Solutions to Final Exam

Transcription

1 Solutios to Fial Exam 1. Three married couples are seated together at the couter at Moty s Blue Plate Dier, occupyig six cosecutive seats. How may arragemets are there with o wife sittig ext to her ow husbad? We do ot require me ad wome to alterate. For full credit, you must use the iclusio-exclusio method. Solutio: If we igore the costrait about who ca sit ext to whom, the umber of arragemets of the six people is 6!. Let A i be the set of arragemets with couple i sittig together (i = 1,, 3). A i = 5!: there are ways to order couple i (ma o the right or woma o the right) ad the there are 5! ways to order couple i with the other 4 idividuals (thik of couple i as oe etity ad the remaiig 4 people as idividuals). Likewise A i A j = 4! ad A i A j A k = 3!. Hece, usig iclusioexclusio, we see that the umber of ways to arrage three married couples so that o wife sits ext to her ow husbad is 6! 3 ( 5!)+3 ( 4!) 3 3! = = 40.. Use the method of characteristic equatios to solve the liear recurrece h = 4h 1 4h with iitial coditios h 0 = 1, h 1 = 4. Solutio: The characteristic equatio is x = 4x 4, or x 4x + 4 = 0, which has as a double root. Hece the solutio ca be expressed i the form h = A + B. Thus 1 = h 0 = (A)( 0 ) + (B)(0)( 0 ) = A ad 4 = h 1 = (A)( 1 ) + (B)(1)( 1 ) = A + B, which yields A = 1 ad B = 1. Therefore h = + = ( + 1). 3. Solve the o-homogeeous liear recurrrece h = 3h with iitial coditio h 0 = 0. Solutio: The geeral solutio to the correspodig homogeeous recurrece h = 3h 1 is h = A3. To fid a particular solutio to the ohomogeeous recurrece, we ca guess a aswer of the same geeral form as the o-homogeeous term 1; that is, we guess h = B for some costat B. Pluggig this ito h = 3h 1 +1 we get B = 3B +1 which requires B = 1. This correspods to the particular solutio h = 1. So the solutio we seek is of the form h = h 1 = A3 1. Sice 0 = h 0 = A3 0 1 = A 1, we get A = 1. Hece h =

2 Alterative solutio: The o-homogeeous term 1 satisfies a liear recurrece with characteristic polyomial (x 1). Sice the homogeeous versio of the origial recurrece has characteristic polyomial x 3, we ca multiply the two ad coclude that the origial sequece h satisfies a liear recurrece with characteristic polyomial (x 1)(x 3). It follows that we ca write h = A+B3 for suitable costats A, B. Puttig 0 = h 0 = A+B ad 1 = h 1 = A + 3B (where h 1 is computed via the origial o-homogeeous recurrece) ad solvig this liear system, we get A = 1 ad B = 1. Hece h = For > 0, let a be the umber of ways to tile a 1-by- strip with 1-by- tiles ad 1-by-3 tiles (so that a 0 = 1, a 1 = 0, ad a = 1). Fid a thirdorder recurrece relatio satisfied by a, ad write the geeratig fuctio f(x) = a 0 + a 1 x + a x +... as a ratioal fuctio of x. Do ot solve for a. Solutio: For 3, a tilig of the 1-by- strip ca begi with either a 1-by- tile (i which case the umber of ways to complete the tilig is a ) or a 1-by-3 tile (i which case the umber of ways to complete the tilig is a 3 ). Hece a = a + a 3 for all 3, which is a recurrece with characteristic polyomial x 3 x 1. It follows that a 0 + a 1 x + a x +... ca be writte i the form (A + Bx + Cx )/(1 x x 3 ), where we get the deomiator by takig the coefficiets of x 3 x 1 ad applyig them to the powers of x i icreasig rather tha decreasig order. There are two ways to solve for A, B, C: Method 1: Multiply both sides of (A + Bx + Cx )/(1 x x 3 ) = a 0 + a 1 x + a x +... by 1 x x 3, obtaiig A + Bx + Cx = (1 x x 3 )(a 0 + a 1 x + a x +...). Equatig coefficiets of successive powers of x, we get A = a 0 = 1, B = a 1 = 0, ad C = a a 0 = 0, so the geeratig fuctio is 1/(1 x x 3 ). Method : Multiply both sides of the recurrece a = a + a 3 by x ad sum over all 3. This yields f(x) a 0 a 1 x a x = x [f(x) a 0 ] + x 3 [f(x)].

3 Pluggig i a 0 = 1, a 1 = 0, ad a = 1, we get f(x) 1 x = x [f(x) 1] + x 3 [f(x)], which becomes (1 x x 3 )f(x) = 1, so f(x) = 1/(1 x x 3 ). 5. For > 0, let a be the umber of legth- strigs of 1 s, s, 3 s ad 4 s i which o two 4 s appear cosecutively. Fid a secod-order recurrece relatio satisfied by a. Do ot solve for a. Solutio: A allowed strig of legth ca begi with a 1, i which case there are a 1 ways to fiish it; or it ca begi with a, i which case there are a 1 ways to fiish it; or it ca begi with a 3, i which case there are a 1 ways to fiish it; or it ca begi with a 4, i which case the ext symbol must be a 1,, or 3 (3 choices), ad the there are a ways to fiish the strig, givig 3a strigs of legth that start with a 4. Hece the total umber of possibilities satisfies a = a 1 + a 1 + a 1 + 3a = 3a 1 + 3a. 6. (a) Fid the polyomial p() of degree whose differece table is (where the leftmost etry i the top row is p(0)). Express your aswer i the form A + B + C, ad check it for = 0, 1,, ad 3. (b) Fid a formula for the sum k=0 p(k). Express your fial aswer i the form A 3 + B + C + D, ad check it for = 0, 1,, ad 3. Solutio: (a) Readig dow the left diagoal, we see that p() ca be writte as 1 ( ( ( ) 0) + 1) + 6 = ( 1) = (b) Usig those same coefficiets, we get 1 (+1 ) ( ) ( ) This becomes ( + 1) + ( + 1) + ( + 1)( 1) = Let h be the umber of legth- strigs of 1 s, s, ad 3 s i which 1 s occur a eve umber of times ad s occur a eve umber of times. Fid a exact formula for the expoetial geeratig fuctio of the sequece h 0, h 1, h,..., ad use it to fid a exact formula for h (valid for all 0). 3

4 Solutio: The expoetial geeratig fuctio for strigs cosistig of a eve umber of 1 s (ad othig else) is (e x + e x )/. The expoetial geeratig fuctio for strigs cosistig of a eve umber of s (ad othig else) is (e x + e x )/. The expoetial geeratig fuctio for strigs cosistig of a arbitrary umber of 3 s is e x. Hece the expoetial geeratig fuctio for strigs cosistig of a eve umber 1 s, a eve umber of s, ad a arbitrary umbers of 3 s is (e x +e x )/ (e x +e x )/ e x, or 1 4 (e3x +e x +e x ). The coefficiet of x /! i this expasio is 1 4 ( ( 1) ), so the umber of such strigs is is 1 4 (3 + + ( 1) ). 8. Express S(, 1) (for 1) as a polyomial i, ad prove that your formula is valid by usig the combiatorial iterpretatio of Stirlig umbers of the secod kid. Solutio: S(, 1) is the umber of ways to divide objects ito 1 disjoit subsets. Exactly oe of these subsets must be of size, with the rest beig of size 1. Hece, such a divisio of the objects is equivalet to choosig of the objects to form a set of size (ad lettig the other elemets form sets of size 1 by themselves). Therefore S(, 1) = ( ) = ( 1)/. 9. A bato is divided ito + 1 cylidrical bads of equal legth ( 1). I how may differet ways ca the + 1 bads be colored if 3 colors are available, assumig that o two adjacet bads may be give the same color? (Two colorigs cout as the same if oe of them ca be coverted ito the other by turig the bato aroud.) Solutio: Here the group actig o the set of colorigs is just the - elemet group cosistig of the idetity elemet (do t tur the bato) ad a o-idetity elemet (do tur the bato). By Burside s Lemma, the aswer ca be writte as 1 (A + B), where A is the umber of of allowed colorigs that are fixed by the idetity operatio (that is, the total umber of allowed colorigs), ad B is the umber of of allowed colorigs that are fixed by the tur operatio. A equals 3, sice (startig at oe ed of the bato) there are 3 possible colors to use at the ed, ad at each subsequet bad there are exactly possible colors to use (amely, the two colors that differ from the oe just used). O the other had, B equals 3, sice (startig from the middle of the bato) there are 3 possible colors to use, ad at each subsequet bad there are exactly possible colors to use (where you must 4

5 make sure to use the same color for each bad ad for its mate o the other side of the bato). So the aswer is 3 (4 + ). 10. Use Burside s Lemma to cout the umber of circular 6-permutatios of the multiset { R, W, B}. Solutio: Here the group actig o the set of colorigs is the 6-elemet group cosistig of rotatios by multiple of 60 degrees. By Burside s Lemma, the aswer is 1 (A + B + C + D), where A is the umber of colorigs that 6 are ivariat uder 0 degree rotatio, B is both the umber of colorigs that are ivariat uder 60 degree clockwise rotatio ad the umber of colorigs that are ivariat uder 60 degree couterclockwise rotatio, C is both the umber of colorigs that are ivariat uder 10 degree clockwise rotatio ad the umber of colorigs that are ivariat uder 10 degree couterclockwise rotatio, ad D is the umber of colorigs that are ivariat uder 180 degree rotatio. We have A = 6!/!!! = 90 (these are just ordiary permutatios of the multiset), B = 0 (the oly way a circular 6-permutatio ivolvig R s, W s, ad B s ca be ivariat uder 60 degree rotatio is if all six letters are the same), C = 0 (for the same sort of reaso as B = 0, but more complicated: a circular 6-permutatio ivolvig R s, W s, ad B s ca be ivariat uder 10 degree rotatio oly if either some color occurs six times or two of the colors occur three times each, cotradictig our requiremet that each of the three colors occurs twice), ad D = 3! = 6 (there are 3! ways of assigig the three colors to the three pairs of diametrically opposite positios i the circular 6-permutatio, ad each of these correspods to a way of arragig two R s two B s, ad two W s). So the umber of orbits, i.e., the umber of 6-permutatios, is 1 (90 + 6) =