= z 20 z n. (k 20) + 4 z k = 4

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1 Problem Set #7 solutons (a Fnd the coeffcent of z k n (z + z 5 + z 6 + z 7 + 5, k 20. We use the known seres expanson ( n+l ( z l l z n below: (z + z 5 + z 6 + z (z 5 ( + z + z 2 + z ( z 20 z + 5 z 20 z n 5 + z n+20 Ths s an acceptable but not entrely canoncal representaton of the generatng functon: to be standardzed (and useful for answerng the queston asked we d lke to change the exponent on z to be a sngle varable. So let k n + 20, replacng every n above wth k 20: k 200 ( (k 20 + z k ( k 6 z k so the coeffcent of z k s ( k 6 for z 20 (and for z < 20, the coeffcent s zero, as these terms do not appear n the seres expanson. (b Fnd the coeffcent of z k n (z + z + z 5 ( + z n, k 5. We use the known bnomal expanson (+z n ( n z below. Note that the upper bound can be, snce every bnomal coeffcent ( n k wth k > n s zero; ths s not necessary, but actually makes our seres manpulaton easer, snce we need not worry about the upper lmt of the sum: (z + z + z 5 ( + z n (z + z + z 5 z z + + z + + z +5 Now we need to perform ndex-shfts on each sum to get our z-terms to match; so we ntroduce new ndces k +, k +, and k + 5 to the three sums: z + + z + + z +5 z k + z k + z k k k k 5 k k k5 [( ] [( ] n n [( ( ( ] n n n z + nz z + + n z z k 2 k k k k20 k5 Page of 6 March, 2008

2 Problem Set #7 solutons So despte all the specal-casng necessary for k < 5 (whch need not be calculated as explctly as shown here, we see that the coeffcent of z k for k 5 wll always be ( ( n k + n ( k + n k What s the generatng functon for a n, the number of ntegers from 0 to 99, 999, whose dgts sum to n? How many of these ntegers have dgts that sum to 27? The generatng functon for numbers from zero to 99, 999 categorzed by dgt sum s ( + z + z z 9 5 (cf. problem 7..5(a from Problem Set #6. We shall perform expansons of the generatng functon to determne the z 27 term of ths seres: ( z ( + z + + z z ( z 0 5 ( z 5 ( 5z 0 + 0z 20 0z 0 + 5z 0 z 50 + z n To fnd the z 27 term, we consder all possble products of z 27 that can be produced va a product of the 50th-degree polynomal and seres gven. A z 27 term arses va all of the followng products: (27+ z 27, 5z 0 (7+ z 7, and 0z 20 (7+ z 7. Thus, the total z 27 coeffcent n ths product s: ( ( ( Show that the followng formulas hold: (a z ( + z( + z2 ( + z ( + z 8. We can prove by nducton that (+z(+z 2 (+z 2n +z+z 2 + +z 2n+ : t s clearly true for n, and assumng t for a partcular n, we can derve ts truth for n + : ( + z( + z 2 ( + z 2n + z + z z 2n+ ( ( + z( + z 2 ( + z 2n ( + z 2n+ + z + z z 2n+ ( + z 2n+ ( + z + z z 2n+ ( + z 2n+ + z 2n+ + + z 2n z 2n+ +(2 n+ + z + z z 2(2n+ + z + z z (2n+2 We thus have that n ( + z2 2 n+ z n, so n the lmtng case (dong the sorts of thngs I tell MATH 205 students that we are never, ever allowed to do, ( + z 2 z n z Page 2 of 6 March, 2008

3 Problem Set #7 solutons (b z ( + z + z2 ( + z + z 6 ( + z 9 + z 8. Ths proceeds smlarly to the argument above, only wth n nstead of 2 n everywhere Fnd a generatng functon and formula for a n, the number of ways to dstrbute n smlar jugglng balls to four dfferent jugglers so that each juggler receves an odd number of jugglng balls that s larger than or equal to three. Montorng the number of balls dstrbuted, dstrbuton of balls to a sngle juggler under these constrants can be modeled wth the seres z + z 5 + z 7 + z ( + z 2 + z + z z. Dong so for jugglers yelds the generatng functon 2. z 2 ( z 2 To fnd a formula for a n, we need to put ths generatng functon n the form a n z n n order to make determnng ndvdual coeffcents easy. We use the known seres expanson for wth z 2 standng n for y: ( y k z 2 ( z 2 z2 k 60 ( m + ( m + (z 2 m z 2m+2 m0 Now we use ndex-shftng, lettng k m+6 and replacng all occurrences approprately to clean up the exponent on z and gettng the followng smplfed sum: ( k 6 + ( k z 2k z 2k Thus, the coeffcent of z 2k s ( k, and the coeffcent of any odd pwoer of z s zero; thus a n ( k f n 2k, and an 0 f n s odd In how many ways can a con be flpped 25 tmes n a row so that exactly fve heads occur and no more than seven tals occur consecutvely? Let n keep track of the total number of flps n the followng process: flp from zero to seven tals; flp a head then flp from zero to seven tals fve tmes. Ths process s guaranteed to yeld exactly fve heads and no more than 7 consecutve tals. Ths procedure has dscrete steps, but they fall nto two categores: flps of a sngle con yeldng heads, wth generatng functon z; and flps of a con from zero to seven tmes yeldng tals, wth generatng functon + z + z z 7. Thus, our generatng functon n total s k6 m0 ( + z + z z 7 [ z( + z + z z 7] 5 z 5 ( z 8 6 whch can be calculated to be z 5 ( 6z 8 + 5z 6 + z z n 5 ( z 6 We then determne the z 25 term n ths seres by fndng all possble products of terms from the factors yeldng z 25 : z 5 (25 5 z 20, z 5 6z 8 (7 5 z 2, and z 5 5z 6 (9 5 z ; thus the total coeffcent of z 2 5 n the above product s ( ( ( Page of 6 March, 2008

4 Problem Set #7 solutons 7... Fnd a generatng functon for a n, the number of parttons of n nto (a Odd ntegers. A partton of n nto odd numbers can specfcally be consdered as a partton nto x ones, x threes, x 5 fves, etc.; then, these varables conform to the equaton x +x +5x 5 + n. The generatng functon descrbng the number of solutons to ths equaton s the product of the generatng functons for the ndvdual choces of x, x, x 5, etc. Thus, we have the product (+z+z 2 +z + (+z +z 6 +z 9 + (+z 5 +z 0 +z 5 ( z( z ( z 5 (b Dstnct odd ntegers. Ths s as above, but wth each x constraned to be zero or, so we have fnte generatng functons for each ndvdual choce, yeldng: ( + z( + z ( + z 5 ( + z Fnd a product whose expanson can be used to fnd the number of parttons of (a 2 wth even summands. For even summands we are determnng the number of nonnegatve solutons to the equaton 2x 2 + x + 6x 6 + 8x 8 + 0x 0 + 2x 2 2. The generatng functon for each of these can, n the nterest of yeldng a fnte product, be capped at the z 2 term, and ther product s thus (+z 2 +z +z 6 +z 8 +z 0 +z 2 (+z +z 8 +z 2 (+z 6 +z 2 (+z 8 (+z 0 (+z 2 whose product s an enormous 66th-degree polynomal, n whch the only term of nterest to us s z 2 (so there are even-term parttons of 2; unsurprsng snce each s the double of one of the free parttons of 6. (b 0 wth summands greater than 2. We are determnng the number of nonnegatve solutons to x +x +5x x 0 0; we do ths by multplyng the generatng functons for each summand, cappng each seres wth the z 0 term to get polynomals nstead of seres, and thus get ( + z + z 6 + z 9 ( + z + z 8 ( + z 5 + z 0 ( + z 6 ( + z 7 ( + z 8 ( + z 9 ( + z 0 whose product s a 67th-degree polynomal, n whch we are nterested n the term 5z 0, denotng 5 parttons of 0 nto parts or larger; we can, knowng there are so few, actually explctly enumerate them: + +, + 7, + 6, 5 + 5, and 0. (c 9 wth dstnct summands. We are determnng the number of solutons to x + 2x 2 + x + + 9x 9 9 wth each x {0, }. Thus the generatng functons for ndvdual summands are bnomals, and we multply them as such: ( + z( + z 2 ( + z ( + z ( + z 5 ( + z 6 ( + z 7 ( + z 8 ( + z 9 Page of 6 March, 2008

5 Problem Set #7 solutons to get a 5th-degree polynomal whch contans the term 8z 9, ndcatng 8 parttons of 9 nto dstnct summands. The complete lst of these s +2+6, ++5, 2 + +, + 8, 2 + 7, + 6, + 5, and (a Show that the number of parttons of n s exactly equal to the number of parttons of n whose smallest summand s. One could do ths wth generatng functons: we know that ( + z + z 2 + ( + z 2 + z + ( + z + z 6 + p(nz n but f we force our parttons to contan at least one, the frst factor n the above product becomes (z + z 2 + z + z( + z + z 2, yeldng the generatng functon z p(nz n p(nz n+ p(n z n So we see that forcng at least one n a partton of n elements s possble n p(n ways. We could also accomplsh ths wth an explct bjecton. Any partton of n wth a n ts expanson can be mapped to a partton of n by smply removng the ; ths s clearly a bjectve map snce t can be reversed by takng any partton of n and addng a sngle to t. Snce ths s a bjecton, the two sets nvolved must be of equal sze. (b Descrbe the parttons of n that are counted by the expresson p(n p(n. We know p(n counts the parttons of n; and from the above concluson we know that p(n counts the parttons of n whch have a n ther expanson. Thus, ther dfference counts the parttons of n whch do not contan a n ther expansons. Let P n be the set of parttons of n, so that P n p(n. Then p(n p(n would be the sze of P n wth some set bjectvely mapped to P n removed. There s a smple njecton of P n nto P n : smply add a sngle to every partton n P n to get a partton from P n. The set mapped onto by ths operaton s precsely those parttons of P n contanng n ther expansons, snce ths procedure could be reversed by removng a. The set yelded by removng all these mages of P n from P n s thus those parttons of n not usng the number ; and ths s what s counted by p(n p(n Use the Ferrer s graph to show that the number of parttons of n nto exactly k summands equals the number of parttons of n havng ts largest summand equal to k. All one needs to do s consder the transpose. A Ferrer s graph assocated wth a partton nto k summands wll have heght exactly k, and thus, ts transpose wll have wdth exactly k, meanng that one of ts summands s sze k but none exceed k. Snce the transpose s a bjectve map, the parttons satsfyng these two condtons are equnumerous. n Page 5 of 6 March, 2008

6 Problem Set #7 solutons 7... Fnd a generatng functon for the number of parttons of n nto (a Summands no larger than. We want nonnegatve solutons to x +2x 2 +x 6 +x n, whch can be expressed as a product of the four ndvdual generatng functons for decson of x, x 2, x, and x : (+z+z 2 +z + (+z 2 +z +z 6 + (+z +z 6 +z 9 + (+z +z 8 +z 2 + (b Summands the largest of whch s. Ths s as above, except we requre at least one, so we constran x, whch affects the fourth multplcand n our expresson: (+z +z 2 +z + (+z 2 +z +z 6 + (+z +z 6 +z 9 + (z +z 8 +z 2 + (c At most four summands. By the transposton bjecton on Ferrer s dagrams we know ths s an dentcal enumeraton to part (a, and thus has the same generatng functon. (d Exactly four summands. By the transposton bjecton on Ferrer s dagrams we know ths s an dentcal enumeraton to part (a, and thus has the same generatng functon. Page 6 of 6 March, 2008