus
(cid:0) (cid:0) (cid:0)
integersisdivisiblebyk!followsbythesameargument. nk(cid:0)11.k(cid:0)1/.n(cid:0)k(cid:0)1/crossingsinvolvevn. Thisargumentisvalidforeach
Proof 2 (divisibility). A number is divisible by 120 if and only if it is verDtex,sowecansumovertheverticesanddividebythenumberoftimes
P
divisibleby5,by3,andby23. Thusitsuf(cid:2)cestoshowthateveryproduct each crossing is counted to conclude that the total number of crossings is
of (cid:2)ve consecutive integers has these factors. Since the multiples of an 4n nk(cid:0)11.k(cid:0)1/.n(cid:0)k(cid:0)1/. Weneedthesum
integer t are spaced t apart, (cid:2)ve consecutive integers contain exactly one D
numberdivisibleby5andatleastonedivisibleby3. Theyalsocontainat P nk(cid:0)11.k(cid:0)1/.n(cid:0)k(cid:0)1/D 3n
leasttwonumbersdivisibleby2,andoneoftheseisdivisibleby4. Hence D
P (cid:0) (cid:1)
thereareatleastthreepowersof2intheproduct. Notethatbeingdivisible (also needed in an inductive proof). One can compute this by writing the
by2andby4doesnotimplythatanumberisdivisibleby8. summand as a polynomial and applying Propositions 4.7 and 4.16. Exer›
Proof 3 (induction and divisibility). We prove by induction on n that cise9.11evaluatesamoregeneralsumbyacombinatorialargument.
.n 2/.n 1/n.n 1/.n 2/ is divisible by 120. The product is 0 when
n C1. FoCr the ind(cid:0)uction(cid:0)step, suppose that the claim holds when n m. 5.23. Multiplicitiesofpokerhands.
D D
To show that .m 3/.m 2/.m 1/m.m 1/ is divisible by 120, we show a)Onepair(twocardsofequalrankandnoothersofequalrank). This
thatthisminus.mC 2/.mC 1/m.Cm 1/.m(cid:0) 2/isdivisibleby120. Withthe occurs in 13 4 12 43 ways: pick the special rank, pick two cards from it,
C C (cid:0) (cid:0) 1 2 3
inductionhypothesis,weconcludethattheclaimholdswhenn m 1. pickthethreeotherranks,pickonecardeachfromthoseranks.
D C (cid:0) (cid:1)(cid:0) (cid:1)(cid:0) (cid:1)
The desired difference simpli(cid:2)esto5.m 2/.m 1/m.m 1/. Thus it b)Fullhouse(twocardsofequalrankandthreecardsofanotherrank).
suf(cid:2)cestoshowthattheproductoffourconseCcutiveiCntegersi(cid:0)sdivisibleby Thisoccursin13 12 4 4 ways: pickthetworanks(ordermattersbecause
(cid:1) 2 3
24. Wecouldapplyadivisibilityanalysis(asinProof2)orprovethisstate› thechosenranksaredistinguishedbythenumberofcardstheycontribute),
(cid:0) (cid:1)(cid:0) (cid:1)
mentitselfbyinduction. Theinductionstepwouldreducetothestatement pick2cardsfromthe(cid:2)rstrank,pickthreecardsfromthesecondrank.
that the product of three consecutive integers is divisible by 6. We could c)Straight(cid:3)ush((cid:2)vecardsinsequencefromthesamesuit). Astraight
prove this using divisibility or induction, reducing to the statement that (cid:3)ushisdeterminedbychoosingasuitandchoosingtherankwherethe5›
theproductoftwoconsecutiveintegersisdivisibleby2. Ifweeverswitch card sequence starts. There are four suits and 10 starting values (10 J Q
tothedivisibilityapproach,thenweuseanargumentlikeProof2. KAisthehighest),sothereare40suchhands.
69 PartIISolutions Chapter5: CombinatorialReasoning 70
5.24. Bridgedistributions. Theprobabilityofeachdistibutionisthenum› 5.26. Thebinomialtheorembyinductiononn. Forthebasisstep,wehave
ber of such hands divided by the total number of 13›card hands, 52 . For .x y/0 1 0 x0y0. Now suppose that the expansion formula holds
13 C D D 0
each distribution, we list the number of hands and the rank. To count when the exponent isn. Weconsider thesummation when the parameter
the hands, we (cid:2)rst assign the multiplicities to suits, then we ch(cid:0)oos(cid:1)e the is nC1. The ind(cid:0)u(cid:1)ction hypothesis tells us that .x C y/n D nk 0 nk xkyn(cid:0)k.
speci(cid:2)ednumberofcardsfromeachsuit. Sincewewanttheexpansionfor.xCy/nC1,wemultiplybothPsideDsb(cid:0)y(cid:1).xCy/.
Thenumberofwaystoassignthemultiplicitiesdependsonhowmany To simplify the resulting expression, we want want to combine the
times each multiplicity occurs. With four distinct multiplicities, there are termswheretheexponentson x agreeandon y agree. Therefore,weshift
24waystoassignthemtosuits. Whenthreenumbersarise(onerepeated), theindexinthe(cid:2)rstsummation. WethenusePascal’sFormulatocombine
asin5440,thereare12ways. Withthreesuitsofthesamemultiplicity,as correspondingtermsinthetwosummations. Forthetermsthatdon’tpair
in4333,thereare4ways(thisiswhythisdistributionrankssolow). Since up, we have n 1 nC1 and n 1 nC1 , so these become the top
n D D n 1 0 D D 0
13isodd,therecannotbefoursuitswiththesamemultiplicityortwopairs andbottomtermsofthedCesiredsummation. Thefullcomputationis
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
withequalmultiplicity.
n
n
distrib. #hands rank distrib. #hands rank .x y/n 1 .x y/ xkyn k
C (cid:0)
444344343312 1244(cid:0)1(cid:0)431144(cid:1)332(cid:1)(cid:0)(cid:0)311333311(cid:1)(cid:1)3(cid:0)3123(cid:1) 5110((12(013...560%%%))) 777456210000 22441(cid:0)(cid:0)21177331(cid:1)(cid:1)73(cid:0)(cid:0)1145331(cid:1)(cid:1)63(cid:0)(cid:0)1121331(cid:1)(cid:1)03(cid:0)(cid:0)1120033(cid:1)(cid:1) 23310t9i(e(.00(0.035.616%1%%))) C DD nC nk XkxDk0C(cid:18)1ykn(cid:19)(cid:0)k C n nk xkynC1(cid:0)k
55343222 1122(cid:0)11(cid:0)5533(cid:1)(cid:0)(cid:1)113433(cid:0)(cid:1)2(cid:0)1(cid:1)23123(cid:1)2 24((1150..56%%)) 88232111 1122(cid:0)(cid:0)118833(cid:1)(cid:1)(cid:0)(cid:0)112333(cid:1)(cid:1)(cid:0)2(cid:0)11311(cid:1)3(cid:1)2 2212((00..1192%%)) XnkDC01(cid:18) (cid:19)n xlyn(cid:0).l(cid:0)1Xk/D0(cid:18)n(cid:19) n xkynC1(cid:0)k
5431 24 13 13 13 13 3(12.9%) 8320 24 13 13 13 13 23tie(0.12%) D l 1 C k
5440 1(cid:0)2(cid:0)51(cid:1)53(cid:0)(cid:1)(cid:0)41(cid:1)43(cid:0)(cid:1)(cid:0)23(cid:1)1(cid:0)0(cid:1)32(cid:1) 13(1.2%) 8410 24(cid:0)1(cid:0)883(cid:1)(cid:0)(cid:1)1(cid:0)343(cid:1)(cid:0)(cid:1)1(cid:0)213(cid:1)(cid:0)(cid:1)1003(cid:1) 26(0.045%) XlD1 (cid:18) (cid:0) n(cid:19) n XkD0n(cid:18) (cid:19)
55553201 1122(cid:0)(cid:0)115533(cid:1)(cid:1)(cid:0)22(cid:0)1123(cid:1)33(cid:1)(cid:0)(cid:0)111033(cid:1)(cid:1) 194((30..29%%)) 8500 1(cid:0)2(cid:0)(cid:1)183(cid:0)(cid:1)(cid:0)(cid:1)153(cid:0)(cid:1)(cid:0)(cid:1)103(cid:0)(cid:1)2(cid:1) 31(0.0031%) D xnC1C XkD1(cid:20)(cid:18)k(cid:0)1(cid:19)C(cid:18)k(cid:19)(cid:21)xkynC1(cid:0)k!CynC1
6322 12(cid:0)163(cid:1) (cid:0)133 (cid:1)(cid:0)123 (cid:1)2 6(5.6%) 9211 12 193 123 113 2 27(0.018%) nC1 nC1 xkyn 1 k
6331 12 13 13 2 13 8(3.4%) 9220 12 13 13 2 13 29(.0082%) D k C (cid:0)
6421 24 1(cid:0)36(cid:1)1(cid:0)33(cid:1)1(cid:0)3 1(cid:1)13 7(4.7%) 9310 24 1(cid:0)39(cid:1)1(cid:0)32(cid:1)1(cid:0)3 0(cid:1)13 28(0.010%) XkD0(cid:18) (cid:19)
(cid:0)6 (cid:1)(cid:0)4 (cid:1)2(cid:0) (cid:1)1 (cid:0)9 (cid:1)(cid:0)3 (cid:1)1(cid:0) (cid:1)0
6430 24 13 13 13 13 12(1.3%) 9400 12 13 13 13 2 33(.00097%) This proof is a direct generalization of the computation that obtains the
6511 1(cid:0)261(cid:1)3(cid:0)41(cid:1)3(cid:0)31(cid:1)3(cid:0)20(cid:1) 15(0.71%) 10,1,1,1 (cid:0)4(cid:1)91(cid:0)3 (cid:1)41(cid:0)3 3(cid:1)0(cid:0) (cid:1) 34(.00040%) expansionof.x Cy/3 fromtheexpansionof.x Cy/2,forexample.
(cid:0) (cid:1)6(cid:0) (cid:1)5(cid:0) (cid:1)1(cid:0) (cid:1) (cid:0) 1(cid:1)0(cid:0) 1(cid:1)(cid:0) (cid:1)
6520 24 13 13 13 13 16(0.65%) 10,2,1,0 24 13 13 13 13 32(.0011%) 5.27. Equality of numbers of even and odd subsets, by the Binomial The›
(cid:0)6 (cid:1)(cid:0)5 (cid:1)(cid:0)2 (cid:1)0 10(cid:0) 2(cid:1)(cid:0) 1(cid:1) 0
6610 12 13 2 13 13 25(0.072%) 10,3,0,0 12 13 13 13 2 35(.00016%) orem. Let A be the set of even›sized subsets, and let B be the set of
77232221 24(cid:0)14(cid:0)73(cid:0)(cid:1)61(cid:0)7(cid:1)313(cid:1)3(cid:0)(cid:0)(cid:1)1(cid:0)1231(cid:1)2(cid:1)3(cid:0)3(cid:1)(cid:0)01(cid:1)13(cid:1) 1171((01.5.91%%)) 1111,,12,,10,,00 11(cid:0)22(cid:0)(cid:0)11111(cid:1)03131(cid:0)(cid:1)(cid:1)(cid:0)(cid:0)11(cid:1)31233(cid:0)(cid:1)(cid:1)(cid:0)2(cid:0)1(cid:1)0031(cid:0)0(cid:1)3(cid:1)2(cid:1) 3367((..000000002151%%)) osidzde›ski.zneTdhsuusbjsAejtsD.nThie(cid:21)0bi2nniomaniadljcBojefD(cid:2)cieni(cid:21)t0(cid:0)nk2(cid:1)inCc1o.unWtsetwhaenstutboseshtsowoftnhaotf
77431310 11(cid:0)22(cid:0)11(cid:1)7733(cid:0)(cid:1)(cid:0)11(cid:1)3433(cid:0)(cid:1)2(cid:0)1(cid:1)131(cid:0)03(cid:1)2(cid:1) 2108((00..2369%%)) 1123,,10,,00,,00 12(cid:0)(cid:0)1132(cid:1)(cid:1)(cid:0)(cid:0)4113(cid:1)(cid:1)(cid:0)(cid:0)103(cid:1)(cid:1)2 3398.(6.0(cid:2)001000(cid:0)0130%%)/ Pcani(cid:21)0cIon(cid:0)u2itn(cid:1)ht(cid:0)ethPeexips(cid:21)ae0tn(cid:0)ss2ioiCofP1ne(cid:1)v.De1n0C(cid:0)s.i(cid:1)xz/enpDositivenklyxPkanfrdomt(cid:0)heths(cid:1)eetBsionfoomddiaslizTehneoergeamti,vewlye
P(cid:0) (cid:1)
5.25. I(cid:0)nd(cid:1)u(cid:0)ct(cid:1)i(cid:0)ve(cid:1)proof of n n! . The formula holds for n 0 under bysettingx D(cid:0)1. Thevalueofthesumbecomesthetotalnumberofeven
k D k!.n k/! D subsets minus the total number of odd subsets. Setting x 1 on both
the convention that the (cid:0)(cid:147)f(cid:1)actorial(cid:0)(cid:148) of a negative number is in(cid:2)nite. For sidesyields n . 1/k .1 1/n 0. ThusthenumberofsDubs(cid:0)etsofeach
n >1,weapplyPascal’sFormulaandtheinductionhypothesistoobtain k (cid:0) D (cid:0) D
typeisthesame. Whenn 0,thereisoneevensubsetandnooddsubsets,
nk D n(cid:0)k1 C nk(cid:0)11 = k!..nn(cid:0)11/!k/! C .k .1n/(cid:0)!.1n/!k/! = n(cid:0)nkk!.nn!k/! C nkk!.nn!k/! D k!.nn!k/!. whichmotivPate(cid:0)s(cid:1)theconveDntionincombinatoricsthat00 D1.
(cid:0) (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) (cid:0)
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
71 PartIISolutions Chapter5: CombinatorialReasoning 72
5.28. x1C(cid:1)(cid:1)(cid:1)Cxk (cid:20)n has nCkk solutionsinnonnegativeintegers. times,becauseinanysubsetofthepairswecanswitchthechoiceofwhich
for ePacrhoomf,1t(hseumnummabtieorn)i.s(cid:0)WmeCk(cid:1)hk(cid:0)1a1ve. cSouunmtmedinsgoluthtiiosnosvteorx01C(cid:20)(cid:1)m(cid:1)(cid:1)C(cid:20)xnk Danmd memPbreoroofft4h.eLpeatiranwbaesitnhethdeeosirriegdinvaallsueele.cTtehderseeta.re 2n (cid:0)1 ways to pair
applyingtheSummationIde(cid:0)nt(cid:0)ity(cid:1)yieldsatotalof nCkk solutions. person 1 with another person. Every such choice can be extended to a
aTsinhrdeedPsnusuromomluobntfeiior2sno(sftnrCc.skao.oknrlC1rus/1eft/o(cid:0)s1iorp1nmosnDadttoitonotCnkhkn)e.o.tnIrnnatenrgosafdotuirvcmeeeiadnnteeeqgxuetarrtas(cid:0)ioovlnua(cid:1)trwiioaitnbhslektoxCkPC11ik.vDCa11TrxhiiaeDbdleens›. apdM1aeu(cid:2)rDltntiiti1ptioi)lo.ynninBicngyatnboinybpd1eauidicrnotsintobhenyeionpfnoaarinnmr(cid:0)in1(ogwrinuaDityp1es2r.tiahT=te.ih2vruneensms!a/uanbyiDnisetilin.dt2gusnt2ti(cid:0)honne1(cid:0))f/,oa2arnmn(cid:0)p1DeuwolapQitlcehinl,Daa1wi0.m2hDieicd(cid:0)1h. (1bo/yr.
C (cid:0)
5.29. Theequ(cid:0)ationx1C(cid:1) (cid:1)(cid:1)(cid:0)(cid:1)Cx(cid:1)k Dn has nk(cid:0)11 solutionsinpositiveintegers. a5r.3r2ay..AOcfotmhebsien,antodroitaslhparvoeofththQeastanm2eDro2w2nanCdnc.oAlurmrannignedne2x.dFotosrienaachncnhboyicne
Proof1(transformation). Solutions(cid:0)in(cid:0)p(cid:1)ositiveintegersto ik 1xi Dn ofdistinctindicesi and j,therearetwo(cid:0)d(cid:1)ots: positions.i; j/and.j;i/.
correspond to solutions in nonnegative integers to k y n Dk, since
i 1 i D P(cid:0)
subtracting1turnsapositiveintegerintoanonnegatiDveinteger, andthis 5.33. Summingthecubes.
is invertible. The model of selections with repetitionPthen says that there a) m3 6 m 6 m m. Beginning with the right side, we compute
D 3 C 2 C
are .n(cid:0)k/Ck(cid:0)1 n(cid:0)1 suchsolutions. m.m 1/.m 2/ 3m.m 1/ m m3 3m2 2m 3m2 3m m m3.
(cid:0)Prok(cid:0)o1f2(cid:1)(dDire(cid:0)kc(cid:0)t1b(cid:1)ijection). Inthemodelofdotsandbars,eachxi counts b(cid:0)) in 1(cid:0)i3(cid:0)D(cid:1)C.n.n2C(cid:0)1//(cid:1)2(cid:0). UsiCngpDart(a(cid:0))andtCheideCntity (cid:0)in 0 Cki DDnkC11 ,
the dots between successive bars. If there is at least one dot for each xi, weconcludDethat D C
P P (cid:0) (cid:1) (cid:0) (cid:1)
then the bars must be placed in distinct places between dots. Thus there
are n 1 places in which bars can go, and we choose k 1 of them to n
determ(cid:0)ineasolutioninpositiveintegers. (cid:0) i3 6 nC1 6 nC1 nC1 .n 1/n 6 .n 1/.n 2/ 6.n 1/ 1
D 4 C 3 C 2 D C 24 (cid:0) (cid:0) C 6 (cid:0) C 2
t5h.3e0n.thPerosoufmbyisin0ducti1on,wohninchthisa1t ifkinD0 1ki aDndnk0CC11ifkfo>ra1l.lOn;nkt2heNr.igIhfntsDid1e,, XiD0 .n(cid:0)C(cid:1)1/n (cid:0).n2 (cid:1) 3n(cid:0) 2(cid:1)/ .4n 4/(cid:20) 2 n.nC1/ 2: (cid:21)
k C k P D(cid:0) (cid:1) (cid:0) (cid:1) D 4 (cid:0) C C (cid:0) C D 2
2 is1ifk 1and0ifk >1. Hencetheidentityholdswhenn 1. For (cid:20) (cid:21)
k 1 D (cid:0) (cid:1) (cid:0) (cid:1) D (cid:2) (cid:3)
thCeinductionstep,supposethattheidentityholdswhenn m 1(andall c) Combinatorial proof of part (a) by counting a set in two ways. Con›
(cid:0)k). F(cid:1)orn Dm,wethenhave in 0 ki D nk C in(cid:0)01 ki D nk DC k(cid:0)n1 D nkC11 , sider the 3›tuples of numbers, where each number is in [m]. There are m
where we have used the inductDion hypothesis aDnd then PascalC’s FormCula triplesinwhichweuseonlyonetypeofnumber. There are6 m triplesin
P (cid:0) (cid:1) (cid:0) (cid:1) P (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) 2
(thebinomialcoef(cid:2)cientrecurrence)inthelasttwosteps. whichweusetwodistinctnumbers,becausetherearetwowaystopickthe
(cid:0) (cid:1)
5.31. Thereare.2n/!=.2nn!/waystopartition2n distinctpeopleintopairs. valueusedonlyonceafterpickingthepairused,andtherearethreeposi›
tions in which the value used only once can be placed. Finally, there are
Proof1. Thereare.2n/!waystoputthepeopleinalinearorder,and m
ways to pick three values, and there are 6 orders in which the chosen
when we do this we can form pairs from the (cid:2)rst two, the next two, and 3
so on. However, this counts each partition into pairs exactly 2nn! times, (cid:0)va(cid:1)lues can be written. Thus the right side of part (a) counts the triples,
groupedbyhowmanydistinctvaluesareused.
because wedon’tcare which person ofapair iswritten(cid:2)rst, andwedon’t
carewhatorderthepairsarewrittenin. 5.34. Thenumber ofcubes ofallpositiveintegersizesformed byann byn
Proof 2. There are 22n 2n2(cid:0)2 (cid:1)(cid:1)(cid:1) 22 ways to pick pairs successively, by n assembly of unit cubes is nC1 2. In each of the three directions, the
since the number of ways to pick the next pair doesn’t depend on how the 2
(cid:0) (cid:1)(cid:0) (cid:1) (cid:0) (cid:1) coordinatesofacubewithsidesoflengthn 1 i canbechosenini ways.
previous pairs were chosen. Divide this by n!, since all n! orderings of the Hencethedesiredvalueis n (cid:0)i3 f(cid:1)orcubesCwit(cid:0)hpositiveintegersizes.
pairsyieldthesamepartition. Regroupingthefactorsyields.2n/!=.2nn!/. i 1
ppaerirmsP.urtoTaothifeo3nr.,eiTmnhanei!rnepinoasgrseinb(cid:0)2lnpen(cid:1)ewowpaalyeyss.atrToehpipiscakcirrneedapteewosiptelheactthohebpmearatasicstciiogonnrdeiidnnttgootpdoaisisrtoisnm2cent I.ndenUt1i/tsnyin,lgeabinvDien1soim31[iD.anl6con1eC4/f1(cid:2).ncCiPen62tD/snC,31i34C.Dn 6n(cid:0)C213i1(cid:1)/.CE62x(cid:0)]2,it(cid:1)rwaChcti(cid:0)ci1inh(cid:1)g.eBqthuyeatlchsoe.mnSmumo1nm/nfaa=tc4it.oonr
C P 4 (cid:0)(cid:0) (cid:1) (cid:0) (cid:0) C(cid:1) (cid:0)(cid:0) (cid:1)C C
73 PartIISolutions Chapter5: CombinatorialReasoning 74
5.35. Trackmeetwithkn contestants. can complete the triple by choosing one smaller element and one larger
a) k 1 divides kn 1. In each group race, k 1 of the k runners are element,inanyway. Sincetherearei 1smallerelementsandn i larger
(cid:0) (cid:0) (cid:0) (cid:0) (cid:0)
eliminated. At the end, kn 1 of the kn contestants have been eliminated elements, the product rule says that the number of triples with i as the
(cid:0)
intheraces. Thusk 1divideskn 1. middle element. Since every triple has some middle element in [n], the
b) The meet has(cid:0).kn 1/=.k 1(cid:0)/ races, since kn 1 runners lose, and ruleofsumthensaysthatthesumontheleftequals n .
(cid:0) (cid:0) (cid:0) 3
k 1ofthemloseineachraceandareeliminated.
w5.(cid:0)h3e6t.hCeromthbeinealetomreianltp2rnooifsoifnc2lnnudDed2. 2Innf(cid:0)(cid:0)1s1o., Cthoeunntnthe1ne›sluembseenttssomf[u2snt]bbye se5ol.e4mm2e.ennPutsmikDfb0reo(cid:0)rmmio(cid:1)(cid:0)fake(cid:0)nsliee(cid:1)mtDoefn(cid:0)mtmsCkCfnr(cid:1)onm. dTtihshteein(cid:2)rcritsgthemltemsaindedenttcsho.euEnrevtmsertay(cid:0)hines(cid:1)iunwcghaeyclshemotoiceecnhtssoeoflsreoecmtks
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0)
addedfromtheremaining2n 1elementstocompletetheset. Ifnot,then thelastn. Thenumberofwaystoselecti elementsfromthe(cid:2)rstm is m ,
(cid:0) i
tno(cid:0)co1mmpolerteeetlheemseenttslemftuosutt.beomittedfromtheremaining2n(cid:0)1elements Hanednctheethneumnubmerboefrwoafysselteoctsieolnecstikn(cid:0)wihfircohmththereelaasrteniieslekm(cid:0)nien,itnsdcehpoesnendefnr(cid:0)otlmy(cid:1).
5E.a3c7h. sIifden;cko;ulnatsrethneatsuertaclonnsuimstbinergsowfaitlhlplo(cid:20)ssikbl(cid:20)ewn,aythsetnofonkrmkl ,fDromnl ank(cid:0)(cid:0)sllet. the(cid:2)rstm is mi k(cid:0)ni . Summingoveri countsallthese(cid:0)lec(cid:1)tions.
of n people, a committee of size k and a subcommittee of(cid:0)si(cid:1)z(cid:0)e(cid:1)l wi(cid:0)th(cid:1)i(cid:0)n it(cid:1). 5.43. in m (cid:0)mrC(cid:1)i(cid:0) n(cid:0)s(cid:1)i D mrCsnC11 . TherightsidecountsselectionsofrCsC1
elementsD((cid:0)without repetitioCnC) from m n 1 distinct elements. With the
Theleftsidegroupsthissetaccordingtothechoiceofthecommittee(pick P (cid:0) (cid:1)(cid:0) (cid:1) (cid:0) (cid:1) C C
m n 1elementslistedinincreasingorder,ther 1thsmallestelement
the committee (cid:2)rst and then pick the subcommittee from it). The right C C C
must occur at some value, say m i 1 for m i n. For each such
side groups it according to the choice of the subcommittee (pick the sub› C C (cid:0) (cid:20) (cid:20)
choice of i, the corresponding term on the left counts the ways to choose
committee (cid:2)rst and then (cid:2)ll out the remainder of the committee from the
ther smallerelementsandthes largerelements.
remaining n l people. Similarly, both sides count the ternary sequences
(cid:0) Equivalently,therightsidecountslatticewalksfromtheorigintothe
oflengthn withl twosandk l ones.
(cid:0) point.r s 1;m n r s/. Theleftsidecountstheseaccordingtothe
t5ehl.e3em8n.eonAntecmko,mpbtbeyicnasauutbsoesreiaatsnlpyorfso[uonbf]s.theTtahtoePfrtenkhDea1r2neku(cid:0)2m1k(cid:0)bD1er2ssunc(cid:0)bhe1lso.uwTbhskeetcrsaignhhabtvesinicdgheolcasoerungnetstost h5.e4ig4h.taCtikw0hCmicChkk(cid:0)tiih(cid:0)e1CstneC(cid:0)pii(cid:0)f1r(cid:0)oDm xmCDnCkrk(cid:0)t1o.xTDhreCrig1hitssmidaedceo.untsselectionsof
k elementDs with(cid:0)repetition allowed from m n types. The left side groups
accompanyit. Hencetheleftsidealsocountsthenonemptysubsetsof[n], P (cid:0) (cid:1)(cid:0) (cid:1) (cid:0) (cid:1) C
thesebyhowmany,sayi,areselectedfromthe(cid:2)rstm types. Thisidentity
groupedbythelargestelementinthesubset.
istheselections›with›repetitionversionoftheVandermondeconvolution.
5ri.g3h9t.cAoucnotmsbthineawtoaryisaltopcrhooofostehaatcomnkDm0ikttenkeDwitnh2nc(cid:0)h1a.irTfhroemfoarmseutlaofonnptehoe› 5.45. A [n] B [n]jA\ Bj D n4n(cid:0)1 . Let S be the set of triples .x;A;B/
P (cid:0) (cid:1) suchthat(cid:18)A;B (cid:18)[n]andx A B. Foreachchoiceof A;B [n],thereare
ple,by(cid:2)rstchoosingthechairandthenchoosingasubsetoftheremaining P P(cid:18) 2 \ (cid:18)
A B waystochoosex tocompleteatriplein S,sothesumcounts S. For
peopletocompletethecommittee. j \ j
Thesumontherightcountsthesamesetbythesizeofthecommittee. eachofthen waystochooseanelementx 2[n],thereare2n(cid:0)1 choicesof A
To form a committee with k people, choose the k from the set of n people, containing x and2n(cid:0)1 choicesof B containing x, soalso jSj D n4n(cid:0)1. Since
bothsidesoftheformulacount S,theyareequal.
andthenchoosethechairfromthecommittee.
5p.a4i0rs. oAfecloemmbeinntastforroimaltphreoosfetth[na]t. WinhD(cid:0)1e1ni tDhe2nla.rgTehreofrtighhettwsiodeelceomuenntstsailsl e5x.4p6a.nsPioSn(cid:18)[onf]Qthii2sSp1r=oidDuctnhCas1.2nCtoenrmsisd,esrinthceetphreodituhctfaQctinoDr1.c1anCc1o=nit/r.ibTuhtee
P (cid:0) (cid:1) 1or1=i. ForeachS [n],thereisaterm 1 1=i. Thusthedesired
i 1,therearei waystocompletethepair. Sinceeverypairhasitslarger (cid:18) i=S i S
elCementintheset 2;...;n ,thesumcountsallthepairs. sumequals in 1.1C1=i/D in 1 iCi1 DnQC12. Q2
f g D D
c5o.4u1n.tsAalclotmribpilneastoofrieallempreonotfstchhaotseninDfr1o.im(cid:0)th1/e.nse(cid:0)t [in/].DW3ne.caTnhgerroiugphtthseidsee 5n.47.m,Ifanfmd:QtNhis!faNilsiswdhee(cid:2)nnneQd bmy fm.1n./TDheremkDa0renk ,nthseunbsfemts.no/fD[n]2nhawvhineng
P (cid:0) (cid:1) (cid:20) D C P (cid:0) (cid:1)k
by the index of the middle element. When the middle element is i, we sizek. Whenthisissummedoverallk with0 k n,wehavecountedall
(cid:20) (cid:20)(cid:0) (cid:1)
75 PartIISolutions Chapter5: CombinatorialReasoning 76
subsetsof[n],andthereare2n suchsubsets. Whenk >n,wehave n 0, 5.52. Forn >1,thenumberofevenpermutationsof[n]equalsthenumber
k D
sincetherearenosubsetsof[n]withmorethann elements. Thus of odd permutations of [n]. Every transposition changes the parity of a
(cid:0) (cid:1)
m n permutation. Thusinterchanging the(cid:2)rsttwoelementsinthewordform
n n
f .n/ 2n when n m of the permutation maps the set of even permutations into the set of odd
m
D k D k D (cid:20)
XkDm0(cid:18) (cid:19) XkDn0(cid:18) (cid:19) permutations. Themapisinjectiveandsurjective,sincewecanobtainan
n n even permutation mapping to a particular odd permutation uniquely by
f .n/ < 2n when n >m
m D k k D transposing the (cid:2)rst two elements of the odd permutation. Thus the map
XkD0(cid:18) (cid:19) XkD0(cid:18) (cid:19) isabijection,andthetwosetshasthesamesize.
5.48. Therearen!chainsofdistinctsubsets A0;A1;...;An of[n]suchthat
dAi0st(cid:26)inAct1s(cid:26)iz(cid:1)e(cid:1)s(cid:1)0(cid:26);1A;n....D;ins.tinTchtunsesesaocfhthseetsaudbdsestasnreeqluemireesntthteomthteohpraevveiothues 5F.o5r3i.frEovmer1y ptoernm(cid:0)ut1atsiuocncecsasnivbeelys,opreterdforumsinthgeattrmanossptons(cid:0)iti1ontrtahnastpsowsiitticohness.
set, and there are n! orders in which this can be done. If repetitions are theelementinpositioni withtheelementi,unlessi isalreadyinposition
i. After step i, all of 1;...;i are in their desired positions, and hence the
allowed,theneachelementcanbeaddedatanystepornotatall,sointhis
casethereare.n 1/n suchchains. later positions are a permutation of fi C1;...;ng. Thus after n(cid:0)1 steps,
C theonlypositionleftforelementn ispositionn.
5.49. Parityandinverseforpermutations. Theparityofthepermutationis
The permutation nn 1 1 needs at least n=2 transpositions for
theparityofthenumberofpairsi; j suchthati < j and j appearsearlier (cid:0) (cid:1)(cid:1)(cid:1) b c
sorting. The number of elements not in their desired positions is n (if n
than i (these are inversions). To determine the inverse of permutation
is even) or n 1 (if n is odd). Since each transposition places at most two
a ;...;a , mapping i to a , we write the pairs .a ;i/ in increasing order of (cid:0)
1 n i i elementsintotheirdesiredpositions,thenumberoftranspositionsneeded
a andthenkeeponlythesecondcoordinate.
i isatleasthalfthenumberofelementsoutofplace,or n=2 .
b c
permutation #inversions parity inverse
987654321 36 even 987654321 5.54. Theminimumnumberofadjacenttranspositionsneededtotransform
135792468 10 even 162738495 apermutation f totheidentitypermutationisthenumberofinversionsin
259148637 15 odd 418527963 f (pairs i; j with i < j but i before j). Over all permutations of [n], the
maximumofthisis n . If f isnotalreadysorted,thensomeadjacentpair
2
5.50. CorrectingthelabelsApples, Oranges, andApples/Oranges. Weare ofelementsisaninversion(n 1increasescanonlyoccurif f is12 n).
(cid:0) (cid:1) (cid:0) (cid:1)(cid:1)(cid:1)
told that all labels are wrong, so a permutation with no (cid:2)xed point has Transposingsuchanadjacentpairreducesthenumberofinversionsby1.
beenappliedtothelabels. Thecorrectactionistoselectonepieceoffruit Hence we can transform f to a permutation with no inversions at most
fromthe binlabeled Apples/Oranges. Since thelabel iswrong, allfruitin by t transpositions, where t is the number of inversions in f. The only
thatbinwillbethesametypeastheselectedpiece;callthistype B. Aper› permutationwithnoinversionsistheidentity,sothissorts f.
mutationof[3]withno(cid:2)xedpointisasinglecycle. SincetheApple/Orange
Transposing two adjacent elementsreduces thenumber ofinversions
labelmovedtoType B,itmustbethattheType B labelmovedtotheother byatmost1,soatleastt transpositionsareneeded.
purebin,andthatlabelmovedtotheApple/Orangebin.
Overpermutationsof[n],thenumberoftranspositionsneededismax›
5.51. When the Drummer Problem is changed by having three drummers imized by maximizing the number of inversions. This occurs when f is
who rotate, the (cid:2)nal drummer cannot be determined from the initial per› nn 1 1,whereitequals n .
(cid:0) (cid:1)(cid:1)(cid:1) 2
mutation. It suf(cid:2)ces to present a permutation so that different ways of
(cid:0) (cid:1)
reachingtheidentitypermutationattheendleadtodifferentanswersfor 5.55. Bijection from the set A of permutations of [n] to the set B of n›
the (cid:2)nal drummer. When there are three couples, we might start with tuples .b1;...;bn/ such that 1 (cid:20) bi (cid:20) i for each i. Each permutation a D
321. If there is only one dance before the end, we reach 123 immediately a ;...;a A isalistofnumbers. Foreachi,theelements1;...;i forma
1 n 2
andendwiththeseconddrummer. Ifweinsteadhavethreedancesbefore sublist of a. Let b be the position of i in the sublist consisting of 1;...;i.
i
the end, we might move to 231 and 213 before reaching 123 with the (cid:2)rst Let f.a/ be the resulting list b ;...;b . By construction, 1 b i, so
1 n (cid:20) i (cid:20)
drummerattheend. f.a/ B.
2
77 PartIISolutions Chapter5: CombinatorialReasoning 78
To prove that f is a bijection, we describe a function g: B A. We suchapermutation,takeapermutationof[k]andinsertk 1immediately
! C
build g.b/ from an empty list by inserting numbers in the order 1;...;n. precedingoneofthek elements. Therearek k!waystodothis.
(cid:1)
Before inserting i, the list consists of 1;...;i 1 . We insert i to have The identity permutation has no element out of place, so it is not
f (cid:0) g
positionbi. Afterprocessingb,wehaveapermutationof[n]. counted in this sum. Any other permutation has an element out of place,
To prove that f and g are bijections, it suf(cid:2)ces to show that they are and k is uniquely de(cid:2)ned for it. Thus each non›identity permutation of
injective (in fact g f(cid:0)1), since A and B are (cid:2)nite and have the same [n 1]iscountedexactlyonceintheset,asdesired.
D C
size. First consider f. Given distinct permutations in A, there is some
least value j such that the subpermutations using elements 1;...; j are 5.58. Permutationsof[4]and[5]without(cid:2)xedpoints. Avoiding(cid:2)xedpoints
meansthatthecyclesinthefunctionaldigraphhavelengthatleast2.
different (by the Well Ordering Property, since they differ when j n.
D Withfourelements,thisrequiresasingle4›cycleortwo2›cycles. There
Sincetheyarethesameearlierbutdifferatthe jthstep,thecorresponding
are six 4›cycles (start at element 1 and visit the other three elements in
valuesofb aredifferent.
j
some order) and there are three permutations consisting of two 2›cycles
Forg,iftwoelementsofBdiffer(cid:2)rstatthe jthindex(bj 6Db0j),thenthe (pickthemateofone(cid:2)xedelement). Theansweris9.
subpermutationsof1;...; j inthetwoimagepermutationsaredifferent.
With (cid:2)ve elements, we require a single 5›cycle or a 3›cycle and a 2›
5.56. Among positive integers, the inequality n! > 2n holds if and only cycle. There are 24 5›cycles (start at element 1 and visit the other four
if n 4. By explicit computation, the inequality fails when n 3, and elementsinsomeorder). Thereare20permutationsconsistingofa3›cycle
(cid:21) (cid:20)
4! 24 > 16 24. This provides the basis for a proof by induction that and2›cycle(pickthe2›cyclein10ways,andforma3›cycleontheremaining
D D
n! > 2n when n 4. For the induction step, suppose that k! > 2k for some elementsintwoways). Theansweris44.
(cid:21)
positiveintegerk. Wethenhave
5.59. If f: A A, and n;k are natural numbers with k < n, then fn
! D
.k 1/! .k 1/k!>2 k!>2 2k 2kC1 fk B fn(cid:0)k. We use induction on n. When n D 2, we have k D 1, and the
C D C (cid:1) (cid:1) D formula f2 f1 f1isthede(cid:2)nitionof f2. Fortheinductionstep,suppose
D B
thattheclaimistruewhenn m;weprovethatitalsoholdsforn m 1.
andtheinequalityholdsalsowhenn k 1. D D C
D C For k 1, again the de(cid:2)nition of iteration yields fmC1 f1 fm. Now
5.57. Forn N, n k k! .n 1/! 1. considDer 1 < k < n 1. Using the de(cid:2)nition of iteratioDn, theBinduction
Proof 12(inducktDi1on(cid:1)onDn). WChen(cid:0)n 1, we have 1 1! 1 2! 1. hypothesis,theassocCiativityofcomposition,andthede(cid:2)nitionofiteration
P D (cid:1) D D (cid:0)
Fortheinductionstep,supposethattheformulaholdswhenn m. When again,wehave
D
n m 1, we separate the last term of the sum and apply the induction
D C
hypothesistoobtain fm 1 f fm f .fk 1 fm 1 k/ .f fk 1/ fm 1 k fk fm 1 k
C (cid:0) C (cid:0) (cid:0) C (cid:0) C (cid:0)
D B D B B D B B D B
m 1 m
C
k k! .m 1/.m 1/! k k! 5.60. ThePennyProblemfunction. Thefunction f takesaunitfromeach
(cid:1) D C C C (cid:1)
XkD1 XkD1 piletomakeanewpile.
[.m 2/! .m 1/!] [.m 1/! 1] .m 2/! 1 a)Thefunctionaldigraphof f whenn 6.
D C (cid:0) C C C (cid:0) D C (cid:0) D
Thustheformulaalsoholdswhenn m 1. 21111
D C
Proof2(combinatorialargument). Theformula.n 1/! 1countsthe (cid:15)
C (cid:0)
permutationsof[n 1]exceptfortheindentitypermutation. Thesumma›
C
tionalsocountsthisset,partitionedbylettingk 1bethehighestvalueof 111111 6 42 321
C
i suchthatelementi isnotinpositioni. Forsuchapermutation,element
(cid:15) (cid:15) 5(cid:15)1 (cid:15) (cid:15)
k 2;...;n 1 are located in positions k 2;...;n 1 (one choice only),
C C C C
andelements1;...;k 1arelocatedinpositions1;...;k 1,formingsome
C C
permutation of [k 1] such that k 1 is not in the last position. To form
C C (cid:15) (cid:15) (cid:15) (cid:15) (cid:15)
2211 411 33 222 3111
79 PartIISolutions Chapter5: CombinatorialReasoning 80
b)Thefunction f isinjectiveandsurjectivewhenn 2,anditisneither there are i 1 pennies below it in its pile. If pi is the number of pennies
(cid:20) (cid:0)
when n 3. When n 1, the set has size 1, and f is the identity. When at leveli (this is the same as the number of piles with at leasti pennies),
(cid:21) D
n 2,thesetis 2;11 ,and f transposesthetwoelements. then p p . Thus the list p p ;p ; is also a partition of n.
D f g 1 (cid:21) 2 (cid:21) (cid:1)(cid:1)(cid:1) D 1 2 (cid:1)(cid:1)(cid:1)
Whenn 3, f mapsboth.n/and.2;1;1;...;1;1;1/intotheelement Applying the same procedure to p yields the original partition l. Thus
(cid:21)
.n 1;1/. Furthermore, thesetwoelementsaredistinctwhenn 3, so f this transformation is a bijection from the set of partitions of n to itself.
(cid:0) (cid:21)
isnotinjective. Thepartitionl hask parts(piles)ifandonlyifthelargestpart(p )inthe
1
When n 3, the element .1;...;1/ is not in the image. Since it has corresponding partition p is k. Thus the bijection restricts to a partition
(cid:21)
only 1s, any element mapping to it has only one pile. The only element fromthesetofpartitionsofn withk partstothesetofpartitionsofn with
withonepileisapileofn,anditsimageis.n 1;1/,whichisnotall1s. largestpartk. Thusthesetwosetshavethesamesize.
(cid:0)
5.61. There is a 3›cycle in the functional digraph of f: R ! R de(cid:2)ned by 5.63. The number of partitions of n into distinct parts equals the number
f.x/ 1=.ax b/forx b=aand f. b=a/ . 1=b/ .b=a/ifandonlyif
D C 6D(cid:0) (cid:0) D (cid:0) (cid:0) of partitions of n into odd parts. Given a partition of n into distinct parts,
a b2 0. Such3›cyclescorrespondtopoints x suchthat f.f.f.x/// x
C D D wede(cid:2)neacorrespondingpartitionofn intooddparts. Thismapwillbea
and that f.x/ x. The solutions to x 1=.ax b/ are the solutions to
6D D C bijection,provingthatthetwosetshavethesamesize. Weusethefactthat
ax2 bx 1 0(giventhat x b=a).
C (cid:0) D 6D(cid:0) each positive integer can be expressed in a unique way as an odd number
Note that f. b=a/has been chosen so that f.f. b=a// b=a. This
(cid:0) (cid:0) D (cid:0) timesapoweroftwo.
isa2›cycleandaslongasx = b=a; .a b2/=.ab/ wecanusetheformula
f.x/ 1=.ax b/. 2f(cid:0) (cid:0) C g Let p D p1;p2;(cid:1)(cid:1)(cid:1) be a partition of n into distinct parts. For each pi,
GDiventhaCt f.x/D ax1Cb,wehave f.f.x//D aCab.xaCxbCb/. Now nweewwpraitretitpiionDby.2iknicClud1/in2gji,,fworitehackhi ai,n2djijpiaurntsiqoufesliyzede2tkeirCm1in.eSdi.ncFeotrhmesae
new parts sum to p , doing this for each i produces a partition of n into
1 a b.ax b/ i
f.f.f.x/// C C : oddparts. Notethatdistinctparts(suchas10and40)maybothyieldodd
D a ax b b D a.ax b/ ab b2.ax b/
a b.aCx b/ C C C C C partsofthesamesize(suchastwo5’sandeight5’s),butthemultiplicities
C C
ofthesepartswillbedifferentpowersof2.
Settingx f.f.f.x///yieldsxa.ax b/ xab xb2.ax b/ a b.ax b/, Toinvertthemapandretrievetheoriginalpartition pfromapartition
D C C C C D C C
whichsimpli(cid:2)estox.a b2/.ax b/ a b2. Ifa b2 0,thenweobtain ofnintooddparts,letl bethenumberofpartsofsize2k 1. Thenumberl
C C D C C 6D
ax2 bx 1 0,whichispreciselytheconditionfor x f.x/. Therefore, C
hasauniqueexpressionasasumofpowersof2. Foreachpowerof2used
C (cid:0) D D
a3›cyclecanoccuronlywhena b2 0. in the binary expansion of l, combine that many copies of k into a single
WhenaCb2 D 0,theformuClaforDf.f.x//simpli(cid:2)esto aaxbCxb. Applying partforthepartition p. Whentheoriginaltransformationisappliedto p,
f againyields thenumberofcopiesof2k 1producedisthesumofdistinctpowersof2,
C
andtheuniquebinaryexpansionofl identi(cid:2)esthesepowers.
1 bx
f.f.f.x/// x:
Therefore,whena b2 0D,eavaeaxrbCyxbpCoibntDnoatxiCn b0C; bb2=xaDliesona3›cycle. r05e.(cid:20)6p4rme.s1eFn<otramnti;2okn<2).(cid:1)N(cid:1)(cid:1),<thmerkeaisndexnacDtlym1o1neCchm2o2icCeo(cid:1)f(cid:1)(cid:1)inCtegmkekrs(cmal1l.e.d.tmhkesku›nchomthiaatl
WeleavetheaCnalysDisfor f.x/D acxxCCdb andafd 6D(cid:0)bctgothereader. Proof 1 (strong induction o(cid:0)n n(cid:1)). (cid:0)Fo(cid:1)r n D 0,(cid:0)th(cid:1)e unique solution is
m i 1for1 i k. Forn >0,letm bethelargestintegert suchthat
5.62.a)PTahretiptiaorntsitioofnisntoefg6erasr.e6,51,42,411,33,321,3111,222,2211,21111, kti D(cid:20)n(cid:0). Weclaim(cid:20)th(cid:20)atcombiningthiswitkhtheunique.k(cid:0)1/›representation
111111. (cid:0)of(cid:1)n0 Dn(cid:0) mkk yieldstheuniquek›representationofn.
partibt)ioTnhseonfunmwbietrholfarpgaersttitpioanrst ko.f nAwpiathrtiktiopnarltsDeqlu1;all2s;(cid:1)t(cid:1)h(cid:1)eonfunmcbaenrboef thanFmirks.t,oO(cid:0)btshe(cid:1)errvweitshea,tnt0h>ereksmuk1lt,inwghmickh(cid:0)1yiineltdhsenre>premskeknCtatimkokn(cid:0)11ofDN imskklCe1ss,
viewed as n pennies in piles, with the size of each pile being one of the contradictingthechoiceofm (cid:0). Thusthenumberswehavech(cid:0)osendoform
k
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
parts. Eachpilehasapennyatthebottomlevel;eachpennyisatleveli it ak›representationofn.
81 PartIISolutions Chapter6: Divisibility 82
For uniqueness, it suf(cid:2)ces to show that mk is uniquely determined, c) If f is any polynomial of degree k with rational coef(cid:2)cients, then
because the induction hypothesis implies that n0 has a unique .k (cid:0) 1/› f c an be written in the form kj 0bj xj , where the bj’s are rational . Let
representationtocompletethek›representationofn. Ifwechooseavalue f k c xj. Ifk 0,then f Dc x . Fork >0,wecompletetheproofby
larger than mk, then by the choice of mk we already total more than n. If indDuctiojDn0onj k. LetDb c k!,sPoDtha0t0f(cid:0).x(cid:1)/ b x isapolynomialofdegree
wechooseavaluelessthanmk,thenthemostwecangetis mkk(cid:0)1 fromthe atmoPstk 1. BytheiknDducktionhypo(cid:0)th(cid:1)esis,(cid:0)wekcaknchooserationalnumbers
tXkioD(cid:0)p01 (cid:18)temrkmk(cid:0)(cid:0)a1ni(cid:0)diP(cid:19)ikDD(cid:0)11XkiD(cid:0)(cid:0)m01kk(cid:18)(cid:0)(cid:0)1mmi(cid:0)kki(cid:1)(cid:0)(cid:0)fr11o(cid:0)(cid:0)mkit(cid:19)h(cid:20)eljoDmXwmk(cid:0)ke(cid:0)1rk(cid:18)temrkm(cid:0)sj.1T(cid:0)hku(cid:19)sDth(cid:18)emt(cid:0)okmt(cid:0)kalk(cid:1)(cid:19)is(cid:0)a1t<m(cid:18)osmtkk(cid:19) bpra0a;tri.to.dn(.b);a)blf,knw(cid:0)2(cid:0)u1emIskiobfnetaorhnw,adwtteohfnah.ltxay/vaien(cid:0)fyefbx.fkpxu(cid:0)/rxkne(cid:1)DcstDsioePndkjDokfj0(cid:0)Df.10bxthj/bijDxjs(cid:0)xj,f(cid:1)Pow.r(cid:0)hkSmjDei(cid:1)0nrbebceejtl(cid:0)hoxbjen(cid:1)kgbiansjs’sttdhoaeerI(cid:2)ed.nienCestdoierngievsederafrslos.sreoBmlyay,.
P (cid:0) (cid:1)
SincePrno(cid:21)ofm2kk.,(tohrdeibnoaurnydinadbuocvteiopnroohnibni)t.sCalotnersnidaetrivaeke›xrperpersessieonntsa.tion of n sfu.pxp/oDse fkj20Ib,j axjn,dwlheterkebtehethbej’sdeagrreeeraotfiofn.alBnyumpabretr(sc.),Wweepkrnoovwe tthhaatt
using m1;.(cid:0)..(cid:1);mk. Let j be the maximum index such that m1;...;mj are each bj iPs inDfac(cid:0)t(cid:1)an integer, by induction on j. For j D 0, evaluate f at
consecutive;either j Dk ormjC1 >mj C1. Wehave 0b;.thHeernescueltb, wmhuicsht ibseaanninintetgeegrerb.ecFaoursethfe2inId,uacltsiooneqsuteapls, asskjuD0mbej t0jhaDt
XiDj1 (cid:18)m1Cii(cid:0)1(cid:19)DXiDj1 (cid:18)m1mC1(cid:0)i(cid:0)11(cid:19)D(cid:18)m1mC1 j(cid:19)(cid:0)1D(cid:18)m1jC j(cid:19)(cid:0)1D(cid:18)mj jC1(cid:19)(cid:0)1 0ff0..<rr//DDr (cid:20)PkrjkjD00a0bbnjjd(cid:0)rrjj(cid:1)w.,eoTrhhbearvoeDnlpyfrn.orvo/ne(cid:0)dzetrhorajt(cid:0)te10rbbm0j;sr.ja..r.e;Tbthrh(cid:0)eo1rsieagrwhetit›ihhnatjeng(cid:20)desrrPis,d.seoWiwseea(cid:0)hhsaa(cid:1)uvvmee
sTehtutisnwgemobttoaiina1k›froerp1reseint<atijo.nAosfinnCth1eby(cid:2)rrsetpplarcoionf,gwmej mwiuthstmajlsCo1shaonwd ofproduPctsDofin(cid:0)t(cid:1)egers,sobr isanPintDeger(cid:0),w(cid:1)hichcompletestheinduction.
i
(cid:0) (cid:20)
uniqueness;thisalsocanbedoneonthebottomend.
6. DIVISIBILITY
5.65. Polynomials with rational coef(cid:2)cients that map integers to integers.
Let I bethesetofpolynomialswiththisproperty.
6.1. Thephrase(cid:147)Letnberelativelyprime(cid:148)makesnosensebecauserelative
a)If p;q I andn Z,then p q I andn p I. Polynomialsare
2 2 C 2 (cid:1) 2 primalityisconsideredonlyforpairsofnumbers.
real›valued functions, and we have de(cid:2)ned the sum f g of real›valued
C
functionsbyaddingtheirvaluesateachpointofthedomain. Hence p q 6.2. When pisaprimenumber,theintegersarerelativelyprimeto pareall
C
mapstheintegermtotheinteger p.m/ q.m/,butwealsomustverifythat integersthatarenotdivisibleby p. Since phasnofactorsotherthanitself
C
p q isapolynomial! Fortunately,if p.x/ a xi andq.x/ b xi,then and 1, the integer multiples of p are the only numbers having a common
i i
C D D
the polynomial h.x/ .a b /xi has the same value as p q at every factorwith p otherthan1.
i i
D C P CP
point x,so p q isthispolynomial. Similarly,np.m/isaninteger,andthe
functionn pCisthesamPeasthepolynomial .n a /xi. 6.3. Thenumbersrelativelyprimeto0are(cid:6)1. Since0isdivisiblebyevery
(cid:1) (cid:1) i integer,thenumberswhosegreatestcommondivisorwith0is1are 1.
xj Db)1j!Txh.xe(cid:0)po1ly/n(cid:1)(cid:1)o(cid:1)m.xia(cid:0)lsj(cid:0)Cxj(cid:1)1a/nidsaPpkjoDl0ynnjo(cid:0)mxj(cid:1)iable;PlownegctaonIoibftfanijngt(cid:18)heZc.oNefo(cid:2)tceiethnatst 6.4.PIfrogcodf.a1;(bin/tDeg1e,rthcoenmgbcinda.ntiao;nnsb)/. DByn.Theorem 6.12, the set of in(cid:6)teger
by multiplying out the factors. If j 0, the value of the product with
(cid:0) (cid:1) D combinations of na and nb is the set of multiples of gcd.na;nb/. If k is
no factors is 1, and this is the polynomial whose value is 0 everywhere.
an integer combination of na and nb, then k rna snb n.ra sb/ for
Notethatwhen x isaninteger, x isaproductofintegersandhenceisan D C D C
j some integersr;s. Thus all integer combinations of a and b are multiples
x
integer. Thismakessense,becau(cid:0)s(cid:1)ewerecognize j asthenumberofways of n. Conversely, each integer multiple tn is an integer combination of
to choose j elements from x distinct elements, when x is an integer. Now a and b, since gcd.a;b/ 1 yields r;s such that ra sb 1, and then
kj 0nj xj is a sum of integer multiples of polyn(cid:0)o(cid:1)mials in I. By part (a) tn t.ra sb/n .rtn/aD .stn/b. ThusthesetofinteCgercDombinationsis
(anDdinductiononk),theresultingpolynomialisalsoin I. theDsetofCmultiplDesofn,aCndwehaven gcd.a;b/.
P (cid:0) (cid:1) D