represented by 4! different arrangements of boxes, divide by 4! to get ways


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1 Problem Set #6 solutios A juggler colors idetical jugglig balls red, white, ad blue (a I how may ways ca this be doe if each color is used at least oce? Let us preemptively color oe ball i each color, so that the 9 remaiig balls may be colored ay way we like This is a questio of partitioig 9 idetical items ito distict classes, alteratively, it could be cosidered as a oegative iteger solutio to the problem x + x + x = 9 I either case, the correct eumerator is ( ( 9+ = = 55 (b I how may ways ca this be doe if each color is used at least three times? We proceed above, but istead we preemptively color three balls i each color, leavig oly balls left over to be freely colored; here our eumerator will thus be ( ( + = 5 = 0 We could i fact eumerate these maually: there are three colorigs i which there are balls of two colors ad 6 of the remaider, six colorigs i which there are,, ad 5 balls i three distict colors, ad oe colorig i which there are balls of each color 8 (a I how may ways ca idetical cois be distributed to four differet persos if each perso receives at least oe coi? This is much as i the previous problem: preemptively had out oe coi to each perso, ad the partitio the remaiig 8 cois freely This is couted by ( 8+ ( = = 65 (b I how may ways ca differet books be distributed to four idetical boxes if each box receives exactly three books? There are several approaches to this problem: oe is to partitio the books amog distiguishable boxes i (,,, ways, ad the, realizig that each partitio is! represeted by! differet arragemets of boxes, divide by! to get ways!!!!! of arragig the books Alteratively, oe may umber the books (for istace, ad cosider the boxig of the books as a selectio process We select book ad two other books to accompay it ay of ( ways; the we take out the lowestumbered book remaiig, ad associate it with two of the remaiig 8 books i ( 8 ways; the we agai take the lowestumbered book ad box it with two of the remaiig 5 books i ( 5 ways; ad lastly, we take the remaiig lowestumbered book, ad box it up with the remaiig books i ( ways! These two approaches, of course, yield the same result: = ( ( 8 5!!!!! ( ( = 500 (c Compute the umber of ways to partitio a set of differet objects ito four oempty subsets The partitio is ot explicitly ordered ad partitio of a set ito uordered subsets is the precise defiitio of the Stirlig umber, so here the quatity desired is simply S(, =! ( ( i i i=0 ( i = ( 0 ( + ( ( + ( 0 = 650! Page of 7 February 8, 008
2 Problem Set #6 solutios 0 Explai the followig formulas: (a S(, = ( There are two possible explaatios: either a algebraic demostratio of equality or ( costructio of a bijectio betwee quatities eumerated by S(, ad The algebra is possible but difficult; the combiatorial costructio is actually far simpler Note that S(, eumerates the partitios of {,,,, } ito oempty, uordered subsets Such a partitio would cosist of sets of size ad oe set of size Thus, the partitios are uiquely determied by the choice of twoelemet set, sice the rest of ay give partitio ca be recostructed by placig every elemet of {,,, } ot appearig i the twoelemet set i a sigleto set by itself Thus, oempty partitios of a elemet set ito ( subsets, which are eumerated with S(, are i a bijective correspodece to choices of a sigle twoelemet subset of a elemet set, which are eumerated with ( ( Thus, S(, = (b S(, = ( ( + The combiatorial proof of this follows closely the surjectio argumet from problem (appearig o problem set # Here S(, eumerates partitios of {,,,, } ito oempty, uordered subsets This ca occur i two differet ways: either we partitio {,,,, } ito sigletos ad a triplet, or ito sigletos ad two pairs Let A ad B be sets cosistig of these two types of partitios respectively : it is clear that A B = ad that A B is the set of all partitios of {,,,, } ito subsets Thus, S(, = A B = A + B, ad our aim is to bijectively map A ad B oto sets eumerated with ( ( ad A partitio i A is uiquely determied by its triplet, sice the remaider of the partitio ca be recostructed by placig every elemet of {,, } ot appearig i the triplet ito a sigleto Thus, elemets of A ca be bijectively mapped to elemet subsets of {,, }, demostratig that A = ( A partitio i B is uiquely determied by the two pairs occurrig therei, sice, as above, the partitio ca be recostructed by placig every elemet ot occurrig i the pairs i a sigleto We may eumerate B thus by coutig the umber of pairs of pairs If we choose four elemets to belog to these pairs, we may do so i ( ways; but the, the assigmet of elemets to two pairs may occur i ay of ways: {a, b} ad {c, d}; {a, c} ad {b, d}; or {a, d} ad {b, c} Thus, there are ( ways to choose two pairs as extraordiary parts of a partitio 6 (a I how may ways ca idetical objects be distributed to five differet boxes if the first two boxes receive o more tha two balls each? The assigmets of idetical objects to 5 boxes is equivalet to the umber of oegative iteger solutios to x + x + x + x + x 5 = Page of 7 February 8, 008
3 Problem Set #6 solutios which is ( + Now, we must exclude those solutios i which x or x We ca eumerate the solutios where, for istace, x by lettig x = y + ad fidig oegative iteger solutios to (y + + x + x + x + x 5 = y + x + x + x + x 5 = of which there are ( + A similar argumet fids the umber of solutios i which x The, to fid cases where both exceed, we use trasformatios o both x ad x to fid that these solutios are idetically the oegative solutios of (y + + (y + + x + x + x 5 = y + y + x + x + x 5 = 6 of which there are ( Sice we ow kow the total umber arragemets, ad the umber of arragemets violatig our coditios sigly ad i pairs, we may use iclusioexclusio to fid the umber of arragemets violatig oe of our coditios: ( ( ( ( = (b I how may ways ca differet objects be distributed to five differet boxes if the first two boxes receive o more tha two balls each? The most straightforward approach to this (that I ca devise is a casewise cosideratio: there are ie possibilities for the first two boxes cotaiig aywhere from zero to four objects, ad the remaider of the objects are distributed amog the remaiig boxes A table summarizig calculatios associated with these ie cases (which ca be reduced to 6 usig symmetry is below: Box Box # selectios # remaider placemets Total cofiguratios ( 0 0 ( ( ( ( 0 0 ( ( 0 ( ( ( 0 ( 0 ( ( 0 ( ( ( 0 ( ( ( ( ( ( ( ( ( ( ( ( Page of 7 February 8, 008
4 Problem Set #6 solutios which has a total of + + ( + ( ( + This ca be cleaed up to become ( ( ( ( R Amog all digit itegers, how may of them cotai the digits 0 ad but ot the digits 8 ad 9? Let our uiverse set X cosist of all digit itegers cosistig of digits 0 7 Clearly X = (7(8(8 (8 = 7 8 Now let A ad B be the sets of digit itegers cosistig of digits 0 7 respectively without 0 ad without Havig both a 0 ad a is thus idetical to membership i A B (ie lackig either a 0 or a Clearly A = (7(7(7 (7 = 7 ad B = (6(7(7 (7 = 6 7 ; ad furthermore, A B = (6(6(6 (6 = 6 Thus, our desired eumeratio A B ca be calculated to be X A B + A B = = R How may ways are there to distribute 0 idetical balls to six distict boxes if box ad box each receive fewer tha 0 balls? This is much as i the first part of problem 6: the assigmets of 0 idetical balls to 6 distict boxes is equivalet to the umber of oegative iteger solutios to x + x + x + x + x 5 + x 6 = 0 which is ( 5 5 Now, we must exclude those solutios i which x 0 or x 0 We ca eumerate the solutios where, for istace, x 0 by lettig x = y + 0 ad fidig oegative iteger solutios to (y x + x + x + x 5 + x 6 = 0 y + x + x + x + x 5 + x 6 = 0 of which there are ( 5 5 A similar argumet fids the umber of solutios i which x 0 The, to fid cases where both exceed 0, we use trasformatios o both x ad x to fid that these solutios are idetically the oegative solutios of (y (y x + x + x 5 + x 6 = 0 y + y + x + x + x 5 + x 6 = 0 Page of 7 February 8, 008
5 Problem Set #6 solutios of which there are ( 0 5 Sice we ow kow the total umber arragemets, ad the umber of arragemets violatig our coditios sigly ad i pairs, we may use iclusioexclusio to fid the umber of arragemets violatig oe of our coditios: ( ( ( = R5 A bridge had is a subset of cards chose from the stadard deck How may bridge hads cotai cards of the same rak? Let A i be the set of bridge hads which cotai is (eg A has hads with aces, A cosists of those with twos, ad so forth up to A cosistig of hads with four kigs Clearly each A i = ( 8 9, sice cards are predetermied ad the remaiig 9 cards of the had are chose freely from the remaiig 8 cards of the deck Likewise, A i A j = ( 5 ad Ai A j A k = ( 0 Itersectios of more tha sets are obviously empty, sice o had of cards cosists of cards i each of differet raks choices of i for listig all Ai, ad ( Sice there are ( choices of i ad j to list all possible A i A j, ad ( choices of i, j, ad k to list all possible Ai A j A k, we may see by iclusioexclusio that: A A A = i = ( A i A i A j + i,j i,j,k ( ( ( A i A j A k ( ( 0 Icidetally, the likelihood of this occurrece is approximately % = Costruct a geeratig fuctio for a, the umber of distributios of idetical jugglig balls to (a Six differet jugglers with at most four balls distributed to each juggler We wat our expoet o z to record the umber of jugglig balls distributed by every step of our process Let the idividual steps be the act of givig balls to a sigle juggler Sice our balls are idetical, our choices i a sigle step are: give o balls to the juggler (which ca be doe oly oe way, represeted by the polyomial z 0 (or just ; give oe ball (which ca be doe i oly oe way, represeted by the polyomial z, ad so forth up to givig balls to a juggler Thus, the act of givig a juggler up to balls is represeted by the polyomial + z + z + z + z Doig so for each of six jugglers, we get ( + z + z + z + z 6 as our geeratig fuctio (b Five differet jugglers with betwee three ad seve balls (iclusive distributed to each juggler As above, but the act of givig each juggler a collectio of balls is rtepreseted by the polyomial z + z + z 5 + z 6 + z 7 ; so doig so for each of five jugglers has geeratig fuctio (z + z + z 5 + z 6 + z 7 5 Page 5 of 7 February 8, 008
6 Problem Set #6 solutios 75 Fid a geeratig fuctio for the umber of itegers whose digits sum to amog (a Itegers from 0 to 9999 A easy represetatio for these is as fourdigit umbersequeces with leadig zeroes allowed; so we represet 7 as 007, for istace (ote that this has o effect o the digit sum We break our selectio of a umber dow ito the four steps of pickig idividual digits Sice the iformatio we most wat to record with each step is the digit sum, we use the value of a digit as the expoet of z Thus, for each digit, we have te choices zero through ie, ad for each choice, we wat to record the umber of ways to make that choice (of which there is oly oe as the coefficiet, ad the value of that choice as the expoet We thus associate with choice of a digit the polyomial z 0 + z + z + + z 9 ; doig so four times, we get the geeratig fuctio ( + z + z + + z 9 (b Fourdigit itegers This is as above, but with the restrictio that our first digit caot be zero; thus the firstdigit selectio polyomial omits the z 0 term, givig us the geeratig fuctio (z + z + + z 9 ( + z + z + + z 9 76 Fid a geeratig fuctio for the umber of ways to select balls from a ifiite supply of red, white, ad blue balls subject to the costrait that the umber of blue balls selected is at least, the umber of red balls selected is at most, ad a odd umber of white balls are selected Sice we wat to record the umber of balls selected, we shall use that as the expoet of z Our selectio ca be broke up ito idepedet procedures: selectio of red balls, white balls, ad blue balls Our selectio of red balls demads o more tha, so there are 5 possibilities: we ca select zero, oe, two, three, or four This step is thus represetable by the polyomial + z + z + z + z The selectio of white balls demads a odd umber, so we may select oe ball, or three, or five, ad so forth: this yields the ifiite series z + z + z 5 + z 7 +, sice there is o costrait placed o how may white balls we take Likewise, we may take ay umber of blue balls three or higher: we may take three, or four, or five, ad so forth, yieldig the series z +z +z 5 + to describe this step Multiplyig all our steps gives us the geeratig fuctio for the process as a whole: ( + z + z + z + z (z + z + z 5 + z 7 + (z + z + z Fid a geeratig fuctio for the umber of positive iteger solutios of (a x + x + x + 5x = Let the expoet of z represet the total sum o the left side of the equatio; the choice of x as,,, ad so forth would icrease the left side by, or, or 6, ad so forth, so the cotributio of x selectio to the geeratig fuctio is z + z + z 6 + z 8 + Likewise selectig x to be a positive iteger icreases our total by thrice its value, so selectio of x may yield a icrease of ay multiple of i the total leftside sum, so x selectio is associated with the series Page 6 of 7 February 8, 008
7 Problem Set #6 solutios z + z 6 + z 9 + z + Likewise, x ad x 5 are associated with the respective series z + z 8 + z + z 6 + ad z 5 + z 0 + z 5 + z 0 + (b x + x + x + x =, where each x i satisfies x i 5 Here we ca freely choose x to be,,, or 5, which results i additio of,,, or 5 to the total sum beig kept track of; thus, our decisio procedure for x is represeted by the polyomial z + z + z + z 5 ; likewise for x, x, ad x, so our geeratig fuctio is (z + z + z + z 5 Page 7 of 7 February 8, 008
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