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C h a p t e r 4 : C o y n t i n g M e t h o d sL e s s m i 4 1: C - M n i i o g P r i n c i p l e s ,p a g e 2 3 5khaki blackred red/khaki red/blackblue biue/khaki blue/blackgreer green/khaki green/blackEach X represents a different combination. There are6 xs; therefore, there are six different vanations of theoutfit to choose from,b) The number of outfit variations, O, is related to the-"r of shirts and the number of shorts;O .number of shirts) • (number of shorts)2Ml I are six different variations of the outfit tochoose from. This matches the part a) result.2. a)UpholsteryleatherclothColourred•.il.-. rred1.1,1 It•.V-xt.-8 uThereto".• thf ro 0 Mpholstery-colour choices thatare available.b) The number of upholstery-colour choices, U, isrelated to the number of colours and the number ofkinds of upholstery;U - (Ml mber of colours) • (number of upholstery)iJ 1 28There are 8 upholstery-colour choices that areavailable. This matches the part a) result.3. a) The Fundamental Counting Pnnciple does notapply because tasks in this situation are related bythe word OR.b) The Fundamental Counting Pnnciple does applybecause tasks in this situation are related by the wordA N D .c) The Fundamental Counting Pnnciple does notapply because tasks in this situation are related bythe word OR.d) The Fundamental Counting Pnnciple does applybecause tasks in this situation are related by the wordA N D .4. a)Game 1winGame 2 Game 3 Series Resultwinlosswinlosswinwinlosswinlossloss lossb) By looking at the tree diagram. I can see there are2 ways in which Kims team can win the seriesdespite losing one g a m e .5. The number of colour-size vanations, C, is relatedto the number of colours and the number of sizes;C = (number of colours) • (number of sizes)C = 5 - 4C - 20There are 20 colour-size variations that are available.6. The number of computer systems, S, theemployees can build for their customers is related tothe number of desktop computers (dc), the number ofmonitors (m), the number of printers (p), and thenumber of software packages (sp);- (# of y ) (# of m) • (# of p) • (# of sp)^ S 4 f. 3.5 -• .}uOrhfciufo;u, Ihe employees can build 360 differentcomputer systems for their customers.7. The number of possible meals. M, is related to thenumber of soups (s), the number of sandwiches (sw), thenumber of drinks (dr), and the number of desserts (P):M = (# of s) - (# of sw) • (# of dr) • (# of d)/W = 3 5 • 4 2M= 120Therefore, there are 120 different meal possibilities.8. Event A: Selecting a rap C DOREvent B: Selecting a classic rock C Dn{A u B) = n{A) + n{B)n{A ;„„ B) = 8 + 10n{A u B) = 18Therefore, Chadene can select from 18 C D s to play inher car stereo that will match T o m s musical tastes.9. a) The number of different PIN combinations, C. isrelated to the number of digits from which to select foreach digit of the PIN. F:C = P l • P2--P3- P4- PsC = 9 9 9 9 9C = 59 049There are 59 049 different five-digit PIN combinations.F o u n d a t i o n s of Mathemati. s «/ solutions Manual 4-1
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b) llv: niirnlHjro! <iilferont PIN cfmibinations, N. )sn.lalifl to llu iiiinibcr ol digits Itorn w h i r j i to st;lt)rj foreach diyil ol llir; PIN. DN /)i n, / ) ; Dt DiN M p. / 0 5W 15 17f)IhtjK.- or<: only 1f» 120 difforoni livc-ffigit PINf;urnl)inolif)iis in w h i d i the digits cannot repeal.10. f ho n(jrnt)t;r of different t)ytes that can bo created,N. IS rolatfHl lo Ifie number of digits from which tof.hoosf! for each ditji! of the byte. B:N - th H • / i . Bi Bs • Bf, B, By.N - 7 2 2 2 2 2 2 - 2N • 256Thf.Tutonj. 2b6 different byles can bo creaUid1 1 . a) 1 he fiurntwf of different upper-~caso letterf.ossibilitips. H. IS related to the number of upper-caseletters ftom which to choose for each odd position ofthe countrys postal code, P:W - Pi P ; P .H 26 • 26 26A/-- 1 / 57t>Tlie number of different digit possibilities. D, is relatedto the number of digits from which to choose for eacheven position of the countrys postal code, P;D^P- Pi Pf.D---^10-10 10O 1000The number of different postal codes that are possiblein this country. C. is related to the number of upper-case letter possibilities, N, and the n u m b e r o f digitpossibilities, D:C W • DC = 17 576 • 1000C ^ 17 576 000Tticrefore, 1 7 576 000 postal codes are possible,b) Tlie number of different upper-case letterpossibilities, N. is related to the number of upper-caseletters from which to choose for each odd position ofthe countrys postal code, P;Al = P, Po • Pf,A / - 2 1 - 2 1 - 2 1N = 9261The number of different digit possibilities, D, remainsthe same since all digits can be used. The number ofdifferent postal codes that are possible in Canada, C, isrelated io the number of uppercase letter possibilities,Ay. and the number of digit fK)SSibi!ities, D;C = N D0 = 9261 1000C = 9 261 000Therefore. 9 261 000 postal codes are possible inC a n a d a .12. T o answer this question, I need U) determine howmany digit r-.oml.nnations ttioro arc for the last fourdigits of one ot these two phone n u m l j e i s . and thenmultiply it by 2,The number ot digit combinations. C, is rolnted to thenumber of possible digits for each of the last fourdigits of one of tho phono numbers, Pc p , . p... p.,. f>,C = 10 - 10 - 10 - 10C-^ 10 000The number of phone numtjers is 2C since there an)two given tt.nnplates for the phone numbers in thoquestion.2 C 2(10 000)2 C =^ 20 000Therefore. 20 000 different phone numbers arepossible for this town,13. The number of different codes, C. is related tonumber of positions from which to select for eachswitch of the garage door opener. G:C^Gi- G2- Ga • G,, • Gr, • Gr, • G, - Gr; G:.C = 3 • 3 - 3 - 3 ^ 3 • 3 • 3 - 3 3C - 19 683Therefore, 19 683 different codes are possible.14. Event A; Selecting a pickup truck O REvent B: Selecting a passenger van O REvent C: Selecting a car O REvent D: Selecting a sports utility vehiclen{A UBKJC u D) = n{A) + n(B) + n{C) + n(D)n{A u B u C i.j D) = 8 + 10 + 35 + 12niA uBu G>j D) = 65Therefore, a customer has 65 choices w h e n rentingjust one vehicle.15. a) Multiply the number of sizes of the crust, by thenumber of types of the crust, by the number of typesof cheese, by the number of types of tomato sauce.2 • 2 - 2 • 2 = 16Multiply this number by the number of differenttoppings.1 6 - 2 0 = 320Therefore, there are 320 different pizzas that can bemade with any crust, cheese, tomato sauce, and1 topping.b) Multiply the number of types of cheese by thenumber of types of tomato sauce.2 - 2 = 4Therefore, there are 4 different pizzas that can bemade with a thin whole-wheat crust, t o m a t o sauce,cheese, and no toppings.4-2
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1 i . a | The number of different ypper-^case letterpossibilities, N, is related to the number of upper^caseletters from which to choose for each of the first threepositions of the Alberta licence plate, P:W = 24 - 24 • 24N= 13 824fl... number of different digit possibilities, O, is relatedso In. - number of digits from which to choose for each ofH..- I . r i three positions of the Alberta licence plate. PD - P.-P^^Ps- ••0 - 10 - 10P m o oThe number of different possible Alberta licence plates,I = -I J ! the number of upper^case letter, / ; and the number of digit possibilities, O;^ / i O< , i , I f . 0 01 ; i.ji CO. Alberta licence plates are possible,b) I he number of different upper-case letterpossibilities, N, remains the s a m e since the number ofletters in the plates and the number of letters that canbe used is the s a m e as in a).The number of different digit possibilities, 0 , is relatedto the number of digits from which to choose for each ofthe last four positions of the Alberta licence plate. P;D = P4 • Ps • Fe • P?D = 10 • 10 • 10 - 10D = 10 000The number of different possible Alberta licence plates,C, is related to the number of upper-case letterpossibilities. N, and the number of digit possibilities, D:C = N- DC = 13 824 - 10 000C = 138 240 000138 240 000 13 824 000 = 124 416 000So, 124 416 000 more licence plates are possible,17. e g,, If multiple tasks are related by A N D , it meansthe Fundamental Counting Pnnciple can be used andthe total number of solutions is the product of thesolutions to each task. For example, A 4-digit PINinvolves choosing the 1st digit A N D the 2nd digit A N Dthe 3rd digit A N D the 4th digit. So the number ofsolutions IS 10 10 10 10 = 10 000. O R means thesolution must meet at least one condition so you mustadd the number of solutions to each condition, andthen subtract the number of solutions that meet allconditions. For example: Calculating the number of4-digit PINs that start with 3 O R end with 3. Thesolution IS the number of PINs that start with 3, plusthe number of PINs that end with 3. minus the numberof PINs that both start and end with 3:1000 + 1 0 0 0 - 100 = 1900.18. a) i) Event A Drawing a king O REvent B; Drawing a queenrif/. = n(A) + niB)n{A u e ) = 4 + 4n{A u e ) = 8Likelihood = ^52Likelihood = ^13Therefore, there is a 2 in 13 chance that a king or aqueen will be drawn.ii) Event A: Drawing a diamondOREvent B: Drawing a clubI >•"• P) -•./• I Ip.)n{A u S) = 26Likelihood = ^2Therefore, there is a 1 m 2 chance that a diamond ora club will be drawn,iii) Event A: Drawing an Ace O REvent B: Drawing a spaden(A u B) = n{A) + n(B) - n{A n B)n(Au B) = 4 +13^1n(A u e ) = 16Likelihood = ^52Likelihood =13Therefore, there is a 4 in 13 chance that an ace or aspade will be drawn.b) No. e.g.. because the Fundamental CountingPrinciple only applies w h e n tasks are related by theword A N D .19. e.g.. T o begin, there are 90 two-digit numbers.There are 10 with a 1 in the tens column, 10 with a 2in the tens column, and this pattern continues until Ireach the 10 with a 9 in the tens column. Next. I mustdetermine the numbers that are divisible by either 2 or5. I know that every other number is even and thusdivisible by 2. This means that or 45 of the two-2digit numbers are divisible by 2. The two-digitnumbers that are divisible by 5 can be found bystarting at the first two-digit number, 10, and thencounting by 5 until I get to a three-digit number.By doing this. I can determine that the two-digitnumbers that are divisible by 5 are: 10, 15, 20, 25. 30.35, 40, 45. 50. 55, 60. 65, 70. 75, 80, 85. 90. 95.There are 18 of them. I see that half of themor 9 areeven and thus divisible by 2.F o u n d a t i o n s of Mathemati ^ V -..lutions Manual4-3
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Therefore, there are 9 numbers that are divisible by 5and not by 2. If I add this together with the number oftwo-digit numbers that are divisible by two (45). I seethat there are 54 two-^digit numbers divisible by 2 or 5,Whatever is leftover from the two digit numbers arethe ones that are not divisible by either 2 or 5, Thisamount is:90 54 = 36. Thus, there are 36 two-digit numbersthat are not divisible by either 2 or 5.20. The number of different outcomes for a studentstest. N, is related to the number of possible answersfor each question on the test. T:W = Tl • 12 • Ta • T4 - Ts - Te • T/ • Te • Tg • TnIV = 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2IV = 1024A perfect score is only 1 out of these 1024 outcomes;therefore, there is a 1 in 1024 chance that the studentwill get a perfect score.2 1 . This question is solved by constant application ofthe Fundamental Counting Pnnciple.If an item from each category is selected:0 = 3 5 4 20 = 120If no soup is selected:0 = 5 4 20 = 40If no sandwich is selected:0 = 3 4 20 = 24If no drink is selected:0 = 3 5 20 = 30If no dessert is selected:0 = 3 5 - 4O = 60If no soup or sandwich is selected:0 = 4 20 = 8If no soup or drink selected:0 = 5 20 = 10If no soup or dessert is selected:0 = 5 40 = 20If no sandwich or dnnk is selected:0 = 3 20 = 6If no sandwich or dessert is selected:0 = 3 40 = 1 2If no dnnk or dessert is selected:0 = 3 50 = 1 5If only n <^nur s a n d w i r h drink or dessert is selected:0 • : 5, 4 2f,.,,., - 120 4 40 t ;•-•} t M) . 60 + 8 + 10 + 20 + 6 + 12-I 1 h 4 3 < S i- 4 • 2,1 - 3501 herefure, 359 meals are possible if you do not haveto choose an item from a category.L e s s o n 4 . 2 : I n t r o d u c i n g P e r m u t a t i o n s a n dF a c t o r i a l N o t a t i o n , p a g e 2 4 31. a) 6! = 6 - 5 - 4 • 3 - 2 - 161 = 720b) 9 - 8 ! = 9 - ( 8 - 7 - 6 - 5 - 4 - 3 - 2 - l )9 - 8 ! = 9 - 4 0 3 2 09 8! 3628803 7 I^ 2 1. 5!c) — =• 3!53!5!3!5!3!5!3!8!— = 5 43 2 15 - 4 . M3!^ = 5 - 4 . 1-20a /• b j j 4 J 2 1f b 5 4 3 2 r7 6 <.i 4 3^y 8/ 6717!2 18 18!7!8!7!8!7!7!e) 3 ! - 2 ! = ( 3 - 2 - l ) - ( 2 - l )3 ! - 2 ! = 6 - 23 t - 2 ! = 129! ^ 9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1 413! ^ l 4 ^ 3 ^ 2 l H 3 ^ ^ ^ ^9 ^ 9 8 7 g g 4 3 2 1413! 3 2 1 4 3 2 14 ! 3 ! 3 2 4!413! 3 29!413!9!413!= 3 - 4 - 7 - 6 - 525204-4C h a p t e r 4: C o u n t i n g Methods
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PosstioJ! Ptji^ilmn *^(jsitifmIfirmii.itKtti1P e r m i H a t i o i i2P e r m i i f a t i o r i3K.IPermutattor*4Permutat,ir,n5Permutation6Raj Sarah Kenb) Let L represent the total number of permutations:1 = 3 - 2 - 1L = 3!3. .b)D !c)15 H i j 1. 1 :A < 2 ! 4 :*,IS M I Il!4 J 2 1 l^t-lf98!1 0 0 - 9 9 =100!98!4. Expressions a), c), and d) are undefined becausefactorial notation is only defined for natural numbers,5. a) 8-7-»^d S / ^1 38 - 7 - 6 f H / / X I8 - 7 - 6 ! = 5 6 - 7 2 08 - 7 - 6 ! - 4 0 3 2 0h i 1?1 ^ 1 I 10 y 3 7 fi 5 4 / 110! ^ 1 0 " F 8 " " / 6 5^4 3^2^ 11 2 ! ^ -12 ^-1 10 9 8 7 6 5 4 3 2 110!" 10 9 8 7 6 5 4 3 2 11 ^ = 1 2 1 1 . 1 ^10! 10!12!10!12!10!= 12-11-11328! 8 ^ 6 !2 ! - 6 ! 2 6!2 ! - 6 ! 2: 4 - 728d)8!2 ! - 6 !8!21-6!/ • r ,5! " •)5!5! 5!7 ^5!7 - 6 !5!42e)9! 91 i;1 .11 % 4 <2 12 i^1 ;6!2 ! - 2 !6!2 ! - 2 !6!= 4 ( 3 - 5 - 4 - 3 )= 4(180)4 1 = 7202 ! - 2 ! ;f| 4 ! + 3 ! + 2 ! + 1 ! = ( 4 - 3 - 2 - l ) - f ( 3 - 2 - l ) + ( 2 - l ) + 14 ! + 3 ! + 2 ! + 1 ! = 24 + 6-f 2 + 14!-f 3 ! + 2 ! + 1 ! = 33e . a ) ^ . M ! L - ; ) ( " - ^ ) ( " f ) ; - ( ^ ) ( ^( n - 1 ) ! ( n - l ) ( n - 2 ) ( „ - 3 ) . . . ( 3 ) ( 2 ) ( l )n!( n - 1 ) !F o u n d a t i o n s of Mathematics V/ Solutions IVIanual
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b) , •• + 2|!1,1 41b; . ,.({n + 2){n + i)(n)in^i}...(3)i2}{tin-. 41( o . 1)1 ( / n n i n p 1)(» ? | (3)(2)(1)m in)(n ~1](n ?] l 3 ) ( 2 ) ( l )p i i 1!c)/: ^ 1 ^ l i y ^ P 4 | ( n ^ 5 ) . . . ( 3 ) ( 2 ) ( l )i|! (/) 3)(r, 4 H r r : i ^ 3 ) ( 2 ) ( i r,1 ^ ^ ( n ) ( r i ^ l ) ( / i 2)(,/ rfn-Z]Ie)ini, i 5) n I ^ 4 J ( f n 3 } ( f ) j 2 ) ( u + l).-.(3)(2)(l)( , 1 - 3 ) ! " " ^ " ! o , 3 ) ( i i r 2 ) ( o i 1 i (3)(2){1)( . 0 , 5 ) {n , 5)(/j , 4)(// ,(/H 111 - 3.ffl 5)1f n .4)!^ = (ri + 5)(/i + 4)= f)2 + 9o + 20[n I f . { V i ^ 2 ) ( n - ^ 3 ) ( 0 ^ 4 ) . . . { 3 ) ( 2 ) ( 1 )^> ( . r i , r ( , r i ) ( ; ^ r 2 ) ( ^ ^(rt 1)! ( n - l ) { n - - 2 ) !( » ^ 2 ) ! ^ 1(/? 1 ) ! ^ n - - 17. There are nine students in the lineup, so there arenine possible positions. Let L represent the totalnumber of permutations:L = 9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1L = 9!l = 9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1L = 72 • 7 • 30 • 4 • 6L = 72 • 210 • 241 = 362 880There are 362 880 permutations for the nine studentsat the Calgary Stampede.8. There are five students in the club and there arefive possible positions. Let L represent the totalnumber of permutations:L = 5 4 3 2 1// - -b 4 3 • 2 • 1- 2 0 - 6/ - 120There are 120 different w a y s to select m e m b e r s forthe five positions.9. There are six activities to do and there are six days.Let L represent the total number of permutations:hi. H 5 - 4 • 3 - 2 • 1I • : i . 4 6• 1 / 0 - 6/ - 7-01 ht;ru are 720 different w a y s they can sequenceihos(< activities over the six days.10. There are 28 movies, so there are 28 possiblespots for the movies to go. Let L represent the totalnumber of permutations:L = 28!L = 3.048... X 10^®There are about 3.05 x 10^® possible permutations ofthe movie list.fl!(n + l ) ( n ) ( i i ^ l ) . . . ( 3 ) ( 2 ) ( l )(Vr+l)(n!)nl1 1 . a) 1010- - 1010Check n = 9LS RS(y 1 l l ! 109110!9!10 99!10There is one solution. n = 9.4-6 C h a p t e r 4" C o u n t i n g Methods
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b)(11 + 2)! = 9n!( i i K o ^ i ) : . ( 3 p ) ( i )(o + 2)(ii + l)(ii!)( n + :+ n + 2n + 2 = 6n + 3n + 2 = 6n^ + 3 r i - 4 = 0(n + 4 ) ( f i - l ) = 0ri + 4 = 0 or f i - 1 = 0n = - 4 II = 1Check n =LS"-_4 2 ] !s undefinedCheck n = 1LS RS< i 1 2)1 (.1!3!113 - 2 - 1 !1!3 26There is one solution, ri = 1.c ,t z 3 ^ " - ) ( ^ ^ 3 ) . . . ( 3 ) ( 2 ) : 1 i{n 2](n J)...(3)(2)(1)( n ^ 2 ) in-^1 = 8n = 9RS126126- 1 2 68!71i l l :7!8There is one solution, n = 9,CJ) ±7li . 1,:3 ( o + l ) ( r i ) ( n ^ l ) ( o ^ 2 ) , „ ( 3 ) f 2 ) ( l ). - h { / . ^ 2) { 3 ) ( 2 i r i )3 ( o + l ) ( o ) ( i T ^3 { i i + l ) ( n ) = 1263 ( n + n ) = 1263(f,2 + n)-^ 1 2 6 - 03 ^ ( n ^ + n ) ^ 4 2 l = 03 ( f i + n - 4 2 ) = 03 ( n + 7 ) ( r 7 ^ 6 ) - 0fi + 7 - 0 o r n - 6 = 0n = -7 n - 6Check n = - 7LS RS3 ( ^ 7 + 1)! 126( ^ 7 - 1 ) 18- j - ^ y - IS undefinedC h e c k n = 6LS RS3 ( 6 + 1)1 126( 6 ^ 1 ) !3(7!)5!3 - ( 7 - 6 - 5 ! )5!3 - 7 - 6126There is one solution, n = 6.F o u n d a t i o n s o f M a t h e m a t i c s 12 S o l y t i o n s M a n u a l 4-7
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L • r p i e s u iil till niI a / (, L 4I1 - 8 / h H 1L 8 A? 2012. Ihcu: .Ko »nht more players left to organize soIt (OH- c>ifih! iriore spots left in the batting order. Letmber of permutations:3 2 13 - 2 - 1L - :VAb 120L = 40 320There are 40 320 possible batting orders.13. There are 7 possible digits to use and there are 7digits in each serial number. Let L represent thenumber of permutations:1 = 7 - 6 - 5 - 4 - 3 - 2 - 11 = 7!There are 7! possible serial numbers. This makes sensebecause, e.g., the integer in the factorial (7 in this case)for the number of permutations is normally equal to thenumber of spots in which there are things to place. Thereare seven spots in the serial number so this means thatthe number of permutations should be 7! which matchesthe answer that w a s found.14. There are 5 cars to be arranged between theengine and the caboose so there are 5 spots in whichthe cars can be lined up. Let L represent the numberof permutations:1 = 5 4 3 2 11 = 5!1 = 5 4 - 3 2 11 = 20 61 = 120There are 120 w a y s for the cars to be arrangedbetween the engine and the caboose.15. There would be 7 c h u c k w a g o n s behind Brants sothere are 7 spots where the other dnvers could finish.Let 1 represent the number of permutations:l = 7 - 6 - 5 - 4 - 3 - 2 - 11 = 7!l = 7 - 6 - 5 - 4 - 3 - 2 - 11 = 42 • 20 61 = 42 - 1201 = 5040If Brants w a g o n wins, there are 5040 different ordersin which the eight chuckwagons can finish.16. a) e.g.. Y K O N U , Y U K N O , Y K N O Ub) There are five letters so there are five spots to put theletters. Let 1 represent the number of permutations:1 = 5 - 4 3 2 11 = 5!There are 5! possible permutations. This makessense because e.g.. the integer in the factohal (5 inthis case) for the number of permutations is normallyequal to the number of spots in which there are thingsto place. There are five spots to place the letters sothis means that the number of permutations should be5! which matches the answer that w a s found.17. a) e.g.. Using tnal and error. I have the followingcalculations:1! = 1,2^ = 2 ; 2! = 2, 2" = 4;3! = 6. 2 = 8; 4! = 24. 2 = 16I notice that for n = 4, nl is greater than 2". Thiscontinues for n > 4 because 2** will keep gettingmultiplied by 2. while 4! will keep getting multiplied bynumbers greater than 2 to obtain the higher factonals.b) e.g.. Using what I have in a), I know that for n < 4,nl IS not greater than 2". The calculations for thesevalues of n are s h o w n in a). Thus for n = { 1 , 2 . 3}. nl isless than 2".18. e.g., First, figure out how m a n y w a y s Dadene andArnold can be placed next to each other in the line.Thin cnn ho found using a t i b l oA r n o l d i234D a r l o n c16From the table I can see that there are 18 differentw a y s for Dadene and Arnold to be placed next toeach other in the line. For every one of those18 ways, there are 8 other dancers to be placed in8 different spots in the line. Let 1 represent thenumber of permutations:6 5 4 3 2- 46)3 - 21)1)1 - 1-3(8 - 7I 18(8!)1 18(8 7 6 5L • 18(8 • 42 20I, = 18(336 • 120)1 - 13(40 320)/ - / 2 5 760There are 725 760 possible arrangements of thedancers for the Red River Jig.4-8 C h a p t e r 4: C o u n t i n g Methods
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L e s s o n 4 . 3 : P e r m u t a t i o n s W h e n A l l O b j e c t sA r e D i s t i n g u i s h a b l e , p a g e 2 5 5i a , P5!^ ^ ( 5 ^ 2 ) !5!5 ^315 4 3!3!, 1 ^ - 2 08!)!8!2!c)1 0 ^10!5!, 1.) / •P = 1 0 - 9 - 8 - 7 - 6^Q.^g 3 b 2 ! 09!° ( 9 ^ 0 ) 1p . 9 1^ ° 9!7!( 7 ^ 7 ) 1P . I !^ ^ 0!, P , = 7!^P^ = 7 - 6 - 5 - 4 - 3 - 2 - 1, P , = 504015!P =15 5 ^Q,( 1 5 - 5 ) !15!1 5 - 1 4 - 1 3 - 1 2 - 1 1 - 1 0 !10!^ ^ P g = 1 5 - 1 4 - 1 3 - 1 2 - 1 1^ g P g - 3 6 0 3 6 02 ;/t <- i:• Pt-rniiitiific»ri Ji ^; 11} 12P r e s i d e n t ^Vice P r e s i d e n t "K,4jiK.itrifiaKntrr vl,lf">rNel/lf_ N.i/irM o t u i m a dM(>umai:yMt.harfiddN j / i iKatrsn.^Jt.hatiirt;,_ Jessb) „ p - ^4!It a president and vice-(4^2)14!2!4 ^ 3 ^2!, P , - 4 . 3, P , = 1 2The formula for „Pr gives an answer of 12. Thismatches my results from part a)3. a) ^ {n-r)lP. =6!( 6 ^ 4 ) !Ph f. 4 3 2!, P , 6 5 4 - 3J] 360There are 360 diOerent w a y s the chocolate bars canbe distributed.b) P = r ^« ^ ( 6 ^ 1 ) !6!5!6 - 5 !5!The chocolate bars can be distributed in 6 different ways.p =F o u n d a t i o n s of Mathematics 12 S o l u t i o n s Manual 4-9
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4. loFs is larger, e.g., I know this by looking at theformula for „Pr. The numerator is the s a m e for bothvalues since n is the s a m e . The denominator will besmaller for the first value since it has a greater r. W h e nyou divide a numerator by t w o different denominators,the final value is greater for the one with the smallerdenominator. Based on this. I know that WPB has thelarger value since its expansion has the smallerdenominator.9 IJ ci . As^-unimg that any o f t h e 10 digits can be putin any i, tbp o remaining spots for the SINs, let Sfct.ff.M-(it tin- number of social insurance numbers:S - 10 10 HJ !(i 10 1(1 10 10- KlS 1f(J DUO 000ri»M-f. .iro 100 000 000 different SINs that can ber(;fji..t;r<!<l in cMch of these groups of provinces andlerri!f)nf);5. ,/JP9]6!fd/ L 0 /,P MThere are 504 different w a y s the positions can be filled,M.O 4).10. a) }P« - 11!1 5 ^ =15-14-13-12-11t11!^gP^ = 1 5 - 1 4 - 1 3 - 1 2, , P , - 3 2 7 6 0There are 32 760 possible executive committees.7. 3P38!^ ( 8 - 8 ) 18!^ 0 !P31P tJP H 7 b, P, - 40 520Therefore, 40 320 different signals could be created. ( 5 0 0 0 - 3 ) !p ^ 5000!5000 3 4gg7,5000^3 =5 0 0 0 - 4 9 9 9 - 4 9 9 8 - 4 9 9 7 !4997!5 0 0 0 - 4 9 9 9 - 4 9 9 85Qj,j,P3 = 124 925 010 000There are about 124 925 010 000 different w a y s thetickets can be drawn.1 1 / f)H12/•12 11 10 q 7!/ I12 n Ui 0 f-0504(1There are 95 040 w a y s the coach can select thestarting five players.b l AP^(11^4)111!7!1 1 - 1 0 - 9 - 8 - 7 !7!„ F ^ = 1 T 1 0 - 9 - 8, , P , - 7 9 2 0There are 7920 w a y s the coach can select thestarting five players, if the tallest student must start atthe centre position.p . 1 ^10^3 7,p ^1 0 - 9 - 8 - 7 !7!^ j , P 3 = 1 0 - 9 - 8, „ P 3 = 7 2 0Multiply by 2, since Sandy and Natasha can play theguard positions in either order. (720)(2) = 1440There are 1440 w a y s in which the coach can selectthe starting five players, if Sandy and Natasha mustplay the two guard positions.1 1 . a ) n > 0 a n dn - 1 > 0n> 1Therefore, the expression is defined for n > 1,where n e I.b) n + 2 > 0f i > - 2Therefore, the expression is defined for n > - 2 ,where n e I,4-10 C h a p t e r 4 I o u n t i n g M e t h o d s
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c) ri + 1 > 0 A N D n >0n> 1Therefore, the expression is defined fcw h e r e n e Ld) n + 5 > 0 A N D n + 3 > 0n > - 5 n > - 3Therefore, the expression is defined for n > ^ 3 ,w h e r e n e I.12. a) ^,P,6 ^6!6 ^ =( 6 ^ 4 ) 16!2!6 - 5 - 4 - 3 - 2 !2!6 - 5 - 4 - 3360There are 360 w a y s to draw the four marbles if you donot replace the marble each time.b | Let L represent the number of ways:L = 6 • 6 - 6 - 6L = 6L = 1296There are 1296 w a y s to draw the four marbles if youreplace the marble each time.c) e.g., Yes; if you replace the marble, there are morepossibilities for the next draw.13. a)2 0 - 1 9 - 1 8 - 1 7 - 1 6 - 1 5 !15!2„P5 = 2 0 - 1 9 - 1 8 - 1 7 - 1 620 Pg = 1 8 6 0 4 8 0There are 1 860 480 different w a y s to award thescholarships.b) Let L represent the number of ways:1 = 20 - 20 - 20 - 20 - 20L = 20^1 = 3 200 000There are 3 200 000 different ways to award thescholarships.14. a) ^„P,10!1 0 ^10^4 =( 1 0 - 4 ) !10!6!1 0 - 9 - 8 - 7 - 6 !6!^qP^ = 1 0 - 9 - 8 - 7^pP^ = 5 0 4 0b) Subtract the total possible n u m b e r s by the answerto part a).104 = 10 00010 000 - 5040 - 4960There are 4 9 6 0 different phone numbers.15 u) • I eed to solve . ^ = 20( n - 2 ) !fi r ^tiO 1 1 ^ 2 > 0r i > 2nlTherefore,( n ^ 2 ) !{n){n^i):n - j p ; ,h h )(2)(1"i^ ](;(t){ „ ) ( „ :20 is defined for n > 2. where n e I.T- = 2020^ = 20(11^2)!( n ) ( n ^ l ) = 200 ^ - 0 = 20n " ^ f i ^ 2 0 = 0( n + 4 ) ( n - 5 ) = 0/? > 4 0 or n - 5 = 0n -4 n = 5T h e root n = -4 is not a solution to n > 2Check fl = 5LS RS5P2 205!( 5 ^ 2 ) 15!3!5 - 4 - 3 !3!5 - 420There is one solution, n == 5.b) 1 need to sovo{n tin 1 112)1n - f 1 > 0 A N D n -Z -? 0n>-1 11 1 - 0n > 1Therefore,n e I.(n + 1 ^ 2 ) !72 IS defined for n > 1, whereThere are 5040 different phone numbers possible.F o u n d a t i o n s of Mathemati utions Manual 4 t <
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:.. I- >)<(r, + l)!C h e c k f = 2727272= 72(11^1).( r i + 1 | ( / / l ( » 1i|u- ••) t . J ) i 2 | i l )2 ) " V ) ( 2 | ( 1 )(ii + l ) ( n ) = 72+ n = 72n ^ + n - 7 2 = 0(o + 9 ) ( f i - - 8 ) = 0n + 9 = 0 o r f i - ^ 8 = 0n = - 9 fl = 8The root n = 9 is not a solution to n > 1.Check n = 8LS RS8 +1P2 729P2(9 2)19!7!9 - 8 - 7 !JI9 - 872There is one solution, n = 8.1 e. a l The equation I need to solve = 30 .^ 0 - r)l6 - r > 0r < 6Therefore. = 30 is defined for 0 < r < 6, where ( 6 - - f ) lre I..e «6 J ^ 4 3 2 116 r i !720M30= 30( 6 ^ r ) . = I 2 030( 6 - f ) ! = 24= 4r 2LS RS6P2 306!( 6 ^ 2 ) !6!4!6 - 5 - 4 !4!6 - 530There is one solution, r = 2.b) I h<- ....luation I need to solve is 27 r 07!420Th<.i.-f.,rf; 2where r e l .7!420 IS defined forO < r < 7.2 -if71>/ ,n7 6 - 5 - 4 3 2 1, 1 !5040( 7 ^ r r( 7 ^ f ) i :4202102102105040210( 7 - f ) ! = 247 r - 4r 3Check r = 3LS RS2(^P^ 4 2 0( 7 - 3 )7 - 6 - 5 - 4 !4!2 ( 7 - 6 - 5 )2 ( 2 1 0 )420There is one solution, r = 3.4-12 C h a p t e r 4 : C o u n t i n g M e t h o d s
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1?I : RSnfn nPn ^ 1nl fl!nl n!0! [fl + 1nl n!1 1!n! nl1n!LS = RS18. a) e.g., The formulas for both „P„ and r,Pr have anumerator of nl. However, the formula for „ P „ has adenominator of 1 and the formula for „Pr has adenominator of ( o ^ r ) Lb) e.g., A group of friends each order a differentflavour of ice cream from a shop with 12 flavours.How many possibilities are there if the group is 12people? If the group is 7 people?19. a) n = 52 and r = 5P ^ 52! ( 5 2 ^ 5 ) !5 ^52 5^.^:,^l:.50.,:*.9,:48-47!47!52P5 = 5 2 - 5 T 5 0 - 4 9 - 4 8ggPg^ 311875200There are 311 875 200 possible arrangements,b) n = 26 and r = 526!( 2 6 ^ 5 ) 126!2« 2 1 !As =2 6 - 2 5 - 2 4 - 2 3 - 2 2 - 2 1 !21!^ePg = 2 6 - 2 5 - 2 4 - 2 3 - 2 226P5 = 7 8 9 3 6 0 0Likelihood =7 893 600•100%311 875 200Likelihood = 0.025...-100%Likelihood = 2.531...%Therefore, there is about a 2.53% chance that anarrangement contains black cards only.1 -•p - i ^ - : 11 l u yP r U l lUKel,hood= J ^ ^ . . . 1 0 0 %311875200Likelihood = 0 , 0 0 0 , , . - 1 0 0 %Likelihood = 0.049...%Therefore, there is about a 0.05% chance that anarrangement contains h o n d ^ only.20. e.g.. „ . , P , ^ ^ /n - 1 - n^ n - o - - 1 - n - ^ 2 . ( f f ; 1 -11= n{n)= n^2 1 . e.g., „ P . , =(7- 157nl(n7 I j inl( n - r - l ) ! n-r{n^rjnl= ( r , ^ r ) „ PM a t h i n A c t i o n , p a g e 2 5 7a) e.g., January 5, April 23, July 24, and October 15would be 5, 113, 205, and 288.b) 365 • 365 • 365 • 365 = 17 748 900 630c) i) = 0.983... or about 98.4%365ii) 1 ^ ^ ^ = 0.016,., or about 1.6%365F o u n d a t i o n s o f M a t h e m a t i c s 12 S o l u t i o n s M a n u a l 4^13
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d) e.g.: i| ^ = 0 J 6 6 . . . or 9 6 . 7 %30ii) 1 ^ ^ = 0.033... or 3.3% 30e) For example, they w e r e close but not the s a m e .M i d - C h a p t e r R e v i e w , p a g e 2 5 91. The number of subs to choose from, S, is based onthe number of buns (b). the number of cold cuts (cc),the number of cheeses (c), the number of toppings (f),and the number of sauces (s):S = (# of b) • (# of cc) - (# of c) • (# of t) - (# of s)S = 3 - 5 • 3 - 12 - 3S = 1620So, Mario can choose from 1620 different subs.2 «; (J Vou < ;ii< lis* on»- ot K W and C 1r)f the firstJi.ir.-K.tr-i oil.; nj VC ,j(>perooM lottrjrs lor thes.-cond .md third oh.mjctoc;, nod one id thoyj] u()f«to.iso loUfMs Ol o blank for tfio lost character.l u o m this I g.;l Itio lollowmg calc utationft o .lotion f.omos - :< 2() 26 ?fli i-l .tdtion i.nmoh 04 / u bI hotoforo, .station names are poissible.3. t vent A Rolling a 2 O Rhvenl B- Rolling 101 1i 2_ 3 4 5 6~2 )%j4 5^ 6 7[ 2 3 6 7 8i 3 4 b " e 7^ 8 9r 4 ^ " h i 7 8 " 9 "10[ 5 6 f ^ 8 9 J O ^ 11 "7 " 8 10^ U ~ 12 "From the table above, there is one w a y to roll a s u mol I wilh a pair f)f d i r e a n d three w a y s to roll a s u m of10 with a pai! of dice.fi(A >B)- niA) < 0(8)n(A . 8 ) 1 + 3niA B) - 4There arc 4 ways that a s u m of 2 or a s u m of 10 canbe rolled with a pair of dice.4, 1 0 - 9 - 8 = 720There are 720 w a y s to select 3 horses to c o m e first,second, third in a 10-horse race.5. a) 8!8!b) 6! • 3!6! • 3!6! • 3!8 - 7 - 6 - 5 - 4 - 3 - 2 - 140 320(6 • 5 • 4 • 3 • 2 • 1) • (3 • 2( 7 2 0 ) • ( 6 )43201)^ = 9 - 8 - 7 - 1d)9!6!9!6!9!6!916!9!_6!Ill510E115-l i5105-105-f: / h ) 4 3 2 1fi f- 4 3 2 1r. f. 4 3 2 1= b il {(i ) 4 3 2 150410 if) 8 / fi f) 4 3 2 IIU (8 / b 5 4 3 2 111 ^ 95 8!3 7 ab / 68!4 3 z4 3 2^ 1 0^ 52 - 9189 16. There are nine players on the team so there are 9different positions. Let L represent the number of(>ormutations:I - 9 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1L -L 9 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1£ - / 2 - 42 • 20 • 6L - 362 880There are 362 880 different lineups that can beformed by nine players on a softball t e a m .7. a) (fi + 5 ) ( n + 4 ) != (f, + 5 ) [ ( r i + 4)(r) + 3)(ri + 2 ) . . . ( 3 ) ( 2 ) { l f= (n + 5)(r, + 4 ) { i i + 3 ) ( n + 2)...(3)(2)(l)= (n + 5)i( n i 4 ) ( i i + 3 ) ( n + 2 ) ( o - M ) ( i i ) . . . ( 3 ) ( 2 ) ( l- ~ , n . 2 ) ( n ^ ( n f i 3 ) m= (n + 4 ) ( n + 3)n r 4n I 3n r 1 2= n" + 7 n + 124-14 C h a p t e r 4 : C o u n t i n g M e t h o d s
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1 (.d)Inl{n + 2}inl{n + 2]inlnlI j u / A ^ - j - > v 2 ) ( i )2)(r, + l )= + n + 2n + 2= i i " + 3 o + 28. a) = 72( r i ) { / i ^ l ) ( i i - 2 ) ( i i - 3 ) . . . | a n 2 H l |( n ^ 2 ; ( 2 ; ( r , l 2 ) f i :( i i ) ( r , ^ l ) ( r , ^ 2 ) !72n + 8 = 0 o r / i - 9 = 0n = - 8 fl = 9Check n = 8I s RS( Bj- 72t 8}!r - , is undefinedf 10)!Check 17 = 9LS RS9! 72( 9 ^ 2 ) !9!7!9 - 8 - 7 !7!9 872I111;.< , -J - I-Lc1-4 1,11-4{ Oj.Check A? = 7" j> i ) h - -1) ( d ( 2 i i : i j( f i - l ) ( n ^ 2 ) = 30n - 2 f i ^ n + 2 = 30n ^ - 3 / i + 2 = 30f i ^ - 3 f i - 2 8 = 0(fi + 4 ) { f i ^ 7 ) = 0/. f _ 0i RSundefined30LS RS( 7 - 1 ) ! 30( 7 - 3 ) !6!4!6 - 5 - 4 I4!6 530There is one solution, n = 7.i . a , P . 9!9!9 ^ 2 7!9 - 8 - 7 !7!, P , = 9 - 8A = 724!1 2 - 1 1 - 1 0 - 9 - 8 - 7 - 6 - 5 - 4 !4!^.Pg = 1 2 - 1 1 - 1 0 - 9 - 8 - 7 - 6 - 5^^Pg = 1 9 9 5 8 4 0 0There is one solution, n = 9.4^15
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12!( 1 2 - 1 0 ) !p . 1 2 !« 2!IV 11 If) iJ a 7 6 0 4 .5 >iP !> I 1 If) 9 p. / C, 0 1 i12 10^2P,o = 2 3 9 5 0 0 8 0 010. a) i: f. • 5 0 ANIJ /i 4 (I! f; _ Irhs- »:i|jr*;. lois I • fl«fin(id lot n - 1 Afiere n e I.t.- / I 4 C A N ! 1 / 1 2 0/!_ -I /* 2r i i f cxpt ;sMi-!i I. dufitioO for n - where n e !.r,; r» 1 - 0 A N [ ) tt 5 0n > 4 n > 5The expression is defined for n > 5, where n g I,d: n + 2 > 0 A N D n s 0f i > - 2The expression is defined for n > 0, w h e r e ne Lb ) a : n > 0 A N D n ^ 2 > 0n>2The expression is defined for n > 2, where n g I.b: r7 ---1 > 0 A N D n - 3 > 0n > 1 n > 3The expression is defined for n > 3. w h e r e n e I.11. n = 20 and r = 620!( 2 0 - 6 ) !p . 2 ^20 e 14!2 0 ^ =20 19 IB 17 16 15 14!~l4!2oPg = 2 0 - 1 9 - 1 8 - 1 7 - 1 6 - 1 5„,P^, 2 7 9 0 7 2 0 0Rennie can load his C D player in 27 907 200 different ways.12. n = 14 and f = 214!P - 13?T h e n ; aio 182 ways that Manny and 2 other players(,;in line up to receive the championship trophy.13. A.jtfH o g . T h e number of w a y s to choose apr..!>i!irnt .ir.d a vice-president from a group of five5!students is 20 . I could also use theFundamental Counting Pnnciple because there arefive choices for president and four choices remainingfor vice-president: 5 • 4 = 20.L e s s o n 4 . 4 : P e r m u t a t i o n s W h e n O b j e c t s A r eI d e n t i c a l , p a g e 2 6 6l . a )b).! 7 6 5 4 .3 2 13!2! (3 2 1) l 2 1)71 7 3 2J 4Ji3197 " ( 3 2 I I7!l 2 l7!3!2!8! _ f w 6y 4J 22 2 V :7 44202!2!2!8!2!2!2!8!- ^ 8 M i fs <5040c)2!2!2!10! 10 9 8 7 6 5 4 3 2 J4T3T2"! 1 3 2 1 3 2 1 2 110!4 ! 3 ! 2 !10!10 9 7 4: 12600d)4 ! 3 ! 2 !12! 1 2 - 1 1 - 1 0 - 9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 12!4!5!12!21415!12!2!4!5!2 - 1 - 4 - 3 - 2 - 1 - 5 - 4 - 3 - 2 - 1= 1 2 - 1 1 - 1 0 - 9 - 7831604 16 C h a p t e r 4 : C o u n t i n g Methods
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— * •ie arrangement of 6 flags: rrangements:7 4-7-•i riuic i j i ^ UU Giiloicnt signals that can be m a d e fromthe 6 flags hung in a vertical line,3. Let C represent the number of ways:6!3 L 3 !C = 20There are 20 different w a y s three coins land as headsand three coins land as tails,4. Let R represent the number of ways:18!RR =10!5!3!1 8 - 1 7 - 1 6 - 1 5 - 1 4 - 1 3 - 1 2 - 1 1 - 1 0 - 9 - 8 - 7 - 6 - 5 - 4 - 3 !1 0 - 9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1 - 5 - 4 - 3 - 2 - 1 - 3 !l? = 1 7 - 1 4 - 1 3 - 1 T 9 - 8R 2 450 448There are 2 450 4 4 8 ways that this record could haveoccurred5. Let C represent the number of ways:2!3!4!? : 3 • 1 1 V f" 0 ( {, !C 1260There are 1260 ways that Norm can distnbute1 cookie to each grandchild,6. a) Let A represent the number of arrangements:A = 5!/ = 5 4 • 3 2 1A = 120There are 120 different arrangements that can bemade using all the letters.b) Let A represent the number of arrangements:2!^ 7 6 - 5 - 4 - 3 - 2 !A ^ 7 6 - 5 4 - 3A 25201 here are 2520 different arrangements that can bemade using all the letters.8 - 7 - 6 - 5 - 4 - 3 - 2 !/• ., i finu.fif,, ,,i , . li ments that can bed ; i • 7i ••[)! • <>nr m.• ..iifnU.-- •„ .rrangements:2-1-3 2 1A 3 9 9 1 6 8 0 0There are 39 916 800 different arrangements that canbe made using all the letters.7. a) Let A represent the number of arrangements:5!5!5!, 11 n f 1 1 "f •. f. ^ ^ I . 2-% .1 - • 4 . V : c . ll o T14 r. ! ! » / CA 756756There are 756 756 different ways he can arrange thebooks on the shelf.h i ^.r.-iJi. llu sets of 5 together.A - 2 A = Brays he can arrange the books,8. e g,, A shish kabob skewer has 4 pieces of beef,2 pieces of green pepper, and 1 piece each ofm u s h r o o m and onion. How many differentcombinations are possible?RR9. a) Let R represent the number of routes:9f54"!9 HJ 6 - 5 4 3• 2• 15 4 3 2 ^i 4 3 2 [R - 7 6 4 3R 126There are 126 routes travelling from point A to point Bif you travel only south or east.4-17
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b) Let R represent the number of routes:13!7!6!1 i 1 11 If) u .". . (•) u 4 ; / 1R ~ - - - - -I f, 4 2 1 n 0 1 A ? 1L i 1 ! 4H I / l b!h»,-ic. ;iir- i r i r . loutes travelling from point A to|ic)ini H ll /HI iravel only south or east.10. Let R represent the number of routes:„ 13!8!5!Il IV 1 1(8 u .3 / h J 4 :^ 2 1f, { ] -. 4 / T 1 5 4 3 / 1/< !:•• 11 t«R = 1287There are 1287 routes travelling from the house ofJess to the house of her friend if she travels onlynorth or west.11 - a) o.g I dl .vw th« following diagram to show thenumber of ways to get to each intersectionB,:^.::;HJ..!s:.e4^ 4--1 ; t !iT h e s u m of the numbers on the top right and bottomleft corners of each block is equal to the number ofroutes to the top left corner of each block. There are560 different routes from A to B. if you travel onlynorth or west.b) I need to go north twice and west four times, for atotal of 6 moves, to travel the first 2 by 4 block of theroute. I need to go north once and west once, for atotal of 2 moves, to travel the next 1 by 1 block of theroute. I need to go north twice and west twice, for atotal of 4 moves, to travel the last 2 by 2 block of theroute.Let R represent the number of routes:R 61 „ 4!4 2 2 0 "4 5 V I 2 I 2 1 2 1 R -15 | 2 i i l ) !P - 180There are 180 different routes from A to B. if youtravel only north or west.12. Let P represent the number of permutations:r53i8 V. r. 4 i ^ 1^ r. 4 3 2 1 .3 2p -i 2P - 5 6There are 56 different permutations of answers thatthe teacher can create.13. a) Let P represent the number of permutations:p = 7lP = 7 - 6 - 5 - 4 - 3 - 2 - 1P = 5040There are 5040 different arrangements possible forthe new totem pole.b) Let P represent the number of permutations:7!2!2I/ R .h • 4 3_ ^ 12 1 . ]P - / 6 •P 1260f here are i2bO difteient arrangements possible forthe new totem pole.14. e.g., nPn will be too high; it gives the number ofarrangements of all n items, but some of thearrangements will be identical because of thea identical items in the group.15. a) e.g., I a m assuming that the coins of the samedenomination are considered identical objects. Let Arepresent the number of arrangements:9!PP4 ! 3 ! 2 !9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 14 - 3 - 2 - 3 - 2 - 2A 1260There are 1260 ways the 9 coins can be arranged in aline.i - i 3 C h a p t e r 4 : C o u n t i n g Methods
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11 i g that the coins of the sameflsidered identical objects. Let Aif of arrangements:1 •.A = 35There are 35 w a y s ns can be arranged in aline.1S. The number of ways to divide the 8 remainingfreezies amongst the other 8 children is what I want.Let P represent the number of permutations:215!^ 2 - 1 - 5 - 4 - 3 - 2 - 1F = 7 - 6 - 4P 168There are 168 w a y s to distribute the 10 freezies1 /, t; ifW mber of permutations:p ^ y f : •<P 560There are 560 permutations possible if you must startwith A and end with C.b) e g., If you start by putting the Is in the first andsecond positions, and then in the second and thirdpositions, and so on and so forth up until you put t h e m inthe ninth and tenth positions, there are 9 differentarrangements of the Is just on their o w n . The number ofdifferent arrangements of all the letters in each of these9 arrangements is the number of ways to organize theother 8 letters. Since the other 8 letters are always thesame, the number of permutations of the letters for eacharrangement of the Is is the same. Let P represent thenumber of permutations:P 9I 313!P = 9• 7 - 6 - 6 - 4 - 3 - 2 - 13 - 2 - 1 - 3 - 2 - 1P 9(8 7 5 - 4 )P 9(1120)P 10080There are 10 080 permutations possible if the two Ismust be together.18. e.g., B A N D I T S has 7 different letters, so thenumber of permutations is 7! B A N A N A S also has 7letters, but there are 3 As and 2 Ns so you mustdivide 7! by 3! - 2! = 12,19. The shortest possible route contains 3 movesdiagonally to tho right, 3 moves diagonally to the left,and 3 moves down Let R represent the number ofroutes:f:U U..fU .• , t. •.i:3 ! .P :. i ; /^11 •» ii d»..u j . ri the top rear vertex of the• . t". Ml. < >.-j •. •••ll of the cube,20. a) e.g., This is the same as arranging the20 players then dividing by 2! ten times because theorder of pairs does not matter. Let T represent thenumber of pairs:2 Pr = 2,375..,x10^There are about 2.38 x 10^^ w a y s to assign 20players to 10 double rooms.b) e.g., This is the same as arranging the 20 playersthen dividing by 4! five times because the order ofpairs does not matter. Let T represent the number ofpairs;- fr = 3 . 0 5 5 . , . x l O "There are about 3.06 x 10^^ w a y s to assign 20players to 5 guadruple rooms21. a) e.g., I can make a table to show all o f t h earrangements that could be made. Position 1 in thetable below is the leftmost position, and position 4 isthe rightmost position.Position2RRKWWRRNWwPosition3 l~.WwRPosition4WRRwVVRRWWRR"vvwwwRWRwwutions Manual 4-19
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b)I kfvom 1•.Ml l.l! tf l.ivcuf 2• IV,. {.I.lf«.-rKjttcm. ot( hliie huftiUcI Ul l/vU-lldVOUi OUIllblhalluUS buUdUSeeach two-flavour combination can be written in twodifferent ways2 « ;sent the number of committees:3. Let T represent the number of possible teams:(,(11/b I 1 1 . i:i: •] i 2 1*, 4 ;•- 2 IJ / 1. c /1 lif.u- i!if H,^4 w a y . h people can be selected from ayruup ot i z lo fonn a dodge-ball team.4. a) C =5!There are 10 possible committees.)resent the number of committees:2 1 ( 5 ^ 2 ) !2 «L 0 2C - 1 0 There are 10 possible committees,c) e.g., My answers for parts a) and b) are the s a m e .This occurred because the s u m of 2 and 3 is 5. 3 ! ( 5 - 3 ) !^ milC = l ± l is " 312 1c,03 = 5 - 2b) 9C39!8 ! ( 9 - 8 ) !9!811!9 8!8!191,.C, =A =A =6!4 ! h6!4 ! 2 !6 - 5 - 4 !4 ! 2 - 16^52-1e C , = 3 . 5X , = 1 50!10!i o C o = 110!01(10--0)110!12!61112- Oi!12!« 6!6!C ^ 1 2 : 1 1 : 1 0 - 9 - 8 - 7 - 6 !"^2 6 ^ b 5 4 3 2 I 6!C ^ 2 - 1 l 1 0 - 9 - 8 - 7" ^ " 6 - 5 - 4 - 3 - 2 - 1, 2 C g = 2 - 1 1 - 5 - 3 - 2 ., , C , = 9 2 48!1!(8-1)!8!1!7!8 - 7 !1-7!1a = BF o u n d a t i o n s o f Mathemati^ i l y f i o n s M a n y a l 4-21
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5. Let C represent the number of combinations:i I -10I Cu"4<1IJ U / l.lf . f 0 ? 110 U H /1 ? iI [, .. -J ^210: li«>re are 210 w a y s 6 players can be chosen to start,1 volleyball g a m e from a team of 10.6. Let C represent the number of combinations:C = 55C5C =c55!55!5!-50!5 1 5 i 52 C1 50!5 i 2 2 fo-O!55 04 52 :.J ,}5 4 2. 2 1c: - 1 • 27 (.3 13 1 /C 3 478 761There are 3 478 761 different combinations of hip-hopsongs you can download for free.7. Let H represent the number of hands:^ = 5 2 ^ 8HHH52!B ! ( 5 2 - 8 ) !52!8 44!5 2 - 5 T 5 0 - 4 9 - 4 8 - 4 7 - 4 6 - 4 5 - 4 4 !B 7 h 5 4 3 T I 4 4 5 2 - 5 1 - 5 0 - 4 9 - 4 8 - 4 7 - 4 6 - 4 58 - 7 - 6 - 5 - 4 - 3 - 2 - 1H = 1 3 - 1 7 - 1 0 - 7 - 4 7 - 4 5 - 2 3W = 752 538150There are 752 538 150 different 8~card hands thatcan be dealt.b) Let L represent the number of different lineups,n = 14 and r = 8 because Connie must be the pitcherof the starting lineup.L= ..C„14!8f(l4-8)!141HIU!14 L i 12 11 10 9 8!o n ; S 4 2 I11 H 7 .1 10-90 i> 4 3 2 1i:-. I I .53003There are 3003 w a y s that the coach can choose hisstarting lineup of 9 players, if Connie must be thepitcher.9. a) Yes, I do agree.e.g.,L =LS RS6C26! 6!2 1 ( 6 ^ 2 ) ! 4 ! l 6 4 l i6! 012! 4!0 f; i 2 1 / " 1 2 16 52 13 515 15LS - K Sb) e.g., S o m e other cases with the same relationshipas part a) are aCi = gC?, eCo = eCe, and 12C7 = 12C5.I notice that if you have two combinations with thes a m e n, and the s u m of the 2s for those combinationsis equal to n, then the value of the combinations willbe the s a m e .c) e.g.,nn - r8. a) The problem involves combinations e.g.,because it does not state that the order of the startingline matters.4-22 Chapi unting Methods
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10.Let T represent the Let S represent thenumber of combinations number of combinations • ..M= hers. for the students;• I8!3!5!8 / ••• i8 / Csent the numcS - 4S 56committees;; - 103I l.t,tu are 560 graduation committees that theprincipal has to choose from. ai Let C represent the number of committees;c10!5 ! ( l 0 - 5 ) !10!5!5!1 0 - 9 - 8 - 7 - 6/ ,C 252There are 252 committees that can be formed if thereare no conditions,b)Let W represent thenumber of combinationsfor the w o m e n ;Let M represent the numbeof combinations for the merM =4!MM21(4-^2)!4!2!-2!4 - 3 - 2 !M2 - 1 - 2 !i l l2-1M = 2 - 36M =Let C represent the number of committees:C=W-MC = 20 6C= 120There are 120 committees that can be formed if theremust be exactly 3 w o m e n .c) Let C represent the number of committees:6!1!(6-1)!6!_1!-5!6 5!1-5!61C = 6There are 6 committees that can be formed if theremust be exactly 4 men.d) Let C represent the number of committees;c = . acc =cc = -c =6!c =c =c5 1 ( 6 - 5 ) !6!5!-1!6 5!5!-161C 6There are 6 committees that can be formed if therecan be no men.e) e.g., C a s e 1 : 3 m e n and 2 w o m e n4! 6!3 ! - 1 ! 2 ! - 4 !60C,C a s e 2: 4 men and 1 w o m a n 4!-0! II 5!, q - , ^ = 1-64 C , - e q = 6Number of committees = 60 + 6Number of committees = 6666 5-person committees can be formed if there mustbe at least 3 men.F o u n d a t i o n s o f M a t h e m a t i c s =2 S o l u t i o n s M a n u a l 4-23
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12. e . g . Lets say I want to assign students to the 0! 1!room with 5 beds firsL Let A represent the number of «l o H ^ ^^j, " I i H oi(l-^0)«w a y s to assign the 12 students to the 5 beds: n ^5!M: .dt 0 ^ 0 - ^ 1 ^ 0 - ^12! C = 1 n = l/A 0 05!-7!I M I In I d ,C, = b A M1!1!1L0!A-,> I I 2 2 4 ^C, = ;JC=1N(»w H i c i f ,m- 12 - s or 7 students left to assign. Lets ^.isMfjh stuojt;!. to hu room with 4 beds n e x t Let B iCo, i C i = 1, 1-111 tlir 1 IMI ll KM of ways to assign the ., ^ -^^ n ..... ...^AL^7 s t H d e n 1 s h - i ! H ; 4 h r d s : "^ ^ ° 0 ! ( 2 ^ 0 ) ! " " " t i l - i ) !cJ ; / - 4 H 1 ^ 2 J !4 1 2 X = 1 ^ 241 3 2 i ,C, 22 I " 2 2 2 1 ( 2 - 2 ) 1B - 35 2 2 2!-0!Now there are 7 - 4 or 3 students left to assign to theroom with 3 beds. Since all of these students will be A =assigned to the room, there is only one combinationfor them. Let C now represent the number of different 2 M = assignments: 2C0, 2C1, 2C2= 1, 2, 1C = 792 . 35 0 1 ( 3 ^ 0 ) ! ^"^^ ^ 1 ! ( 3 ^ l ) !C = 27 720 01 31There are 27 720 w a y s the 12 students can be C = 3 = assigned to these rooms.0!-3! 1!-2!13. a | i ) 5 objects, 3 in each combination s M ^ 3^1 ,|,2iii) 10 objects, 2 in each combination ^ ^ .| 3iii) 5 objects, 3 in each combination ^ ° a ^ Mb) e.g., i) How many w a y s can you choose 3 coins ^ ^ ofrom a bag containing a penny, a nickel, a dime, a 3 ^ ^ ^guarter, and a loonie24-24 C h a p t e . 4 t . c u n t i n g M e t h o d s
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3^23!3C33!3^22 ! ( 3 ^ 2 ) !3C33^23!2!-1!3!3!-0!^ 3 - 2 !"""mi 3^3^ 1^ 1^ 3- ^3^3= 1:p2 = 33C0, 3C1,3C2, 3C3= 1. 3, 3, 14! „ 4 !01(4^0)! ^^^•^^11(4--1)1C ^C, q = 44! 4!2 1- ^ ^ ^ 2 . 1, q = 2 - 34 C M 6^ ^ ^ ^ 4 ! ( i r 4 ) ,C" * 4!-0!c ^ J L" 3! 14Co, 4C1, 4C2, 4C3, 4C3 = 1 , 4 , 6 , 4 . 1c) e.g., The numbers on the left and right sides are all1s; every other number is the s u m of the two numbersabove it.d) sixth row: 1, 5, 10, 10, 5,1seventh row: 1, 6, 15, 20, 15, 6, 1e) e.g., The number in each square of PascalsTriangle is equal to the number of pathways to it fromthe top square.« a l The equation I need to solve isnl2 ! ( o ^ 2 ) ! II • 4 j D n - 2 > 0n > 215 IS defined for n > 2, whore n e N.152 ! ( n ^ 2 ) !nlV / - r;;nl; c - M ) lnlnl= 15= 1 5 - 2 != 15(2)= 30( n ^ 2 ) !n(n~^i){n^2,y J i j i 2 n i .(n^2]{„A] n}(2){fn ( i i - 1 ) = 30n " - / i = 30(o + 5 ) ( o - 6 ) = 011 + 5 = 0 OR n - 6 = 0n = - 5 n = 6Based on the restrictions, n = -5 cannot be a solution.Therefore, n = 6 is the solution to the equation.F o y n d a t i o n s o f M a t h e m a t i c s 12 S o l u t i o n s M a n y a !4^25
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b) T h e equation I need to solve isn!4 ! i p 4)1= 3 5 .I) ANI ^ n 4 fAn,! .}3:3 c, dufiiied for n > 4. where n e N.3535 41840840840840.,(.•,. mn ?){n i)(n-4i(n^5).„{3)(2)m4in -51 . ( i l p f l )M n - l ) ( n - 2 ! i r / 3)n* - 5n + 6n - r f + bit - 6 n = 840„ M +11ll" ^ 6 f i - 8 4 0 = 0W n t e out all of the factors of - 8 4 0 ; ± 1 , ±2, ±3. ±4. ±5,±6. ±7, ±8. ±10. ±12, ±14, ±15. ±20, ± 2 1 , ±24. ±28,±30, ±35, ±40, ±42. ±56, ±60, ±70, ±84, ±105, ±120,±140, ±168. ±210. ±280. ±420. ±840. These are allthe possible roots of the equation. Substitute theminto the equation and if the equation goes to 0, then Ihave a root of the equation. By trial and error, theroots are -4 and 7. T h e other roots are not real. Since4 is out of the domain, the only real solution is n = 7.c) The egu.ition I need to solve isn! j 1 " >|! _2 { n - / f 3 ! i / < 2 3 ) ! n > 0 kU) 2 0 A N l i /; 2 (:n _ ? // ^ -2A N D r n 2 3 = i01(n + 2)!r^-1n12 ! ( / 7 - 2 ) twhere n e N3!(n i 2 3)!is defined for n > 2,_^ !"/J i2!(n 2i " AH,; :> iH/2 i n ^ 2)12(f7 21 6 ( » ^ 2 312n! «n .-21!( n - 2 , ) ! " 6 ( u - r 2 - 3 ) !2n! I n 1 2)1( n ~ 2 H " " 6 l n - l H2n(M 1)(/; 1 2 H , 7 f l ) ( n )612/11/ 1) 1-5 • 2)(n + l ) ( n )1 2 ( n - l ) - ( n + 2 ) i / ; +1) 01 2 n - 1 2 - ( n ^ + n + 2 u i 2) <1 2 n - 1 2 - n ^ - o - 2 i ^ 0~ n + 9 n 11 0An-2)iii 7t 0n 2 0 or n - 7 0n 2 11 = 7Both the roots are within the domain, so there are twosolutions, n = 2 and n = 7.6!d) The equation I need to solve isi ( 6 - r ) !15 .r > 0 A N D 66!15 is defined for 0 < r < 6, where r e I.15r ! ( 6 - r ) ! =r ! ( 6 - ^ f ) !r ! ( 6 ^ r ) !6!6!1572015r ! ( 6 - f ) ! = 48By substituting each of the integers r for 0 < r < 6, Iget r = 2 or r = 4.16. a) 1, e.g., the player can only win if the sixnumbers they choose are the same and in the sameorder as the six numbers drawn.66!b)6 1 ( 6 6 - 6 ) !66!6160!6 6 - 6 5 - 6 4 - 6 3 - 6 2 - 6 1 - 6 0 !6 - 5 - 4 - 3 - 2 - 1 - 6 0 !66 65 64 63 62 616 - 5 - 4 - 3 - 2 - 1ggCe = 11-13-16-21-31-619 0 8 5 8 7 6 8There are 90 858 768 different w a y s the player canwin.4-26C h a p t e r 4 : C o u n t i n g M e t h o d s
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c | e , g „ N o , E v e n if e v e r y o n e in t h e c i t y p l a y s , it isvery u n l i k e l y that anyone will win s i n c e each p l a y e r•>-v I !•! K, / 6 7 c h a n c e of w i n n i n g . * t} I .:«• !i,in,in-: ;:t s i d e s in a p o l y g o n is e q u a l t o • • • » ! : l -. f i ttie n u m b e r of v e r t i c e s = rt • " . , (,!• n •..( u / •. . . d i a g o n a l is f o r m e d b y e l i n e-ni:i;r.: nu hiiq .> v e r t e x t h a t is not d i r e c t l ysic.id.-t li.ti^ Uu u u m b e r of v e r t i c e s t h a t w i l l m a k e. H j o t , , j i vvifr . II. in ( < - s i d e d p o l y g o n is n - 2.t ryiug i - lm „ W >Nii the v a l u e s from t h e p o l y g o n so n t h e s i d e of t h e t e x t b o o k p a g e , t h e r e is a p a t t e r n ;(d = n u m b e r of d i a g o n a l s ). a =I 6 I 2 I 15 I 9• U4H < 15 fl /» -iRearranging, d = „C2- n. Thus, the n u m b e r ofdiagonals for an n-sided polygon can be determinedusing nC-i-n.18. a) C a s e 1: 2 boys and 3 girtsC c - ^ " " ^ 2 ! 5 ! 3110!7 ^ 2 - 1 3 ^ 3 = 2 1 - 2 8 6, C , - „ C 3 = 6006C a s e 2: 3 boys and 2 girts 2 3 ! 4 ! 2!11!X 3 - „ q = 3 5 - 7 8, 0 3 - , 3 C , = 2730C a s e 3: 4 boys and 1 giric c = ^ i l L " 4 ! 3 ! 1!12!, q - „ C , = 3 5 - 1 3, C , - „ C , = 455C a s e 4 : 5 boys and 0 girisC C" 5 ° 5!2! 0113!, C , . „ C „ = 2 1 . 1, q . „ C „ = 2 1N u m b e r of groups = 6006 + 2730 + 455 + 2 1N u m b e r of groups = 9212There are 9212 different groups of 5 students with atleast 2 boys to choose from.b) N u m b e r of groups with no conditions;° 5!-15!C a s e 1 : 1 boy and 4 girls1 1 6 ! 4 1 9 1. X , - „ C , = 5005C a s e 2 : 0 boys and 5 girts7! J 3 ! ^5!8!.C. =• " 0 ! 7 !, C „ - „ C , = 1287N u m b e r of groups with at least two boys;2 0 C 5 - 7 C 1 • 1 3 C 4 - 7Co - 1 3 C 5 = 9 2 1 2There are 9 2 1 2 different groups of 5 students with atleast 2 boys to choose from.c) e.g.. I prefer indirect reasoning because fewercalculations are needed.19. a ) e.g.. Combinations and permutations bothinvolve choosing objects from a group. Forpermutations, order matters. For combinations, orderdoes not matter. For example, a be and bac aredifferent permutations, but the s a m e combination.b) Divide nPr by rf to get „Cr. For example. e C = iand oP.i 3 6 0 ; 1520. First, determine the total n u m b e r of o u t c o m e spossible 111 a s s u m e that once a song is selected, itcannot be selected again. T h e n u m b e r of outcomes, O.is;0 = ^5166!O , 1 3 0 1 9 9 0 9a) N u m b e r of times the event could occur;* 5 ! 2 1 !3,,C, 6 5 7 8 0Probability (P);p 6 5 7 8 01 3 0 1 9 9 0 9P = 0.505...%There is about a 0 . 5 1 % chance that the five songs wilbe from C D 2 and C D 4 .b) N u m b e r of times the event could occur;12 14 15 12 18 5 4 4 3 2 0Probability (P);„ 5 4 4 3 2 0x 1 0 0 %1 3 0 1 9 9 0 9x 1 0 0 %P = 4.180...%There is about a 4 . 1 8 % chance that one of the fivesongs will be from each C D .F o u n d a t i o n s of Mathematics 12 S o l u t i o n s Manual 4-27
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c) There is only o n e time where your favourite songfrom each of the 5 C D s will be played.Probability (P):p = 1 x 1 0 0 %1 3 0 1 9 9 0 9P = 0.000008%There is about a 0 , 0 0 0 0 0 8 % or 1 in 13 0 1 9 9 0 9chance that your favourite song from each of the5 C D s will be played.2 1 . A + rP2 + Anl nl nl3!|fi- ^ 2 ! { n - 2 ) ! 1.l(n-l)lnl nl nl6 ( n M 3 ) ! ^ 2 { n - ^ 2 ) ! ^ ( / i 1)n! 3 n ( i i ^ l ) ( n ^ 3 ) !- + ^ - - 7 — ^ M + 6(n^3}l 6(n^3)l 6(/)- 3),OII3P(II 1)(/I 3 ) ! + 6 r t ( n - - 3 ) !6{rA3)li ( n 3 ) n ( n - 1 | ( u 2 ) i 3 r t ( u i| , On6 " ~ ^n{f)2-^2u I M 2 t 3// 3 I 6)n ( f r I 5)622. e.g.,LS_^pin 1 r ) !RSC^ + Arnl nlr { n ^ r f { r ~ ^ l ) { n ^ ( r ^ i ) }" n - ^ ( f - l ) ] f i l r{n)rt> (/ 1)] rl[n Jf(r) + M f ) f i ! + r ( r i ! )nl{n i 1 t I r )f)l(n + l )H i n f l r)lLS - R STherefore. „ + i C r = „C, t .C,L e s s o n 4 . 7 : Solwlng Coynting Problems,page 2 8 81. a) This situation involves combinations because theorder of the 3 toppings o n the pizza does not matter.b) This situation involves permutations because thethree spots for the candidates w h o are selected areall different so order matters.c) This situation involves permutations because for agroup of 3 numbers, there are different w a y s to rollthose three numbers because of the different coloursof the dice,d) This situation involves combinations because the5 children w h o are selected are all in the s a m eposition. N o information is stated in the questionabout positions the children m a y play, s o I c a n onlyassume that they are not playing in specific positions,2, e.g., Situation A involves combinations a n dsituation B involves permutations. For situation A,order does not matter since the 3 people w h o areselected will all be considered equals. For situation B,this is not the case. Each of the 3 people w h o areselected will have a different position with a differentamount of power and different roles.3. a) , C =3!3!-0!3 ^ 3 = 1There is 1 w a y that Maddy can bid on 3 items if shebids o n only her 3 favourite items.b) A = ^^ « 3!-5!There are 56 ways that Maddy c a n bid o n 3 items ifshe bids o n any 3 of the 8 items.13!( , 3 q f = 2 8 5 6 1There are 28 561 different four-card hands with o n ecard from each suit.200!5 a) P200 ^ 52 0 0 - 1 9 9 - 1 9 8 - 1 9 7 - 1 9 6 - 1 9 5 !195!2„Pg = 200-199-198-197-1962ooPg = 304 278 004 800There are 304 278 004 800 w a y s that the top fivecash prizes can be awarded if each ticket is notreplaced w h e n drawn.4-28 C h a p t e r 4 : C o u n t i n g M e t h o d s
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b) ( 2 0 0 f = 320 000 000 000There are 320 000 000 000 w a y s that the top fivecash prizes can be awarded if each ticket is replacedw h e n drawn.6.1./- i)i ilO= 180There are 180 w a y s that the 5 starting positions onthe basketball team can be filled.10!7.2!-2!-2!-2!-2!10!y 2 2 2 •= 1 0 - 9 - 7 - 6 - 5 - 3 2 - 11134002!-2!-2!-2!-2!_ 10!2 ! - 2 1 2 1 ^ 2 ! ^ !There are 113 400 ways that the five different pairs ofidentical teddy bears can be arranged,8. C a s e 1 : 3 flags are used5!^5 I h5 ^5 4 3 2 15 P 3 = 5 - 4 - 360C a s e 2: 4 flags are used^ ^ ^ ^ ^ ( 5 ^ 4 ) .5 ^A 1205! 1!5!C a s e 3: 5 flags are used•P., = 5!5P5= 120Let S represent the number of different signals thatcan be sent using at least three of the flags;5 = 60 + 120 + 120S = 300There are 300 different signals that can be sent usingat least three of the flags.9 e.g., First make a table to show the number ofways the two cabin cruisers can be arranged next toeach other.CC 1 CC• Af i,uyiMncnt 1 i A r r a t i y e m e i i t 2A r r a n g e m e n t 3 A i r a r i g o i i i c n t 4 AA r u m g c - m e u t 5 hArramnmnmt G i i i! A r r . m j e m o n l / : p I J - y - Arr;inc|.,inent 8 _ i ; V. " i: A r r . i n g o m e r i t 9 f, " 4; A r i a n g e m e i i t 10 f. i 7iFor each of these arrangements, the number of w a y sthe SIX boats can dock is the number of ways that theother four boats can dock. Let D represent the totalmTbjr of ways that the boats can dock;: 4 24n ..p.3 240 ways that the six boats can dock.I i . e.g.. Each row of seats is different, and within arow, the seats are a s s u m e d to be different. Therefore,there are 10 different people being seated in10 different spots. Let A represent the number ofseating arrangements;A = 10!71 = 3 628 800There are 3 628 800 ways that the 10 players can sitin the van.1 1 . 2!60There are 60 different arrangements that are possiblefor the letters if there are no conditions.b) 3! = 6There are 6 different arrangements that are possiblefor the letters if each arrangement must start and endwith an N.12. e g . W h e n there is an even a m o u n t of numbers,half of them will be o d d . In this case there are100 possible numbers that each number can be.Therefore. ^ , or 50 of them are o d d . Since I wanteach number to only be 1 of these 50 odd numbers,the number of sequences S is;S = 50 50 • 50S= 125 000There are 125 000 completely odd sequences.13.11!5h&.= 462 You can take 462 different routes.F o u n d a t i o n s of Mathemati dutions Manual4-29
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14. e.g., Lets assign people to ttie 5-person car first.Let J represent the nunnber of ways to assign thepeople to this car:5!11!J = 4368Now there are 16 - 5 or 11 people left to assign to theremaining two vehicles. Lets assign people to the4-person car next. Let K represent the number ofways to assign the people to this car:17. e.g.,K11!4!7!K = 330Now there are 11 - 4 or 7 people left to assign to theremaining vehicle. There is only 1 way to assign thesepeople to the 7-passenger van because all of themare going to be assigned to it. Now let T represent thetotal number of assignments:r = J K-17 = 4 3 6 8 • 3307 = 1 441 440There are 1 441 440 w a y s the 16 people can beassigned to the 3 vehicles.Top of Board2 6 10 615.StartNumber of Paths = 2 + 6 + 10 + 6Number of Paths = 24There are 24 paths that the red checker can follow.16. C a s e 1: 0 hearts and 5 non-hearts: 13C0 • 39C5C a s e 2: 1 heart and 4 non-hearts: 13C1 39C4C a s e 3: 2 hearts and 3 non-hearts: 13C2 39C3C a s e 4: 3 hearts and 2 non-hearts: 13C3 • 39C2Let H represent the number of hands with at most 3hearts:H = 13C0 • 39C5 + 13C1 • 39C4 + 13C2 • 39C3 + 13C3 • 39C2H = 1 575 757 + 13 82 251 + 78 9139 + 286 741H = 2 569 788There are 2 569 788 different five-card hands thatcontain at most three hearts that can be dealt.ords-rin.ilii l ryesU-.,i peiuiiit-itionv, .,r, Civ !)!nl>iii.ni)iKlfiUlCfll:yestoiinHiti.HiiJivici, by 1!. -.vf.r-i. t h e n u n i l H i 1ir.lenUCtll itt.TTlAND ORiJbu l-UMdiinu;ntdl CtHuUinqPrinc iple. multiply the numberof ways ejir.h tHsk (.rin ocnirof way; ej< htask f.an occuf,C„ =18. Number of Total Outcomes:13!6!-7!i3Ce = 1716Number of Outcomes Where 3 Boys and 3 Gids C a n Go:e 3 T ^ 31.31 31-4!^ ^ 3 ^ = 700Probability (P):700P = x 1 0 0 %1716P = 40.792...%There is about a 40.8% chance that there will be threeboys and three giris on the trip.19. e.g.. If I have an A a s the first letter, there are 4possibilities for the second letter: A, L, S , or K.If A is the second letter:4 possibilities for the third letter: A, L, S , or KE a c h one of these has 3 possibilities for the fourthletter. 4 ( 3 ) = 12If the second letter is L, S , or K:3 possibilities for the third letter: A and 2 of L, S , andK (depending on which letter is second)The As have 3 possibilities for the fourth letter, andthe other two letters have 2 possibilities for the fourthletter. 3 + 2(2) = 7Total for all three second letters that are L, S , or K:7(3)= 214-30C h a p t e r 4: C o u n t i n g Methods
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Total if A i s ttie first letter;2 33Therefore, if the first letter is A, there are 33 possiblearrangements.If the first letter is L, S. or K, there are three possibilitiesfor the second letter; A, and 2 of L, S, and K (the onesthat are not the first letter).If A is the second letter;3 possibilities for the third letter: A, and 2 of L. S. and KThe A has 3 possibilities for the fourth letter and theother two letters have 2 . 3 + 2(2) = 7If A is not the second letter;/ .sibiiities for the third letterThe A has 2 possibilities for the fourth letter and theother letter has 1. 2 + 1 = 3Total for both second letters that are not A;3(2) = 6Total for one of three times where first letter is L, S. orK:7 + 6 = 1 3Total w h e n first letter is L, S, or K;3(13) = 39Total arrangements:39 + 33 = 72Therefore. 72 four-letter arrangements can be madeusing all of the letters in the word A L A S K A .20. If I have an O as the first letter, there are4 possibilities for the second letter, each of which has3 possibilities for the third letter. 4(3) = 12Therefore, there ar^; 1 / uo^..sible arrangements whenO is the first letter.If the first letter is B, K, or S;There are 3 possibilities for the second letter; O, andtwo of B, K. and S (the ones that are not the firstletter). O has 3 possibilities for the third letter whilethe other 2 have 2. 3 + 2(2) = 7Total if the first letter is B K, or S:3 ( 7 ) = 21Total arrangements:21 + 12 = 33Therefore, 33 threedetter arrangements can be madeusing all of the letters in the word B O O K S .H i s t o r y C o n n e c t i o n , p a g e 2 9 0A. Yes. Each number from 0 to 127 is assigned adifferent character or symbol on the keyboard. Sincethe numbers already have an established order, thecharacters and symbols assigned to these numbersdo, as well.B. Yes. Each number in ASCII (pronounced "askey")must be converted into a stnng of Os and I s to createthe binary code, so order matters. Each 0 or 1 isassociated with a position in the stnng. A differentpermutation of Os and I s represents a differentnumber in the ASCII code system.C. There are 128 numbers in ASCII that must berepresented by a string of Os and I s . You need todetermine the length of the stnng needed to create128 different arrangements of Os and I s . You canbegin by thinking about a stnng of length of 5.A b o x d i a g r a r j . , f ;an help youdetermine the .,i..:it)fr <-AS(,i| numbers you canrepresent.Within each box you can place a 0 or a 1. There aretwo choices for each box. since repetition of Os andI s is allowed. So for a stnng length of 5, there are2 - 2 - 2 • 2 - 2 = 2^ or 32 ASCII numbers that can berepresented. Obviously, the stnng must be longer for128 numbers. If n represents the string length, and128 numbers must be represented, then 2" = 128. Bytrial and error, n = 7.A binary stnng of length 7 is needed to representeach ASCII code.C h a p t e r S e l f - T e s t , p a g e 2911 nl Let N represent the number of different serialnumbers:IV = 26 - 26 • 10 - 10 - 10 - 3IV = 2 028 000There are 2 028 000 different serial numberspossible, if repetition of characters is allowed,b) Let N represent the number of different senalnumbers:IV = 2 5 - 2 4 - 1 0 - 9 - 8 - 3IV = 1 296 000There are 1 296 000 different senal numberspossible, if no repetition is allowed.2. Event A: Drawing a spadeEvent B: Drawing a diamondn(A m = iiiA) + n(B)niA H) - 13 + 13niA •• li) ^- ?uTherefore, there are 26 ways to draw 1 card that is aspade or a diamond.3. a) n + 9 > 0/ 7 > - 9(n + 10)(n + 9)! is defined for n > - 9 . where n e I.(n + 10)(n + 9)! = (n + 10)[(n + 9)(n + 8)...(3)(2)(1)](n + 10)(n + 9)1 = (n + 10)(n + 9)(n + 8)...(3)(2)(1)( n + 10)(n + 9)! = ( n + 10)!b) n - - 2 > 0 A N D n > 0n > 2(n^2)^—A is defined for n > 2, where n c I.nl( n - 2 ) ! ( n - 2 ) ( „ ~ 3 ) . . . ( 3 ) ( 2 ) ( l )n ^ ; H „ - 1 ) ( „ - 2 ) ( „ - 3 ) . . . ( 3 ) ( 2 ) ( I )nl n ( n - l )nl /•? - nF o u n d a t i o n s of Mathematics 12 Solutions Manual 4-31
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4 a) hi - 1/0rhnn:foro, there are 120 different ways that the 5 carst.aii Ix; paikod side by side.b) I el B rofJHjsent the number of arrangements;B ./f ./,P ~ 2» 4f i - 2 1 4 3 2 1[i - 481 h(;r(4<)fu, there are 48 different ways the cars can beparked so the 2 black cars are next to each other.There are 126 different four-book selections that canbe m a d e( 9 T 4 ) !0fd9 8 f o b• - Vj -M a / hThere are 3024 different four-book selections can bearranged in order of preference.c) e.g., The order matters in part b). There are still126 ways to choose the four books from the nineoptions, but there are also 4! = 24 ways to arrange thebooks,(126 • 24 = 3024)m P ,^ ^prP6. „ P , = 8 4 ( „ q )( " - 4 ) 1 ( n - 2 jf ) ( n - l ) ( f ] - - 2 ) ( n - 3 ) = 4 2 n ( n ^ 1 )( n - ^ 2 ) ( r , - 3 ) = 42n- - 3A7 - 2r? t 6 42fP 5n 36 -:- 0{ n - r 4 ) ( n - ^ 9 ) = 017 + 4 = 0 or n - 9 0n = - 4 n = 9Check n = --4l A" Ai 4)1( 4 4)1is undefined841842 ! ( - 4 2)!Check n = 9LS9!5!u y, ( h 5!5!9 8 7 63024RS8 4 ( , q848484849!21(9 2)19!^2L7!^g 8 712 1 Pg a 2 184i^< 43024There is one solution, n = 9.6! 8!, C , - , C 3 = 8 4 0There are 840 different ways that a 5-personcommittee can be selected if there must be 2 boysand 3 giris.b) C a s e 1: 2 boys and 3 giris;C (• - ^ 6 2 n 3 2!4I 3!5!ApAS-^QC a s e 2: 3 boys and 2 giris:« « 2 3,3, 216!, C 3 . , C , = 560C a s e 3: 4 boys and 1 giri;IL6 4 - 8 i " ^ 4 ! 2 ! i ! 7 !120C a s e 4 : 5 boys and 0 girts:C . c ^ ^ . ^6 5 8 0 5 , ^ , Q,8|As-sCo = QLet C represent the number of 5-person committeeswith at least 2 boys:C = 840 + 560 + 120 + 6C= 1526There are 1526 different ways that a 5-person committeecan be selected if there must be at least 2 boys.^UA-A,3!9!2204-32 Chapter 4 ..r^unting M e t h o d s
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There are 220 different ways that a 5-person• .i.i< .11. t . fed if David and Susan mustd) C a ^ e 1- i l . V . girtsni H!{ : f :" PX „ X , = 8 4 0f;.:sc 2f r.C. = 420C a s e 3: 0 boys and 5 giris" ° « "^ 0!6! 5!3!56Let C represent the number of 5-person committeeswith more girls than boys:C = 840 + 420 + 56C = 1316There are 1316 different w a y s that a 5-personcommittee can be selected if there must be more giristhan boys,8. - ^ = 302!2!There are 30 different arrangements of the letters inthe word 1 r f 1119. 5! • 4! = 2880There are 2880 different arrangements possible.C h a p t e r R e v i e w , p a g e 2 9 31. e.g., The Fundamental Counting Pnnciple is usedw h e n a counting problem has different tasks relatedby the word A N D , For example, you can use it tofigure out how many ways you can roll a 3 with a dieand draw a red card from a deck of cards.Quarter Toonieheadstailsheadstailsheadsta.L.LoonieIv.-adstailsheadstailsheadstailsheadsrails3 I r-* A : rr : the number of sets of answers:, 4 - 4 • 4 • 4 • 4A = 4^°/•. i .;!., c / c" i - •.In..,: I .•)..live 1 048 576 different sets ofanswers.4 a| /. .• u A N D n > 0, - yf ! • ufined for n > 0, where n c I.in^ 2)1^ = 20(fi)(ii-^4)..,(.iil2)(l,l) = 20n^+n I I / 20 = 0ll . 3// 18 = 0l) = 0n + 6 = 0 or n ^ 3 = 0n = ~6 n 3The root n =• 6 is outside the restnctions on thevariable in the equation, so it cannot be a solution.There is one solution, n = 3.b) The simplified version of the eguation isn + 1>0 A N D n - 1 > 0n > - 1 n > 1In , 1)1132 IS defined for n > 1, where r? e I.// 1 !132132(11-^1)1{n l ) f u 2 , .(3)(2)(1)(o + l ) ( n ) = 132n + n = 132rf * n - 132 0(r7 + 1 2 ) ( n - 1 l ) = 00 + 12 = 0 or n - 11 0,0 - 1 2 n = 11The root n = - 1 2 is outside the restnctions on thevanable in the eguation, so it cannot be a solution.There is one solution, n = 11,The tree diagram shows there are 8 possible waysthat the three coins can land.F o u n d a t i o n s of Mathematic •-. oiutions Manual 4-33
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5. e.g., 6^6 has a larger value, e.g., I know because^ is the factonat expression for the permutation6!expression 8^2. Here, I have more objects than forePe, but I a m not using all of them. This leads to fewerpossible arrangements, or in other words, a lower8!value for6!6. Let O represent the number of orders:0 = 12!O = 4 7 9 001 600There are 4 7 9 001 600 different orders in which thesingers could perform the 12 songs.7. , , P , = 25!^5 3 22!25P3 = 13800There are 13 800 different w a y s a director ofeducation, a superintendent of curriculum, and asupenntendent of finance can be selected.b) Let A mnrf^cppf ^hn number of arrangements:10! i n I 8 / h > 1 4 2 12!2!2!10!/ 1Vi < ! i 4 ^= 4 5 3 6 0 02 ! 2 2 !10!2!2!2!There are 453 600 different arrangements that arepossible if all the letters are used, but eacharrangement must begin with the C.1 1 . a)142 5 2 2 5 2 02 ! 3 ! 4 ! 5 !I IKJHJ are 2 522 520 different w a y s Tina can stack theblocks in a single tower if there are no conditions.b) 2 7 7 2 03!4!5!There are 27 720 different w a y s Tina can stack theblocks in a single tower if there must be a yellowblock at the bottom of the tower and a yellow block atthe top.25P^o=11 861 676 288 0002 5 P „ - 1.18rK..x10"There are 11 861 676 288 000 or about 1.2 x 1 0 "different w a y s the test can be created if there are noconditions.23Pg = 19 769 460 48023Pg=1.976...x10°There are 19 769 4 6 0 480 or about 2.0 x 10^°different w a y s the test can be created if the easiestguestion of the 25 is always first and the most difficultguestion is always last.9 p . 5 2 !.^P^ 311875200There are 311 875 200 different five-cardarrangements possible.10. a) Let A represent the number of arrangements:11! 1 1 - 1 0 - 9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 12!2!2!11!2I2I2!11!2 - 1 - 2 - 1 - 2 - 11 1 - 1 0 - 9 - 7 - 6 - 5 - 4 - 3 - 2 - 14 9 8 9 6 0 021212!There are 4 989 600 different arrangements that arepossible if all the letters are used.12.™ 5 5!5!ioCs = 252C• ^ ^ 7 ! 4 != 3 3 0^5 2 2113!A 105Therefore, nd results in the greatest value.13. C, =•20!4116!20 = 4 8 4 5There are 4 8 4 5 different selections of 4 books thatRuth can choose.14. No. e g., Each combination can be arranged inmany different w a y s to m a k e a permutation, so thereare more permutations than combinations15. a) „ C =19!4!15!,,C^ 3876There are 3876 different w a y s that a committee of4 people can be chosen if there are no conditions.9! 10!2 ! 7 ! 2 ! 8 !.C =36-45b) A ,C =c. 1620There are 1620 different w a y s that a committee of4 people can be chosen if there must be an equalnumber of m e n and w o m e n on the committee. ° 4 ! 6 !. „ C , 210There are 210 different w a y s that a committee of4 people can be chosen if no m e n can be on thecommittee.4-34 Chapter 4- C o u n t i n g Methods
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1S. e.g., Let A represent the n u m b e r of w a y s toassign teachers to the first group of 5:5!10!A = 3003N o w there are 15 - 5 or 10 teachers left to assign.Lci f; ir-ptesent the number of w a y s to assign theremaining teachers to the second group of 5;B•. rs left to assign to1 w a y that this canI number of w a y s to10!5!5!8 252N o w there i-the last grot f i - h. i-.-be done. I • -i- -i !assign the tT=A-BT= 3003 252T = 756 756There are 756 756 different w a y s 15 teachers can bedivided into 3 groups of 5.17. c q.. The first point can be joined with 11 more[soinO- to form straight lines. T h e second point canthen be joined with 10 more points to form straightlines (since it w a s already joined with the first point).T h e third point can be joined with 9 more points toform straight lines (since it w a s already joined with thefirst two points). This pattern continues on until I get tothe seconddast point that can only be joined with thelast point (since it w a s already joined with the other10 points). T h e last point cannot be joined any furthersince It IS already joined to every other point in thecircle. Using this observed pattern. I can calculate then u m b e r of straight lines (L):/. = 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 11 = 66There are 66 different w a y s the points can be joinedto form straight lines.18. a) Since there is one more boy than there aregiris, the line must always follow this pattern:B G B G B G B G B G B G B . T h u s the boys are arranged in7 positions, and the giris in 6 positions.7! 6! = 3 628 800There are 3 628 8 0 0 w a y s in which the children canbe arranged in one row if the boys and giris mustalternate positions.b) Group the tnplets as one. There are 3! w a y s inwhich the tnplets can arrange themselves. Let Brepresent the number of different arrangements:e = 1 1 ! 3!6 = 39 916 800 6B = 239 500 800There are 239 500 800 w a y s in which the children canbe arranged in one row if the triplets must stand nextto each other.19. C a s f 1 face cards and 3 non-face cards: 12C2 • 40C a s e 2: 3 face cards and 2 non face cards: 12C3 • 40C2C a s e 3: 4 face cards and 1 non-face card: 12C4 - 40C1C a s e 4: 5 face cards and 0 non-face cards: 12C5 • 40CoLet H represent the number of hands with at least 2t ^. cards:IlCz • 40C3 + 12C3 • 40C2 + 12C4 • 4oCi + 12C5 - 40Co/ 36 • 9880 + 2 2 0 • 780 + 4 9 5 • 40 + 792 1544 272! Li-.e are 844 2 7 2 different five-card hands with atleast two face cards.C h a p t e r T a s k , p a g e 2 9 5A. Combinations. T h e order in which the dice aretossed does not matter (note that players toss all8 dice simultaneously) nor does the w a y the dice arearranged w h e n they land matter. W h a t is important isthe o u t c o m e of each toss a combination of numberof dice that land with the s a m e side up A N D n u m b e rof dice that land with a different side up, for example,7 dice land with the s a m e face up A N D 1 die with theopposite face up.B. Each o u t c o m e can happen two w a y s . For example.8 with the s a m e side up could occur as 8 of theunmarked sides face up or 8 of the marked sides faceup. That is w h y each calculation is the s u m of twocombination values:8 dice land with the s a m e side up^ f+ or 1 + 1 or 27 dice land with the s a m e side upi f f, 16 dice land with the s a m e side upr i . ^1 i ^ l 7or 8 + 8 or 16or 28 + 28 or 568V21^ 8 ¥ 2^6)l2j"i6ji2^5 dice land with the s a m e side upor 56 + 56 or 1124 dice land with the s a m e side up8 ¥ 4 3 dice land with the s a m e side up+I 4 J Ior 70 + 70 or 140or 56 + 56 or 112F o u n d a t i o n s of Mathematics 12 S o l u t i o n s Manual 4-35
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C . Yes, i think it is fair.O u t c o m e Points Number of W a y sO u t c o m e C a nO c c u r8 dice land s a m eside up10 27 dice land s a m eside up4 166 dice land sameside up2 565, 4, or 3 diceland same side up0 364T h e highest number of beans (10) is awarded for theoutcome that can happen in the least number ofways, 8 dice landing s a m e side up; 4 beans areawarded for the o u t c o m e that can happen in thesecond fewest number of ways, 7 dice landing s a m eside up; 2 beans are a w a r d e d for the outcome thatcan happen in the third fewest number of ways, 6 dicelanding same side up. The most likely outcomes of 5,4, and 3 dice landing s a m e side up ail receive thelowest number of beans (0). So the point systemrewards the least likely outcomes with the most beansand the most likely outcomes with the fewest beans.C h a p t e r 4 D i a g n o s t i c T e s t , T R p a g e 2 6 91.a)Coinheadstails30 i*o>u^b«« OutoEMnetheads and redheads and orangeheads and purpleheads and yeUa<Hheads and greentails irdi redtails and orangetails and purpletails andyellOK^ftails and greengf eenb) b) e.g.. By looking at the tree diagram, there is oneway he could flip a head and spin green, and tenpossible outcomes.P(heads and green) = ^10or 10%.2. a)Child 1 Child 2 Child 3B B BB G BB B GG B BG G BG B GB G GG G Gb) There are wo w a y s all three children will be thesame gender, either all boys or all gids, and there areeight possible outcomes.2P(all boys or all gids) = - or 2 5 % chance, assuming8that having a boy or gid is egually likely.3. a) B = {the set of elements not in S}B = {a, b, c, d, e, i, o, u}b) Au B = {the set of elements in A and B}Au B = {a, b, c, d, e, i, o, u, x, y, z}c) ArB = {the set of elements in both A and B}AnB = {y}4-36 C h a p t e r 4: C o u n t i n g Methods
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5. e.g., Let x be the number of good singers withoutdancing skills. Let A be the set of singers and B theset of dancers.n{A) = X + 6n{B) = 10 + 6 or 16niAnB) =6niAuB) = 2 4niA KJB) =niA) + niB) - niA n B)24 = x + 6 + 1 6 - 624 = x + 168 = xUse a V e n n diagram to help solve the problem.c) ii) the intersection of sets A and Satuditioning gsr Issfericm^^ singers ^10 + 6 + x = 2 41 6 + x = 2 4X = 8There were 8 gids who were good singers but notgood dancers.R e v i e w o f T e r m s a n d C o n n e c t i o n s ,T R p a g e 2 7 21. a) v) disjoint sets A and Bd) i) tree diagramfirst coin Mcofid eofcie) iv) outcome table1 2 3 4 5 6t 2 3 4 5 6 72 3 4 5 6 7 83 4 5 6 7 8 94 5 6 7 8 9 105 6 7 8 9 10 116 7 8 9 10 11 12F o u n d a t i o n s of Mathematics 12 Solutions Manual 4-37
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7. e.g., / is the universal set of integers. E is thesubset of even integers. O is the subset of oddintegers.3. a) n{A uB) = n{A) + n{B) - niA n B)niA)=^2, niB) = 9, niAnB) = 5r)(/u 8 ) = 12 + 9 - 5 = 16b) niA u 8 ) = n ( ^ ) + n(B) - niA n B)niA) = 23, n ( 8 ) = 16, n(> n 8 ) = 1niA uB) = 23+ 1 6 - 1 = 3 84. a) Let S represent the universal set of all students.Let A represent the students w h o attended the firstschool dance, and let 8 represent students w h oattended the second school dance. T h e n niA u 8 ) isthe number of students w h o went to one of the firsttwo school dances.niA uB) = niA) + niB) - niA n 8 )niA) = 420, niB) = 480, niA nB) = 285niA u 8 ) = 4 2 0 + 4 8 0 - 285 = 615615 students went to one of the first two schooldances of the year.b) niA u 8 ) is the number of students w h o did notattend either dance.niA u 8 ) = S - niA u 8 )S = 1200, n { A u 8 ) = 615n ( / u 8 ) = 1 2 0 0 - 6 1 5 = 585585 students did not attend either dance.5. a) B = {set of black face cards in a standard deck ofplaying cards} = { J * , Q * , K * , J * , Q A , K * }b) D = {set of different three-digit numbers using thedigits 1, 3, and 5} = {135, 153, 315, 3 5 1 , 513, 531}c) S = {set of all possible sums w h e n a pair of dice isrolled} = {2, 3 , 4 , 5, 6, 7, 8, 9, 10, 1 1 , 12}d) T = {set of all the days of the w e e k with names thatbegin with T} = {Tuesday, Thursday}6. a) { } , {red}, {blue}, {red, blue}b) { } . {2}, {4}, {6}, {8}, {2, 4}, {2, 6}, {2, 8}, {4, 6},{4, 8}, {6, 8}, {2, 4, 6}, {2, 4, 8}, {4, 6, 8}, {2, 4, 6, 8}c) { } , {Apnl}, {May}, {June}, {AphI, May}, {Apnl, June},{May, June}, {Apnl, May, June}d) { } , {100}8. S = {l, N , T , E, R, S, C O }a) A = {set of vowels in S}A = {I, E, 0 }b) 8 = {set of letters in S E C T I O N }8 = {S, E, C, T, I, O, N}c) A u 8= {set of vowels in S and set of letters in S E C T I O N }= {S, E, C, T, I, O, N}d) A 8 = {elements in both A and 8 }A n 8 = {t, E, 0 }e) A u 8 = {elements not in A and elements not in 8 }AuB= { N , T, R, S, C, E, I, 0 }f) A n 8 = {elements in neither A nor 8 } = { N , T, R}9. a ) S = { 1 , 2 , 3, 4, 5, 6, 7, 8, 9, 10, 1 1 , 12, 13, 14}A = {2, 4, 6, 8, 10}8 = { 1 , 3 , 5, 7, 9, 1 1 , 13}A = {the elements of S not in A}A = { 1 , 3, 5, 7, 9, 1 1 , 12, 13, 14}8 = {the elements of S not in 8 } = {2, 4, 6, 8, 10, 12, 14}S4-38 C h a p t e r 4: C o u n t i n g Methods
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A u B = {the elements of A and B}= { 1 , 2 , 3 , 4 , 5, 6, 7, 8, 9, 10, 1 1 , 13}A n B = {the elements in both A and 6} = { }b) S = { A * , 2 * , 3 * , 4 A , 5 A , 6 A , 7 A , 8 A , 9 A , 1 0 A ,A * , 2 * , 3 * , 4 A , 5 * , 6 * , 7 * , 8 * , 9 * , 1 0 * ,A v , 2 ¥ , 3 ¥ , 4 ¥ , 5 ¥ , 6 ¥ , 7 ¥ , 8 ¥ , 9 ¥ , 1 0 ¥ ,A * , 2 * . 3 4 , 4 * , 5 * , 6 » , 7 * , 8 * , 9 * , 1 0 * }A = { 3 A , 6 A , 9 A , 3 A , 6 * , 9 * , 3 ¥ , 6 ¥ , 9 ¥ , 3 * , 6 * , 9 * }e = { 2 A , 4 A , 6 A , 8 A , 1 0 A , 2 * , 4 * , 6 * , 8 * , 1 0 * ,2 ¥ , 4 ¥ , 6 ¥ , 8 ¥ , 1 0 ¥ , 2 * , 4 » , 6 » , 8 » , 1 0 * }A = {the elements of S not in A}A = { A A , 2 A , 4 A , 5 A , 7 A , 8 A , 1 0 A , A A , 2 * . 4 * ,5 * , 7 * , 8 * , 1 0 A , A ¥ , 2 ¥ , 4 ¥ , 5 ¥ , 7 ¥ , 8 ¥ ,1 0 ¥ , A * , 2 * , 4 * , 5 * , 7 * , 8 » , 1 0 * }Ae = {the elements of S not in B} = { A A , 3 A , 5 A , 7 A ,9 A , A A , 3 A , 5 A , 7 A , 9 A , A ¥ , 3 ¥ , 5 ¥ , 7 ¥ , 9 ¥ ,A * , 3 * , 54, 7 * , 9 * }SA u B = {the elements of A and S} = { 2 A , 3 A , 4 A , 6 A ,8 A , 9 A , 1 0 A , 2 A , 3 A , 4 A , 6 A , 8 A , 9 A , 1 0 A ,2 ¥ , 3 ¥ , 4 ¥ , 6 ¥ , 8 ¥ , 9 ¥ , 1 0 ¥ . 2 * , 3 * , 4 * ,6 * , 8 * , 9 * , 1 0 * }A n B = {the elements in both A and S}A n B = { 6 A , 6 A , 6 ¥ , 6 * }s/I Ai { B ) /10. e.g.,a) S is the universal set of all students in my math class.A is the subset of students w h o are taller than 6 ft.6 is the subset of all students with black hair.A n B (shaded) is the subset of students w h o aretaller than 6 ft and have black hair.b) S is the universal set of all cards in a standarddeck of playing cards.A is the subset of all face cards.B is the subset of all red cards.Au B (shaded) is the subset of all face cards and allred cards.11. Let S represent the universal set of all first-yearstudents.Let C represent the students w h o take calculus.Let A represent the students w h o take algebra. Thenn(C u A) will be the number of first-year studentsw h o take neither calculus nor algebra.n{S) = 200n(C) = 110n{A) = 75n ( C r ^ A) = 60n{C uA) = n{C) + n{A) - n{C n A)n{CuA) = 110 + 7 5 - 6 0n{CuA)= 125n{CuAy = S-n{CuA)r 7 ( C u A ) = 2 0 0 - 1 2 5n{CuAY = 75There are 75 first-year students w h o take neithercalculus nor algebra.F o u n d a t i o n s of Mathematics 12 Solutions Manual 4-39
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Chapter 4 Test, TR page 2821. e.g., T h e table s h o w s the possible wins and lossesof one of the teams. S h a d e d cells indicate g a m e s thatwould not actually be played, since o n e of the t e a m swill have already w o n or lost two g a m e s . That m e a n sthere are only six different outcomes.G a m e 1 G a m e 2 G a m e 3 O u t c o m e sW W W 1 o u t c o m eW Ww L w 1 o u t c o m ew L L 1 o u t c o m eL W W 1 o u t c o m eL W L 1 o u t c o m eL L W 1 o u t c o m eL Lb) There are bNO different w a y s a t e a m can loseexactly one g a m e but still win the championship:W L W or L W W .2. a) T h e sandwich has four elements: egg salad orchicken salad, lettuce or tomato, butter ormayonnaise, and whole w h e a t bun or w h o l e grainbagel or s e s a m e seed bun. Using the FundamentalCounting Pnnciple:2 2 2 3 = 24 w a y s to m a k e a sandwich. I a s s u m e dthat the sandwich has exactly one item from each ofset of choices.b) For each of the four digits, there are six optionsand repetition is allowed. Using the FundamentalCounting Pnnciple:6 • 6 • 6 6 = 1296 w a y s to m a k e a password. Ia s s u m e d starting the password with a 0 w a s allowed.c) There are 13 hearts and 13 clubs in a standarddeck of 52 cards, and the sets of hearts and clubs aredisjoint. There are 13 + 13 = 26 w a y s to draw a heartor club. I m a d e no assumptions.d) There are five letters in T E E T H with both T and Erepeating twice. There are = 30 arrangements ofthe letters. I m a d e no assumptions.e) There are 25 different toppings, and Jim mustchoose 3. Order does not matter. Jim can order252300 different pizzas. I a s s u m e d he wouldchoose exactly three toppings and each would bedifferent.5 ! ( l 0 - 5 ) !1 0 - 9 - 8 - 7 6 5!5 I 5 - 4 - 3 - 2 - 1^ 1 0 9 - 8 - 7 6" 5 - 4 - 3 - 2 - 1^ 30 240120= 2523. a) Since order does not matter: 1 0 1 10!. 5105noi5105105^There are 252 selections that can be made.b) Since order matters:° ( 1 0 - 5 ) !1 0 - 9 - 8 - 7 6 5 !10^5 = 1 0 . 9 . 8 - 7 . 61 0 ^ - 3 0 240There are 30 240 selections if the selections areordered by preference.c) e.g., Order does not matter in part a) but it doesmatter in part b), so part a) involves combinations,whereas part b) involves permutations.d) e.g.. The answer to part a) is 5! or 120 timessmaller than the answer to part b). This is becausethe 30 240 five-novel selections from part b) must bedivided by 5! to eliminate combinations that are thesame, because order does not matter.4. To solve the problem, look at the walk in threesections. In the first section, M a h a has to walk threeblocks east and three blocks south for a total of sixblocks; in the second section she has to walk oneblock east and one block south for a total of twoblocks; in the third section she has to walk two blockseast and hwo blocks south for a total of four blocks.First section:6! _ 6 - 5 4 3!3 ! 3 ! " " 3 ! 3 - 2 16! _ 6 - 5 43 ! 3 ! ~6!3!3!6!3!3!Second section:1 ! 1 ! ^ 1-1_2!^1!1!3 2 15 420= 24-40 C h a p t e r 4: C o u n t i n g Methods
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2 3212!2!2!2!2!Using the Fundamental Counting Principle:20 • 2 • 6 = 2 4 0She can walk 2 4 0 different v/ays.This problem can also be solved using a pathwaydiagram;5. a) Order does not matter, so this is a combinationproblem. There are (5 2) ways to choose t w o boysfrom five, a n d (6 2) ways to choose two giris from six.2 i! / Iff, ,f ,• 4 - 3 ! 6-5--••2 th4 D O- 1 0 0,= 150There are 150 ways to choose a committee with twoboys and two giris.b) At least two giris means two, three, or four giris:6!2 l i6![ 6 ^ 4 ) ! 4 !AWCJ 1; .Ji;«: ,1 ic f: t. 1 •;•>.!• committee, therec i c nine fl. f.pl- ioi ttjifidiimig t w o positions:l i f l111113 ^roi,,^1 / 2 712!1 l l j l 2 j • • 7 ! 2 !1 ¥ 11 1 5i ¥ 9 i12 = 36There are 36 committees that can be made if Jim andNanci must be on the committee.d) More boys than giris means three or four boys:5! 6! 5!1 ; U51 ; u51 / 14517 V4.6 l fs+i i u6^ rs1 J"U6 l (5+1 ; U1( 5 3 ) 1 3 ! (6^^-l)!1! (5 ^^•4)14!5! 6! 5!: . .+213! 5111 1!4!5-4-3! 6 - 5 ! 5 - 4 !2!3! 5 ! 1 ! 1!4!5 - 4 6 52 1 110-6 + 5^1-160 + 5There are 65 committees that can be made with moreboys than giris.F o u n d a t i o n s of Mathematics 12 S o l u t i o n s Manual 4-41
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B. a) 001, C . l60 . " :•«!/} 1):/)(/! - 2 H / 3) 60, -1,1 ? | ( u 3) 30u 5f} - 24 0(// B i ( f i i 3) 0n 8 or /I 3-3 is f x i r r i n o f j u sn = 8b | /? - 8. ./ i- 4„P_, ^ f>0(.,c;,3 5j<j /i r- 4 and n > 2 H U n > 47. if you pluf.o t h r f and K as required. Ihrs uanli.;jp[)on only O O P w a y for oaf:h position. Fhuromaininp jsovon letters can then t)o arranged inbetween, keeping in mind that there are repealedIHtom- two 72s, two Ss. and two Os-1.1 " -iOhlV". 2I22! I 212-2l^222; ^ 2 2U!2!2!j i 41,1^1= 630l 2 ! 2 ! 2 ! jThere are 6 3 0 different w a y s to arrange the letterswith the given conditions4-42 C h a p t e r 4 C o u n t i n g M e t h o d s
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