Math 475, Problem Set #12: Answers

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1 Math 475, Problem Set #12: Aswers A. Chapter 8, problem 12, parts (b) ad (d). (b) S # (, 2) = 2 2, sice, from amog the 2 ways of puttig elemets ito 2 distiguishable boxes, exactly 2 of them result i oe of the boxes beig empty. Hece S(, 2) = 1 2! S# (, 2) = 1 2 (2 2) = Alterative solutio: Let 1,..., deote the objects that are beig put ito the boxes. Oe we choose to put 1 ito a particular box (ad for purposes of coutig we ca igore the fact that we have two choices sice the two boxes are idistiguishable), we have a two-way choice for where to put 2, a two-way choice for where to put 3, etc.; i each case, we ca either put the ew umber i the same box as 1 or i the other box. This would give a total of 2 1 possibilities (sice there are 1 elemets to be assiged to the boxes after 1 has bee put i a box), except for the fact that we are required to leave o box empty. This will happe if all 1 elemets are assiged to be i the same box as 1. Hece the umber of permitted assigmets is ot 2 1 but (d) Whe we divide a set of size ito 2 o-empty subsets, we must either (a) have a set of size 3 ad all other sets of size 1, or else (b) have two sets of size 2 ad all other sets of size 1. Case (a): There are ( ) 3 ways to choose the three special elemets that belog to the set of size 3. Case (b): There are ( ) 4 ways to choose the four special elemets that belog to the two sets of size 2, ad the there are 3 ways to divide those four elemets ito two sets of size 2. So there are ( ( ) 3) ways to divide a set of size ito 2 o-empty subsets, B. Fid a formula for s(, 2), valid for all 3. Whe we divide a set of size ito 2 o-empty circular permutatios (let s call them circles for short), we must either (a) have a circles of size 3 ad all other circles of size 1, or else (b) have two circles 1

2 of size 2 ad all other circles of size 1. Case (a): There are ( ) 3 ways to choose the three special elemets that belog to the circle of size 3, ad 2 ways to arrage them i a circle. Case (b): There are ( ) 4 ways to choose the four special elemets that belog to the two circles of size 2, ad the there are 3 ways to divide those four elemets ito two circles of size 2. (Give two elemets, there is oly oe way to arrage them i a circle of size 2.) So there are 2 ( ( ) 3) ways to divide a set of size ito 2 circles. C. Fix 6. (a) I how may ways ca you partitio a set with (distiguishable) elemets ito 3 distiguishable boxes, so that oe of the boxes cotais fewer tha 2 elemets? Call the boxes red, white, ad blue. Let S be the set of all ways to of assigig the elemets to boxes, let A be the set of all ways of assigig elemets to boxes so that there are fewer tha 2 boxes i the red box, let B be the set of all ways of assigig elemets to boxes so that there are fewer tha 2 boxes i the white box, ad let C be the set of all ways of assigig elemets to boxes so that there are fewer tha 2 boxes i the blue box. We wat to cout the umber of elemets of S (A B C). We use the priciple of iclusio-exclusio. S = 3, sice each of the elemets ca be assiged to ay of the 3 boxes. A = , sice we ca either assig all elemets to the white ad blues boxes or assig oe of the elemets to the red box ad assig the remaiig 1 elemets to the white ad blues boxes. Ditto for B ad C. A B = ( 1)1 2, where the four terms correspod to the ways of havig o elemets i the red ad white boxes, oe elemet i the red box ad o elemets i the white box, o elemets i the red box ad oe elemet i the white box, or oe elemet i the red box ad oe (differet) elemet i the white box. Ditto for A C ad B C. Lastly, A B C = 0, sice (by the pigeohole priciple) there s o way to assig 4 elemets to three boxes without havig some box cotai more tha oe elemet. So S (A B C) = 3 3( ) + 3( ). 2

3 (As a way of checkig this, we ca substitute = 6, obtaiig the aswer 90. We ca compute this a differet way: It s just like the problem of assigig six differet studets to three differet classrooms. The umber of such assigmets is the multiomial coefficiets ( ) 6 2,2,2 = 6! = 90.) 2!2!2! (b) I how may ways ca you partitio a set with (distiguishable) elemets ito ito 3 idistiguishable boxes, so that oe of the boxes cotais fewer tha 2 elemets? Let a be the aswer to part (a), ad b be the aswer to part (b). Each assigmet of the elemets to 3 idistiguishable boxes gives rise to 3! = 6 differet assigmets of the elemets to 3 distiguishable boxes, sice there are 3! ways to pait the three boxes red, white ad blue. Therefore a = 6b. Hece b = a/6 = (3 3( ) + 3( ))/6. D. For each k 0, let σ k (x) = =k S(, k)x, the geeratig fuctio for the kth colum i the table of Stirlig umbers of the secod kid. Thus for istace σ 0 (x) is the power series 1 + 0x + 0x 2 + 0x , also kow as the costat 1. (a) Show that σ 1 (x) = x/(1 x). We showed above that S(, 1) = 1 for all 1, so σ 1 (x) = x + x 2 + x = x/(1 x). (b) Show that σ 2 (x) = x 2 /(1 x)(1 2x). We showed above that S(, 2) = for all 2, so σ 2 (x) = x 2 +3x 3 +7x 4 +15x , whece (1 2x)σ 2 (x) = x 2 +(3 2)x 3 + (7 6)x 4 + (15 14)x = x 2 + x 3 + x 4 + x = x 2 /(1 x), so σ 2 (x) = x 2 /(1 x)(1 2x). (c) Give a geeral formula (valid for all k 1) expressig σ k (x) as a ratioal fuctio of x. (Hit: Multiply the recurrece relatio S(, k) = ks( 1, k) + S( 1, k 1) by x ad sum over all k; the express the resultig equatio i terms of σ k ad σ k 1. This lets you express σ k i terms of σ k 1.) Followig the hit, we get S(, k)x = k S( 1, k)x + S( 1, k 1)x. k k k 3

4 The left-had side is just σ k (x). The first term o the righthad side ca be re-writte as k k+1 S( 1, k)x (sice S(k 1, k) = 0) ad the re-idexed as k k S(, k)x +1 = kxσ k (x). The secod term o the right-had side ca be re-idexed as k 1 S(, k 1)x +1 = xσ k 1 (x). Thus we have σ k (x) = kxσ k (x) + xσ k 1 (x). Re-arragig, we get (1 kx)σ k (x) = xσ k 1 (x), which gives σ k (x) = σ k 1 (x) x. Hece, startig from σ 1 kx 0(x) = 1, we get σ 1 (x) = σ 0 (x) x = x, σ 1 x 1 x 2(x) = σ 1 (x) x = x x = x 2, 1 2x 1 x 1 2x (1 x)(1 2x) σ 3 (x) = σ 2 (x) x = x x x = x 3, etc.; the 1 3x 1 x 1 2x 1 3x (1 x)(1 2x)(1 3x) geeral formula is σ k (x) = x k (1 x)(1 2x) (1 kx). (Note that the formula S(, k) = ks( 1, k) + S( 1, k 1) is ideed valid for = k: it just says 1 = Brualdi oly proves this for k 1, ad ot for the case where k ad are equal. It s a good thig that the formula holds for = k, sice this equatio got used i the precedig proof whe we multiplied the equatio by x k before addig it to ifiitely may other equatios of the same sort!) (d) Use part (c) i the special case k = 3 to fid a formula for S(p, 3). (Hit: Do a partial fractio decompositio of (σ 3 (x))/x 3 rather tha σ 3 (x).) We ca t apply partial fractios right away, because the degree of the umerator is ot less tha the degree of the deomiator. But for all 3, the coefficiet of x i the formal power series expasio of x 3 /(1 x)(1 2x)(1 3x) is equal to the coefficiet of x 3 i the formal power series expasio of 1/(1 x)(1 2x)(1 3x), ad this ratioal fuctio of x ca be decomposed by partial fractios: 1/(1 x)(1 2x)(1 3x) = A/(1 x) + B/(1 2x) + C/(1 3x). Multiplyig both sides by (1 x)(1 2x)(1 3x) ad equatig the the quadratic polyomials that occur o both sides, we get three liear equatios, which we solve to 4

5 obtai A = 1/2, B = 4, C = 9/2. So the coefficiet of x i 1/(1 x)(1 2x)(1 3x) is (1/2)(1) + ( 4)(2) + ( 9/2)(3), ad S(, 3) = the coefficiet of x i x 3 /(1 x)(1 2x)(1 3x) = the coefficiet of x 3 i 1/(1 x)(1 2x)(1 3x) = (1/2)(1) 3 + ( 4)(2) 3 +(9/2)(3) 3 = 1/2+( 4)/82 +(9/2)/27(3) = 1/2 (1/2)2 + (1/6)3. Note that this agrees with the formula we derived i class usig iclusio-exclusio. E. (a) Fid a liear recurrece relatio satisfied by the sequece of umbers S(1, 3), S(2, 3), S(3, 3),... (Hit: Use the geeratig fuctio (σ 3 (x))/x 3 you computed above, expressed as a ratioal fuctio of x, ad derive the recurrece relatio from the form of the deomiator.) Sice the deomiator of 1/(1 x)(1 2x)(1 3x) is the degree- 3 polyomial (1 x)(1 2x)(1 3x), ad the umerator is the costat 1 (a polyomial of degree less tha 3), the coefficiets i the formal power series expasio of 1/(1 x)(1 2x)(1 3x) satisfy the recurrece whose characteristic polyomial is (x 1)(x 2)(x 3) = x 3 6x 2 +11x 2 6; that is, S(, 3) 6S( 1, 3)+ 11S( 2, 3) 6S( 3, 3) = 0 as log as > 3. (Recall that S(0,3) is the value that does t fit the same patter as the subsequet terms, so we must avoid the situatio where, 1, 2, or 3 vaishes; that is, we eed > 3.) Thus we obtai the recurrece relatio S(, 3) = 6S( 1, 3) 11S( 2, 3) + 6S( 3, 3), valid for all > 3. (b) Use this to compute S(8, 3) from earlier terms. Usig the iitial coditios S(1, 3) = S(2, 3) = 0 ad S(3, 3) = 1, we successively derive S(4, 3) = 6(1) 11(0) + 6(0) = 6, S(5, 3) = 6(6) 11(1) + 6(0) = 25, S(6, 3) = 6(25) 11(6) + 6(1) = 90, S(7, 3) = 6(90) 11(25) + 6(6) = 301, ad S(8, 3) = 6(301) 11(90) + 6(25) = 966. (c) Compare this with the value of S(8, 3) obtaied by usig the recurrece relatio S(p, k) = S(p 1, k 1) + ks(p 1, k). Usig values i the table i Brualdi, we derive S(7, 3) = S(6, 2) + 3S(6, 3) = (90) = 301 ad S(8, 3) = S(7, 2) + 3S(7, 3) = (301) =