MATH 260 2002/2003 Assigmet 4 Due Jauary 8, 2003 Late pealty: 5% for each school day. 1. 4.6 #10. A croissat shop has plai croissats, cherry croissats, chocolate croissats, almod croissats, apple croissats ad broccoli croissats. How may ways are there to choose (a) (1 poit) a doze croissats? 12+6 1 12 = 6188 (b) (1 poit) three doze croissats? 36+6 1 36 = 749 398 (c) (2 poits) two doze croissats with at least two of each kid? To buy two from of each kid the buy the other twelve, there are 12+6 1 12 = 6188 ways. (d) (2 poits)two doze croissats with o more tha two broccoli croissats? To buy two broccoli croissats: 22+5 1 22 ; To buy oe broccoli croissats: 23+5 1 23 ; To buy o broccoli croissats: 24+5 1 24. The aswer is 22+5 1 22 + 23+5 1 23 + 24+5 1 24 =52975. (e) (2 poits)two doze croissats with at least five chocolate croissats ad at least three almod croissats? To buy five chocolate croissats ad three almod croissats first, there are 16+6 1 16 = 20 349 ways. (f) (3 poits)two doze croissats with at least oe plai croissat, at least two cherry croissats, at least three chocolate croissats, at least oe almod croissat, at least two apple croissats, ad o more tha three broccoli croissats? To buy o broccoli croissats ad satisfy all the other coditios, there are 15+5 1 15 ways; To buy oe broccoli croissats ad satisfy all the other coditios, there are 14+5 1 14 ways; To buy two broccoli croissats ad satisfy all the other coditios, there are 13+5 1 13 ways; To buy three broccoli croissats ad satisfy all the other coditios, there are 12+5 1 12 ways. The aswer is 15+5 1 15 + 14+5 1 14 + 13+5 1 13 + 12+5 1 12 =11136. 2. 4.6 #12. (3 poits) Howmaydifferet combiatios of peies, ickels, dimes, quarters, ad half dollarscaapiggybacotaiifithas20 cois i it? The umber of ways to select 20 cois from five kids is C(20+5 1, 20) = C(24, 20) = 10 626. 3. 4.6 #16. How may solutios are there to the equatio x 1 + x 2 + x 3 + x 4 + x 5 + x 6 =29 where x i, i =1, 2, 3, 4, 5, 6 is a oegative iteger wuch that (a) (2 poits) x i > 1 for i =1, 2, 3, 4, 5, 6? (b) (2 poits) x 1 1, x 2 2, x 3 3, x 4 4, x 5 5, ad x 6 6? (c) (2 poits) x 1 5? (d) (2 poits) x 1 < 8 ad x 2 > 8? 1
(a) Sice x i 2, welety i = x i 2, y i 0. x 1 + x 2 + x 3 + x 4 + x 5 + x 6 = 29 (y 1 +2)+(y 2 +2)+(y 3 +2)+(y 4 +2)+(y 5 +2)+(y 6 +2) = 29 y 1 + y 2 + y 3 + y 4 + y 5 + y 6 = 17 There are C(17 + 6 1, 17) = 26 334 solutios. (b) Let y 1 = x 1 1, y 2 = x 2 2, y 3 = x 3 3, y 4 = x 4 4, y 5 = x 5 5, y 6 = x 6 6. x 1 + x 2 + x 3 + x 4 + x 5 + x 6 = 29 (y 1 +1)+(y 2 +2)+(y 3 +3)+(y 4 +4)+(y 5 +5)+(y 6 +6) = 29 y 1 + y 2 + y 3 + y 4 + y 5 + y 6 = 8 There are C(8 + 6 1, 8) = 1287 solutios. (c) x 1 + x 2 + x 3 + x 4 + x 5 + x 6 =29has C (29 + 6 1, 29) = C (34, 29) = 278 256 solutios. Amog them, C (23 + 6 1, 23) = C (28, 23) = 98 280 solutios have x 1 6. Therefore, there are C (34, 29) C (28, 23) = 179 976 solutios have x 1 5. (d) First, we cout how may solutios are there such that x 2 9. Let y 2 = x 2 9, we fid that there are C(20+6 1, 20) = C(25, 20) such solutios. The we cout how may solutios are there such that x 1 8 ad x 2 9. Lety 1 = x 1 8 ad y 2 = x 2 9, there are C(12 + 6 1, 12) = C(17, 12) such solutios. Therefore there are C(25, 20) C(17, 12) = 46 942 solutios such that x 1 < 8 ad x 2 > 8. 4. 4.6 #22. (4 poits) How may positive itegers less tha 1,000,000 have exactly oe digit equal to 9 ad have a sum of digits equal to 13? Let the digits be x 1 x 2 x 3 x 4 x 5 x 6. Suppose that x 1 =9, the x 2 +x 3 +x 4 +x 5 +x 6 =4.There are C (4 + 5 1, 4) = C (8, 4) = 70 solutios to that equatio. So there are 70 such umbers with the first digit beig 9. It is the same for the other digits. Therefore the aswer is 6 70 = 420. 5. 4.6 #28. (3 poits)howmaydifferet strigs ca be made from the letters i AARDV ARK, usig all the letters, if all three As must be cosecutive? Sice there are two Rs, oe D, oev,oek ad three As i oe block (therefore the As will be couted as oe object), there are 6! 2!1!1!1!1! =360 such strigs. 6. 4.6 #38. (3 poits) I bridge, the 52 cards of a stadard deck are dealt to four players. How may differet ways are there to deal bridge hads to four players. Notice that the players are distict. (I bridge, they are called East, South, West ad North.) There are C (52, 13) ways to choose a had for East, the C (39, 13) ways to choose a had for South, C (26, 13) ways to choose a had for West ad C (13, 13) ways to choose a had for North. The umber of ways to deal the four hads is 52! 13!39! 39! 13!26! 26! 13!13! 13! 13!0! = 52! = 53 644 737 765 488 792 839 237 440 000 13! 13! 13! 13! 7. 4.6 #40. I how may ways ca a doze books be placed o four distiguishable shelves (a) (2 poits) if the books are idistiguishable copies of the same title? We ca use 12 dots to represet the books ad 3 bars to separate them ito four groups (to be placed o the four shelves). There are C (12 + 3, 3) = C (15, 3) = 455 such arragemets. 2
(b) (4 poits)ifotwobooksarethesame,adthepositiosofthebooksotheshelvesmatter? (Hit: Break this ito 12 tasks, placig each book separately. Start with the sequece 1,2,3,4 to represet the shelves. Represet the books by b i,i=1, 2,..., 12. Place b 1 to the right of oe of the terms i 1, 2, 3, 4. The successively place b 2,b 3,..., ad b 12.) Thereare4 ways to place b 1, 5 ways to place b 2,..., 15 ways to place b 12.Theasweris 4 5 6 7 8 9 10 11 12 13 14 15 = 217 945 728 000 8. 5.1 #10. A employee joied a compay i 1987 with a startig salary of $50,000. Every year this employee receives a raise of $1000 plus 5% of the salary of the previous year. (a) (2 poits) Set up a recurrece relatio for the salary of this employee years after 1987. (b) (2 poits) What is the salary of this employee i 1995? (c) (4 poits) Fid a explicit formula for the salary of this employee years after 1987. (a) a = a 1 +0.05a 1 + 1000 = 1.05a 1 + 1000 for all 1 ad a 0 =50, 000 (b) The salary of this employee i 1995 is a 8 = 1.05a 7 + 1000 = 1.05 (1.05a 6 + 1000) + 1000 = 1. 102 5a 6 +2050 = 1. 102 5 (1.05a 5 + 1000) + 2050 = 1. 157 6a 5 + 3152. 5 = 1. 157 6 (1.05a 4 + 1000) + 3152. 5=1. 215 5a 4 +4310. 1 = 1. 215 5 (1.05a 3 + 1000) + 4310. 1=1. 276 3a 3 +5525. 6 = 1. 276 3 (1.05a 2 + 1000) + 5525. 6=1. 340 1a 2 +6801. 9 = 1. 340 1 (1.05a 1 + 1000) + 6801. 9=1. 407 1a 1 +8142.0 = 1. 407 1 (1.05a 0 + 1000) + 8142.0 =1. 407 1 (1.05 50000 + 1000) + 8142.0 = 83422. (c) This is a ohomogeeous recurrece relatio. For the homogeeous recurrece relatio the geeral solutio is a (h) a =1.05a 1 = k 1 (1.05). Sice F () = 1000, we guess q 0 =1.05q 0 + 1000. = q 0. We have Therefore q 0 = 20000. The geeral solutio for the ohomogeeous recurrece relatio is a = a (h) + = k 1 (1.05) 20000. Whe =0, this is a 0 = k 1 20000 = 50000. So k 1 = 70000. The formula is 9. 5.1 #22. a = 70000 (1.05) 20000. (Whe =8, this formula gives a 8 = 70000 (1.05) 8 20000 = 83422.) 3
(a) (3 poits) Fid a recurrece relatio for the umber of ways to climb stairs if the perso climbig the stairs ca take oe, two, or three stairs at a time. Sice the last step oe ca take oe, two or three stairs, a = a 1 + a 2 + a 3. (b) (1 poit) What are the iitial coditios? a 0 = a 1 =1,a 2 =2.Or,a 1 =1,a 2 =2ad a 3 =4. (c) (1 poit)howmayways cathispersoclimbaflight of eight stairs? a 3 = a 0 + a 1 + a 2 =1+1+2=4 a 4 = a 1 + a 2 + a 3 =1+2+4=7 a 5 = a 2 + a 3 + a 4 =2+4+7=13 a 6 = a 3 + a 4 + a 5 =4+7+13=24 a 7 = a 4 + a 5 + a 6 =7+13+24=44 a 8 = a 5 + a 6 + a 4 = 13 + 24 + 44 = 81 10. 5.1 #38. (6 poits) Show that the Fiboacci umbers satisfy the recurrece relatio f =5f 4 + 3f 5 for =5, 6, 7,..., together with the iitial coditios f 0 =0,f 1 =1,f 2 =1,f 3 =2, ad f 4 =3. Use this recurrece relatio to show that f 5 is divisible by 5, for =1, 2, 3,... The origial defiitio of the Fiboacci umbers gives f = f 1 + f 2 for all 2 ad f 0 =0, f 1 =1. It is straightforward to verify that f 2 =1,f 3 =2, ad f 4 =3. Sice f 3 = f 4 +f 5, f 2 = f 3 + f 4,f 1 = f 2 + f 3, we have f = f 1 + f 2 =(f 2 + f 3 )+f 2 =2f 2 + f 3 = 2(f 3 + f 4 )+f 3 =3f 3 +2f 4 = 3(f 4 + f 5 )+2f 4 =5f 4 +3f 5. Next we prove P () :f 5 is divisible by 5 for all 1 by mathematical iductio. Sice f 5 = f 3 + f 4 =5, P (1) is true. We assume that P () is true, i.e., f 5 =5k for some iteger k. The f 5(+1) = f 5+5 =5f 5(+1) 4 +3f 5(+1) 5 =5f 5+1 +3f 5 =5f 5+1 +15k =5(f 5+1 +3k) which is divisible by 5. Therefore P (N +1)is true. 11. 5.2 #4 (3 poits each part) Solve the followig recurrece relatios toghether with the iitial coditios give. (a) a = a 1 +6a 2 for 2, a 0 =3,a 1 =6. (a) The characteristic equatio is x 2 x 6 = 0 (x +2)(x 3) = 0 The geeral solutio is a = k 1 ( 2) + k 2 3. Solvig k 1 + k 2 =3, 2k 1 +3k 2 =6 we have k 1 = 3 5 ad k 2 = 12 5. So a = 3 5 ( 2) + 12 5 3. (c) a =6a 1 8a 2 for 2, a 0 =4,a 1 =10. (a) The characteristic equatio is x 2 6x +8 = 0 (x 2) (x 4) = 0 4
The geeral solutio is a = k 1 2 + k 2 4. Solvig k 1 + k 2 =4, 2k 1 +4k 2 =10 we have k 1 =3ad k 2 =1. So a =3 2 +4. (f) a = 6a 1 9a 2 for 2, a 0 =3,a 1 = 3. (a) The characteristic equatio is r 2 +6r +9=(r +3) 2. The root is r = 3 with multiplicity two. The geeral solutio is a =(k 1 + k 2 ) ( 3) With the iitial coditios, we have k 1 = 3 (k 1 + k 2 )( 3) = 3. Solvig this system of equatios, we have k 1 =3ad k 2 = 2. Therefore, a =(3 2)( 3). (g) a +2 = 4a +1 +5a for 0, a 0 =2,a 1 =8. (a) The characteristic equatio is x 2 +4x 5 = 0 (x +5)(x 1) = 0 The geeral solutio is a = k 1 ( 5) + k 2 1. Solvig k 1 + k 2 =2, 5k 1 + k 2 =8 we have k 1 = 1 ad k 2 =3. So a = ( 5) +3. 12. 5.2 #8. A model for the umber of lobsters caught per year is based o the assumptio that the umber of lobsters caught i a year is the average of the umber caught i the two previous years. (a) (2 poits) Fid a recurrece relatio for {L }, where L is the umber of lobsters caught i year, uder the assumptio for this model. L = 1 2 L 1 + 1 2 L 2. (b) (3 poits)fidl if 100, 000 lobsters were caught i year 1 ad 300, 000 were caught i year 2. The characteristic equatio is r 2 1 2 r 1 2 = 0 1 (2r +1)(r 1) = 0 2 The roots are r = 1 2, ad r =1. The geeral solutio is a = k 1 1 2 + k2. Cosiderig the iitial coditios, we have µ 1 k 1 + k 2 2 = 100000 1 4 k 1 + k 2 = 300000. Solvig this system of equatios, we have k 1 = 800000 3,k 2 = 700000 3 ad a = 800000 3 1 + 700000 2 3. 5
13. 5.2 #12. (4 poits) Fid the solutio to a =2a 1 + a 2 2a 3 for =3, 4, 5,..., with a 0 =3, a 1 =6ad a 2 =0. The characteristic equatio is r 3 2r 2 r +2 = 0 (r 1) (r 2) (r +1) = 0 The roots are r =1,r=2ad r = 1. The geeral solutio is Cosiderig the iitial coditios, we have a = k 1 + k 2 2 + k 3 ( 1). k 1 + k 2 + k 3 = 3 k 1 +2k 2 k 3 = 6 k 1 +4k 2 + k 3 = 0 Solvig this system of equatios, we have k 1 =6,k 2 = 1,k 3 = 2 ad a =6 2 2( 1). 14. 5.2 #18. (5 poits) Solve the recurrece relatio a =6a 1 12a 2 +8a 3 with with a 0 = 5, a 1 =4,ada 2 =88. The characteristic equatio is r 3 6r 2 +12r 8 = 0 (r 2) 3 = 0 The root is r =2with multiplicity 3. a = k 1 + k 2 + k 3 2 2. Substitute i the iitial coditios, we have k 1 = 5 2k 1 +2k 2 +2k 3 = 4 4k 1 +8k 2 +16k 3 = 88 The solutio is: k 1 = 5,k 2 = 1 2,k 3 = 13 ª 2. Therefore a = 5+ 1 2 + 13 2 2 2. 15. 5.2 #26. What is the geeral form of the particular solutio of the liear ohomogeeous recurrece relatio a =6a 1 12a 2 +8a 3 + F() if (a) (1 poit) F () = 2? We use the otatios as i Theorem 6 of 5.2. t =2ad s =1for F () = 2. The characteristic equatio of the homogeeous relatio is r 3 6r 2 +12r 8 = 0 (r 2) 3 = 0 The oly root is r =2with multiplicity three. Sice 1 is ot a root, the particular solutio has the form = p 2 2 + p 1 + p 0 (b) (1 poit) F () =2? Ithiscaset =0ad s =2, which is the root with multiplicity 3. The particular solutio has the form = p 0 3 2. (c) (1 poit) F () =2? = 3 (p 1 + p 0 )2. 6
(d) (1 poit) F () =( 2)? (e) (1 poit) F () = 2 2? (f) (1 poit) F () = 3 ( 2)? (g) (1 poit) F () =3? = p 0 ( 2). = 3 p 2 2 + p 1 + p 0 2. = p 3 3 + p 2 2 + p 1 + p 0 ( 2). = p 0 16. 5.2 #30. (6 poits) (a) Fid all solutio s of the recurrece relatio a = 5a 1 6a 2 +42 4. (b) Fid the solutio of this recurrece relatio with a 1 =56ad a 2 =278. (a) The characteristic equatio is The roots are r = 2 ad r = 3. Wehave a (h) The particular solutio has the form r 2 +5r +6 = 0 (r +3)(r +2) = 0 = k 1 ( 2) + k 2 ( 3) Substitute it i the recurrece relatio, we have Therefore the geeral solutio is = q 0 4 q 0 4 = 5 q 0 4 1 6 q 0 4 2 +42 4 16 q 0 = 5 q 0 4 6 q 0 +42 16 42 q 0 = 42 16 q 0 = 16 = 16 4 (b) Substitutig i the iitial coditios, we have The solutio is: {k 1 =1,k 2 =2}. Therefore a = k 1 ( 2) + k 2 ( 3) +16 4. 2 k 1 3 k 2 +64 = 56 4 k 1 +9 k 2 + 256 = 278 a =( 2) +2 ( 3) +16 4. 7
17. 5.2 #34. (6 poits) Fid all solutios of the recurrece relatio a =7a 1 16a 2 +12a 3 + 4 with a 0 = 2, a 1 =0, ad a 2 =5. The characteristic equatio for the homogeeous recurrece relatio is r 3 7r 2 +16r 12 = 0 (r 3) (r 2) 2 = 0 The roots are r =3with multiplicity 1 ad r =2with multiplicity 2. a (h) = k 1 3 +(k 2 + k 3 )2 =(p 1 + p 0 )4 (p 1 + p 0 )4 =7(p 1 ( 1) + p 0 )4 1 16 (p 1 ( 2) + p 0 )4 2 +12(p 1 ( 3) + p 0 )4 3 + 4 Divide both sides with 4 3, we have (p 1 + p 0 )4 3 = 7(p 1 ( 1) + p 0 )4 2 16 (p 1 ( 2) + p 0 )4+12(p 1 ( 3) + p 0 )+4 3 64p 1 +64p 0 = (60p 1 +64)+60p 0 20p 1 Comparig the coefficiets of adthecostattermsofbothsides,wehave 64p 1 = 60p 1 +64 64p 0 = 60p 0 20p 1. Solvig that system of equatios, we have p 0 = 80, p 1 =16ad By lettig =0, 1, 2, wehave ad k 1 =61,k 2 =17,k 3 = 39 2. Therefore = (16 80) 4 a = k 1 3 +(k 2 + k 3 )2 +(16 80) 4 k 1 + k 2 80 = 2 3k 1 +2k 2 +2k 3 256 = 0 9k 1 +4k 2 +8k 3 768 = 5 a =61 3 + µ17 + 392 2 +(16 80) 4 8