C h a p t e r 4 : C o y n t i n g M e t h o d s

L 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 5

khaki b lack red red/khak i red/b lack blue b iue/khaki b lue/b lack greer g reen/khak i g reen/b lack

Each X represen ts a d i f ferent comb ina t i on . The re a re 6 x 's ; there fore , there are six d i f ferent vana t ions of the outf i t to choose f r o m ,

b) T h e n u m b e r of outf i t var ia t ions, O, is re la ted to the

' - " r of shi r ts and the n u m b e r of shor ts ;

O . n u m b e r of shi r ts) • ( number of shor ts )

2

Ml I ' a re six d i f ferent var ia t ions of the outf i t to c h o o s e f rom. Th is ma tches the part a ) result .

2. a)

Upholstery

leather

cloth

Colour

red

•.il.-. r

red

1.1,1' It

•.V-xt.-

8 u Thereto" . • thf ro 0 Mpholstery-colour cho ices that are ava i lab le .

b) T h e n u m b e r of upho ls te ry -co lour cho ices , U, is re la ted to the n u m b e r of co lours and the n u m b e r of k inds of upho ls te ry ;

U - ( M l mbe r of co lours) • (number of upho ls te ry )

iJ 1 2

8

The re are 8 upho ls te ry -co lour cho ices that are

ava i lab le . Th is ma tches the part a) resul t .

3. a) T h e Fundamen ta l Coun t ing Pnnc ip le does not app ly b e c a u s e tasks in th is s i tuat ion are re la ted by the w o r d O R .

b) T h e F u n d a m e n t a l Coun t ing Pnnc ip le does app ly because tasks in th is s i tuat ion are re lated by the wo rd A N D .

c ) T h e Fundamen ta l Coun t ing Pnnc ip le does not app ly because tasks in th is s i tuat ion are re la ted by the w o r d O R .

d) T h e F u n d a m e n t a l Coun t ing Pnnc ip le d o e s app ly because tasks in th is s i tuat ion are re la ted by the wo rd A N D .

4. a) Game 1

win

Game 2 Game 3 Series Result

win

loss

win

loss

win

win

loss

win

loss

loss loss

b) By look ing at the t ree d i a g r a m . I can see there are

2 w a y s in wh ich K im 's t e a m can w in the ser ies

desp i te los ing one g a m e .

5. T h e n u m b e r of co lour -s ize vana t ions , C, is re la ted

to the n u m b e r of co lours and the n u m b e r of s izes ;

C = (number of co lours ) • ( numbe r of s izes)

C = 5 - 4

C - 20

The re are 20 co lour -s ize var ia t ions that a re ava i lab le .

6. T h e number of c o m p u t e r sys tems , S, the

e m p l o y e e s can bui ld for the i r cus tomers is re la ted to

the n u m b e r of desk top c o m p u t e r s (dc), the n u m b e r of

mon i to rs (m) , the n u m b e r of pr in ters (p) , and the

n u m b e r of so f tware p a c k a g e s (sp) ;

- (# of y ) (# of m) • (# of p ) • (# of sp)

^ S 4 f. 3

.5 -• .}uO

rh fc iu fo ;u , Ihe e m p l o y e e s can bui ld 360 d i f ferent

compu te r sys tems for thei r cus tomers .

7. T h e n u m b e r of poss ib le mea ls . M, is re la ted to the

n u m b e r of soups (s) , the n u m b e r of s a n d w i c h e s (sw), the

number of dr inks (dr), and the n u m b e r of desser ts (P):

M = (# of s) - (# of sw) • (# of dr) • (# of d)

/W = 3 5 • 4 2

M= 120

There fo re , there a re 120 d i f ferent mea l possib i l i t ies.

8. Event A: Se lec t ing a rap C D

O R

Event B: Se lec t ing a c lass ic rock C D

n{A u B) = n{A) + n{B)

n{A ;„„ B) = 8 + 10

n{A u B) = 18

There fo re , C h a d e n e can se lec t f r om 18 C D s to p lay in

her car s tereo that wi l l ma tch T o m ' s mus ica l tas tes .

9. a) T h e n u m b e r of d i f fe rent P IN comb ina t i ons , C. is re lated to the n u m b e r of d ig i ts f r om w h i c h to se lect for each digi t of the P IN . F:

C = P l • P2--P3- P4- Ps C = 9 9 9 9 9

C = 59 049

There are 59 049 d i f ferent f ive-digi t P IN comb ina t i ons .

F o u n d a t i o n s of Mathemati'. s «/ s o l u t i o n s Manual 4-1

b) llv: n i i r n lH j r o ! <ii lferont P IN c fm ib ina t ions , N. )s

n.' lali ' f l to l l u ' i i i in ibcr o l d ig i ts I torn w h i r j i to s t ; l t ) r j for

each d iy i l o l l l i r ; P IN . D'

N ' / ) i n, / ) ; D t Di

N M p. / 0 5

W 15 17f)

Iht jK.- or<: only 1f» 120 d i f fo ron i l ivc- f f ig i t P IN

f ;urnl) inol i f ) i is in w h i d i the dig i ts canno t repea l .

10 . f ho n(jrnt)t;r of d i f ferent t)ytes that can bo c rea ted ,

N. IS rolatfHl l o I f ie n u m b e r of d ig i ts f rom wh ich to

f .hoosf! for e a c h di t j i ! of the by te . B:

N - th H • / i . B i Bs • Bf, B, By.

N - 7 2 2 2 2 2 2 - 2

N • 256 Thf .Tutonj . 2b6 d i f ferent by les can bo creaUid

1 1 . a) 1 he f iu rn twf of d i f ferent upper-~caso letter

f .ossibi l i t ips. H. IS re lated to the n u m b e r of uppe r - case

let ters f t om wh ich to choose for each o d d pos i t ion of

the coun t ry ' s posta l code , P:

W - P i P ; P .

H 26 • 26 26

A/-- 1 / 57t>

T l i e n u m b e r of d i f ferent digi t possib i l i t ies. D, is re la ted

to the n u m b e r of d ig i ts f rom wh ich to c h o o s e for e a c h

even pos i t ion of the coun t ry ' s posta l c o d e , P;

D'^P'- Pi Pf.

D-- - ^10-10 10

O 1000

The n u m b e r of d i f ferent posta l codes that are poss ib le

in th is count ry . C. is re la ted to the n u m b e r of upper ­

case letter possib i l i t ies, N, and the n u m b e r o f digi t

possib i l i t ies, D:

C W • D

C = 17 576 • 1000

C ^ 17 576 000

T t i c re fo re , 1 7 576 000 posta l codes a re poss ib le ,

b ) T l i e n u m b e r of d i f ferent upper -case letter

possib i l i t ies, N. is re la ted to the n u m b e r of uppe r - case

let ters f r om w h i c h to choose for each odd pos i t ion of

the coun t ry ' s posta l code , P;

Al = P, Po • Pf,

A / - 2 1 - 2 1 - 2 1

N = 9261

T h e n u m b e r of d i f ferent digit possib i l i t ies, D, rema ins

the s a m e s ince all d ig i ts can be u s e d . T h e n u m b e r o f

d i f ferent posta l codes that are poss ib le in C a n a d a , C, is

re lated io the number of uppercase letter possib i l i t ies,

Ay. and the n u m b e r of digit fK)SSibi!ities, D;

C = N D

0 = 9261 1000

C = 9 261 000

There fo re . 9 261 000 posta l codes are poss ib le in

C a n a d a .

12 . T o answe r this ques t i on , I need U) de te rm ine how

many digi t r-.oml.nnations t t ioro a rc for the last four

d ig i ts of one ot these two p h o n e n u m l j e i s . and then

mul t ip ly it by 2,

The number ot digit comb ina t i ons . C, is ro ln ted to the number of poss ib le d ig i ts for each of the last four d ig i ts of o n e of tho p h o n o n u m b e r s , P' c \ p , . p... p . , . f>,

C = 10 - 10 - 10 - 10

C-^ 10 0 0 0

T h e n u m b e r of p h o n e n u m t j e r s is 2C s ince there an)

two g iven tt.nnplates for the phone n u m b e r s in tho

ques t ion .

2 C 2(10 0 0 0 )

2 C =̂ 20 0 0 0

There fo re . 20 000 d i f fe rent p h o n e n u m b e r s are

poss ib le for th is t o w n ,

13 . T h e n u m b e r of d i f fe rent codes , C. is re lated to

n u m b e r of pos i t ions f r o m w h i c h to se lec t for each

swi tch of the ga r age doo r opener . G:

C^Gi- G2- Ga • G,, • Gr, • Gr, • G, - Gr; G:.

C = 3 • 3 - 3 - 3 ^ 3 • 3 • 3 - 3 3

C - 19 683 There fo re , 19 683 d i f ferent c o d e s a re poss ib le .

14 . Even t A ; Se lec t ing a p ickup t ruck O R

Event B: Se lec t ing a p a s s e n g e r v a n O R

Event C: Se lec t ing a car O R

Event D: Se lec t ing a spor ts uti l i ty veh ic le

n{A UBKJC u D ) = n{A) + n(B) + n{C) + n (D )

n{A u B u C i.j D) = 8 + 10 + 35 + 12

niA uBu G>j D) = 65

There fo re , a c u s t o m e r has 65 cho ices w h e n rent ing

jus t o n e veh ic le .

15 . a) Mul t ip ly the n u m b e r of s izes o f the crust , by the

n u m b e r of t ypes o f the crust , by the n u m b e r of t ypes

of cheese , by the n u m b e r of t ypes of t oma to sauce .

2 • 2 - 2 • 2 = 16

Mul t ip ly th is n u m b e r by the n u m b e r of d i f ferent

topp ings .

1 6 - 2 0 = 320

There fo re , the re a re 3 2 0 d i f fe ren t p i zzas that can be

m a d e wi th any crust , c h e e s e , tomato sauce , and

1 topp ing .

b ) Mul t ip ly the n u m b e r of t y p e s of c h e e s e by the

n u m b e r of t ypes of t o m a t o s a u c e .

2 - 2 = 4

There fo re , there are 4 d i f ferent p i zzas tha t can be

made wi th a th in w h o l e - w h e a t crust , t o m a t o sauce ,

cheese , a n d no t opp ings .

4-2

1 i . a | T h e n u m b e r of d i f ferent ypper-^case letter

possib i l i t ies, N, is re lated to the n u m b e r of upper^case

let ters f r o m w h i c h to choose for each of the f irst th ree

pos i t ions of the A lber ta l i cence plate, P:

W = 24 - 24 • 24

N= 13 824

f l . . . n u m b e r of d i f ferent digit possib i l i t ies, O, is re la ted

so In. - number of digi ts f r om w h i c h to c h o o s e for e a c h of

H..- I . r i th ree pos i t ions of the A lber ta l i cence p late. P

D - P.-P^^Ps

- ••0 - 10 - 10

P' m o o

T h e n u m b e r of d i f ferent poss ib le A lber ta l icence p la tes,

I = -I J ! the n u m b e r of upper^case letter

, / ; and the n u m b e r of digi t possib i l i t ies, O; ^ / i O

< ' , i ' ' ' , I f . 0 0

1 ; i.j'i CO.' A lber ta l i cence p la tes are poss ib le ,

b) I he n u m b e r of d i f ferent upper -case letter

possib i l i t ies, N, rema ins the s a m e s ince the n u m b e r of

let ters in the p la tes and the n u m b e r of let ters that can

be used is the s a m e as in a) .

T h e n u m b e r of d i f ferent digi t possib i l i t ies, 0 , is re lated

to the n u m b e r of d ig i ts f r o m w h i c h to c h o o s e for each of

the last four pos i t ions of the A lber ta l icence p late. P;

D = P4 • Ps • Fe • P?

D = 10 • 10 • 10 - 10

D = 10 000

T h e n u m b e r of d i f ferent poss ib le A lber ta l i cence p la tes,

C, is re la ted to the n u m b e r of uppe r - case letter

possib i l i t ies. N, and the n u m b e r of digi t possib i l i t ies, D:

C = N- D

C = 13 824 - 10 0 0 0

C = 138 240 000

138 240 000 13 824 000 = 124 4 1 6 000

So , 124 416 000 more l icence p la tes a re poss ib le ,

17 . e g, , If mul t ip le tasks are re la ted by A N D , it m e a n s

the F u n d a m e n t a l Coun t ing Pnnc ip le can be used and

the total n u m b e r of so lu t ions is the p roduc t of the

so lu t ions to each task. For e x a m p l e , A 4-dig i t P IN

invo lves choos ing the 1st digi t A N D the 2nd digi t A N D

the 3rd digi t A N D the 4 th digit . So the n u m b e r of

so lu t ions IS 10 10 10 10 = 10 000 . O R m e a n s the

so lu t ion mus t mee t at least one cond i t ion so you mus t

add the n u m b e r of so lu t ions to each cond i t ion , and

then subt rac t the n u m b e r of so lu t ions that mee t all

cond i t ions . For examp le : Ca lcu la t ing the n u m b e r of

4-dig i t P INs that star t w i th 3 O R end wi th 3. T h e

so lu t ion IS the n u m b e r of P INs that start w i th 3, p lus

the n u m b e r of P INs that end wi th 3. m inus the n u m b e r

of P INs that both start and end wi th 3:

1000 + 1 0 0 0 - 100 = 1900.

18. a) i) Even t A D raw ing a k ing O R

Event B; D raw ing a q u e e n

r i f / . = n(A) + niB)

n{A u e ) = 4 + 4

n{A u e ) = 8

L ike l ihood = ^ 52

L ike l ihood = ^ 13

There fo re , there is a 2 in 13 c h a n c e that a k ing or a q u e e n wi l l be d r a w n .

ii) Even t A: D r a w i n g a d i a m o n d

OR

Event B: D raw ing a c lub

I >'•'"• P) -•.'/• I I''p.)

n{A u S) = 26

L ike l ihood = ^ 2

There fo re , there is a 1 m 2 c h a n c e that a d i a m o n d or a c lub wi l l be d r a w n ,

iii) Even t A: D raw ing an A c e O R

Event B: D raw ing a s p a d e

n(A u B) = n{A) + n(B) - n{A n B)

n(Au B) = 4 +13^1

n(A u e ) = 16

L ike l ihood = ^ 52

L ike l ihood = 13

There fo re , there is a 4 in 13 c h a n c e that an ace or a

s p a d e wil l be d r a w n .

b) No . e.g. . because the F u n d a m e n t a l Coun t i ng

Pr inc ip le on ly app l ies w h e n tasks are re la ted by the

wo rd A N D .

19. e .g. . T o beg in , there are 90 two-d ig i t n u m b e r s .

The re are 10 wi th a 1 in the tens c o l u m n , 10 wi th a 2

in the tens c o l u m n , and th is pat tern con t inues unti l I

reach the 10 w i th a 9 in the tens c o l u m n . Next . I mus t

de te rm ine the n u m b e r s that are d iv is ib le by e i ther 2 or

5. I know that eve ry o ther n u m b e r is even and thus

d iv is ib le by 2. Th is m e a n s that or 4 5 of the t w o -2

digi t n u m b e r s are d iv is ib le by 2. T h e two-d ig i t

n u m b e r s that are d iv is ib le by 5 can be f ound by

star t ing at the f irst two-d ig i t number , 10, and then

coun t ing by 5 unti l I get to a three-d ig i t number .

By do ing th is . I can de te rm ine that the two-d ig i t

n u m b e r s that are d iv is ib le by 5 a re : 10, 15, 20 , 25 . 30 .

35 , 40 , 45 . 50 . 55 , 60 . 65 , 70 . 75 , 80 , 85 . 90 . 95 .

The re are 18 of t h e m . I see that half of t hem 'o r 9 are

even and thus d iv is ib le by 2.

F o u n d a t i o n s of Mathemati ' ^ V - . . lu t ions Manual 4-3

There fo re , the re are 9 n u m b e r s tha t a re d iv is ib le by 5

a n d not by 2. If I add th is toge ther w i th the n u m b e r of

two-d ig i t n u m b e r s that a re d iv is ib le by two (45) . I see

that there are 54 two-^digit n u m b e r s d iv is ib le by 2 or 5,

W h a t e v e r is le f tover f r o m the two digi t n u m b e r s a re

the ones that a re not d iv is ib le by e i ther 2 or 5, Th is

a m o u n t is:

90 54 = 36 . T h u s , there a re 36 two-d ig i t n u m b e r s

that are not d iv is ib le by e i ther 2 or 5.

20. T h e n u m b e r of d i f fe rent o u t c o m e s for a s tudent ' s

test . N, is re la ted to the n u m b e r of poss ib le a n s w e r s

for each ques t ion on the test . T:

W = Tl • 12 • Ta • T4 - Ts - Te • T/ • Te • Tg • Tn

IV = 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2

IV = 1024

A per fect score is on ly 1 ou t of t hese 1024 o u t c o m e s ;

there fo re , there is a 1 in 1024 chance that the s tuden t

wi l l get a per fect sco re .

2 1 . Th is ques t ion is so l ved by cons tan t app l ica t ion of

the F u n d a m e n t a l Coun t i ng Pnnc ip le .

If an i tem f rom each ca tegory is se lec ted :

0 = 3 5 4 2

0 = 120

If no soup is se lec ted :

0 = 5 4 2

0 = 40

If no sandw ich is se lec ted :

0 = 3 4 2

0 = 24

If no dr ink is se lec ted :

0 = 3 5 2

0 = 30

If no desser t is se lec ted :

0 = 3 5 - 4

O = 60

If no s o u p or s a n d w i c h is se lec ted :

0 = 4 2

0 = 8

If no soup or dr ink se lec ted :

0 = 5 2

0 = 10

If no soup or desser t is se lec ted :

0 = 5 4

0 = 20

If no sandw ich or d n n k is se lec ted :

0 = 3 2

0 = 6

If no sandw ich or desser t is se lec ted :

0 = 3 4

0 = 1 2

If no dnnk or desser t is se lec ted :

0 = 3 5

0 = 1 5

If on ly n <^nur s a n d w i r h dr ink or desser t is se lec ted :

0 • :\ 5, 4 2 f,.,,., - 120 4 40 t ;•-•} t M) . 60 + 8 + 10 + 20 + 6 + 12

-I 1 h 4 3 < S i- 4 • 2

,1 - 350

1 he re fu re , 3 5 9 mea ls a re poss ib le if y o u d o not have

to c h o o s e an i tem f r o m a ca tegory .

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 d

F a c t o r i a l N o t a t i o n , p a g e 2 4 3

1 . a) 6! = 6 - 5 - 4 • 3 - 2 - 1

61 = 720

b) 9 - 8 ! = 9 - ( 8 - 7 - 6 - 5 - 4 - 3 - 2 - l )

9 - 8 ! = 9 - 4 0 3 2 0

9 8! 3 6 2 8 8 0

3 '7 I

^ 2 1

. 5! c ) — =•

' 3 !

5

3!

5!

3!

5!

3!

5!

3!

8 !

— = 5 4 3 2 1

5 - 4 . M 3!

^ = 5 - 4 . 1

- 2 0

a /• b j j 4 J 2 1

f b 5 4 3 2 r

7 6 <.i 4 3

^y 8

/ 6 71

7!

2 1

8 1

8!

7!

8!

7!

8!

7!

7!

e) 3 ! - 2 ! = ( 3 - 2 - l ) - ( 2 - l )

3 ! - 2 ! = 6 - 2

3 t - 2 ! = 12

9 ! ^ 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 1

413! 3 2 1 ' ' 4 3 2 1

4 ! 3 ! 3 2 4 !

413! 3 2

9!

413!

9!

413!

= 3 - 4 - 7 - 6 - 5

2520

4-4 C h a p t e r 4: C o u n t i n g Methods

Posst ioJ ! Ptji^ilmn *^(jsitifm

I ' f i rmi i . i tK t t i

1

P e r m i H a t i o i i 2

P e r m i i f a t i o r i

3 K.I

P e r m u t a t t o r * 4

Permuta t , i r , n

5

Permuta t ion 6

Raj Sarah Ken

b) Let L rep resen t the total n u m b e r of pe rmuta t ions :

1 = 3 - 2 - 1

L = 3!

3. .

b)

D !

c) 15 H i j 1 ' . 1 :

A < 2 ! 4 :*,

IS M I I ' l !

4 J 2 1 l^'t-lf

98 !

1 0 0 - 9 9 = 100!

9 8 !

4. Exp ress ions a) , c ) , and d) a re unde f ined because

factor ia l no ta t ion is on ly de f ined for natura l n u m b e r s ,

5. a) 8-7-»^d S / 1̂ 3

8 - 7 - 6 f H / / X I

8 - 7 - 6 ! = 5 6 - 7 2 0

8 - 7 - 6 ! - 4 0 3 2 0

h i 1?1 ''^ 1 I 10 y 3 7 fi 5 4 / 1

10! ^ ' 1 0 " F 8 " " / 6 5^4 3^2^ 1

1 2 ! ^ -12 ^-1 10 9 8 7 6 5 4 3 2 1

10! ' " 10 9 8 7 6 5 4 3 2 1

1 ^ = 1 2 1 1 . 1 ^ 10! 10 !

12!

10!

12!

10!

= 12-11-1

132

8 ! 8 ^ 6 !

2 ! - 6 ! 2 6!

2 ! - 6 ! 2

: 4 - 7

28

d)

8!

2 ! - 6 !

8 !

21 -6 !

/ • ' r ,

5! " •)

5!

5! 5! 7 ^

5!

7 - 6 !

5! 42

e)

9! 91

\ ' i ;

1 '.

11 % 4 '<

2 1

2 i

^1 ;

6 !

2 ! - 2 !

6 !

2 ! - 2 !

6 !

= 4 ( 3 - 5 - 4 - 3 )

= 4 ( 1 8 0 )

4 1 = 720 2 ! - 2 ! ;

f | 4 ! + 3 ! + 2 ! + 1 ! = ( 4 - 3 - 2 - l ) - f ( 3 - 2 - l ) + ( 2 - l ) + 1

4 ! + 3 ! + 2 ! + 1 ! = 24 + 6-f 2 + 1

4!-f 3 ! + 2 ! + 1 ! = 33

e . 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 Mathemat ics V/ S o l u t i o n s IVIanual

b) , •• + 2 | !

1,1 41b; . ,.({n + 2){n + i)(n)in^i}...(3)i2}{t

in-. 4 1 '

( 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)(,/ rf

n-Z]\

I

e )

in

\i, 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.'

f f l 5)1

f n .4)!

^ = (r i + 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 - - 1

7. The re a re n ine s tuden ts in the l ineup, so there are

n ine poss ib le pos i t ions . Let L represent the total

n u m b e r of pe rmuta t ions :

L = 9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1

L = 9!

l = 9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1

L = 72 • 7 • 30 • 4 • 6

L = 72 • 210 • 24

1 = 362 880

The re are 362 8 8 0 pe rmuta t i ons for the n ine s tuden ts

at the Ca lgary S t a m p e d e .

8. The re a re f ive s tuden ts in t he c lub and there a re

f ive poss ib le pos i t ions . Let L rep resen t the total

n u m b e r of pe rmu ta t i ons :

L = 5 4 3 2 1

/

/ - -b 4 3 • 2 • 1

- 2 0 - 6

/ - 120 T h e r e are 120 d i f ferent w a y s to se lec t m e m b e r s for the f ive pos i t ions . 9 . The re are six act iv i t ies to d o and there are six days . Let L represen t the total n u m b e r of pe rmuta t ions :

h i

.' H 5 - 4 • 3 - 2 • 1

I • : i . 4 6

•' 1 / 0 - 6

/ - 7-'0

1 ht ;ru a re 720 d i f ferent w a y s they can s e q u e n c e

ihos(< act iv i t ies ove r the six days .

10. The re a re 28 mov ies , so the re a re 28 poss ib le

spo ts for the mov ies to go . Let L represen t the total

n u m b e r of pe rmu ta t i ons :

L = 28 !

L = 3 .048 . . . X 10^®

The re are abou t 3.05 x 10^® poss ib le pe rmuta t i ons of

the mov ie list.

f l !

(n + l ) ( n ) ( i i ^ l ) . . . ( 3 ) ( 2 ) ( l )

(V r+ l ) (n ! )

nl

1 1 . a) 10

10

- - 10

10

Check n = 9

LS RS

(y 1 l l ! 10

91

10!

9!

10 9 '

9!

10 The re is one so lu t ion . n = 9.

4-6 C h a p t e r 4" C o u n t i n g Methods

b) (11 + 2 ) ! = 9

n!

( i i K o ^ i ) : . ( 3 p ) ( i )

(o + 2 ) ( i i + l ) ( i i ! )

( n + :

+ n + 2n + 2 = 6

n'' + 3 n + 2 = 6

n^ + 3 r i - 4 = 0

(n + 4 ) ( f i - l ) = 0

ri + 4 = 0 or f i - 1 = 0

n = - 4 II = 1

Check n = LS

"'-_4 2 ] !

s unde f ined

Check n = 1

LS RS

< i 1 2)1 (.

1!

3!

11

3 - 2 - 1 !

1!

3 2

6

The re is one so lu t ion , ri = 1 .

c ,

t z 3 ^ " - ) ( ^ ^ 3 ) . . . ( 3 ) ( 2 ) : 1 i

{n 2](n J ) . . . (3 ) (2) (1)

( n ^ 2 ) i

n-^1 = 8

n = 9

RS

126

126

- 1 2 6

8! 71

i l l : 7!

8

The re is one so lu t ion , n = 9,

CJ) ±7l i . 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 ) = 126

3 ( n ' + n ) = 126

3(f,2 + n ) - ^ 1 2 6 - 0

3 ^ ( n ^ + n ) ^ 4 2 l = 0

3 ( f i ' + n - 4 2 ) = 0

3 ( n + 7 ) ( r 7 ^ 6 ) - 0

f i + 7 - 0 o r n - 6 = 0

n = -7 n - 6

Check n = - 7

LS RS

3 ( ^ 7 + 1)! 126

( ^ 7 - 1 ) 1

8 - j - ^ y - IS unde f ined

C h e c k n = 6

LS R S

3 ( 6 + 1)1 126

( 6 ^ 1 ) !

3 (7 ! )

5!

3 - ( 7 - 6 - 5 ! )

5!

3 - 7 - 6

126

The re is o n e so lu t ion , 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

L • r p i e s u i il t i l l ni

I a / (, L 4

I

1 - 8 / h H 1

L 8 A? 20

12. Ihcu: .Ko '»nht more p layers left to o rgan ize so

It (OH- c>ifih! i r iore spo ts left in the bat t ing order . Let

mber of pe rmu ta t i ons :

3 2 1

3 - 2 - 1

L - :VAb 120

L = 40 320 The re are 4 0 320 poss ib le bat t ing o rders .

13. The re are 7 poss ib le dig i ts to use and there are 7

dig i ts in each ser ia l number . Let L rep resen t the

n u m b e r of pe rmuta t ions :

1 = 7 - 6 - 5 - 4 - 3 - 2 - 1

1 = 7!

The re a re 7! poss ib le ser ia l n u m b e r s . Th is m a k e s sense

because , e .g . , the in teger in the factor ia l (7 in th is case)

for the n u m b e r of pe rmuta t i ons is norma l l y equa l to the

n u m b e r of spo ts in w h i c h there are th ings to p lace. The re

a re seven spo ts in the ser ia l n u m b e r so th is m e a n s that

the n u m b e r of pe rmuta t ions shou ld be 7! w h i c h ma tches

the answe r that w a s f o u n d .

14. The re are 5 cars to be a r ranged b e t w e e n the

eng ine and the c a b o o s e so there are 5 spo ts in wh i ch

the cars can be l ined up . Let L rep resen t the n u m b e r

of pe rmuta t ions :

1 = 5 4 3 2 1

1 = 5!

1 = 5 4 - 3 2 1

1 = 20 6

1 = 120

The re are 120 w a y s for the cars to be a r ranged

be tween the eng ine and the c a b o o s e .

15. The re w o u l d be 7 c h u c k w a g o n s beh ind Brant 's so

there are 7 spo ts w h e r e the o ther dnve rs cou ld f in ish .

Let 1 represen t the n u m b e r of pe rmu ta t i ons :

l = 7 - 6 - 5 - 4 - 3 - 2 - 1

1 = 7!

l = 7 - 6 - 5 - 4 - 3 - 2 - 1

1 = 42 • 20 6

1 = 42 - 120

1 = 5040

If Brant 's w a g o n w ins , there are 5040 d i f ferent o rders

in w h i c h the e ight c h u c k w a g o n s can f in ish .

16. a) e .g. . Y K O N U , Y U K N O , Y K N O U b) The re are f ive let ters so the re a re f ive spo ts to put the

let ters. Let 1 represen t the n u m b e r of pe rmuta t ions :

1 = 5 - 4 3 2 1

1 = 5!

The re are 5! poss ib le pe rmuta t ions . Th is m a k e s

sense because e.g. . t he in teger in the fac toha l (5 in

th is case) for the n u m b e r of pe rmuta t ions is normal ly

equa l to the n u m b e r of spo ts in w h i c h there are th ings

to p lace. The re a re f ive spo ts to p lace the let ters so

th is m e a n s that the n u m b e r of pe rmuta t i ons shou ld be

5! wh i ch ma tches the a n s w e r that w a s f o u n d .

17. a) e .g . . Us ing tna l and error . I have the fo l lowing

ca lcu la t ions:

1 ! = 1,2^ = 2 ; 2 ! = 2 , 2" = 4 ;

3! = 6. 2 ' = 8; 4 ! = 24 . 2'' = 16

I not ice that for n = 4 , nl is g rea te r than 2". Th is

con t inues for n > 4 because 2** wi l l keep get t ing

mul t ip l ied by 2. wh i le 4 ! wi l l keep get t ing mul t ip l ied by

n u m b e r s g rea te r t han 2 to ob ta in the h igher fac tona ls .

b) e .g . . Us ing w h a t I have in a ) , I k n o w that for n < 4,

nl IS not g rea te r t han 2". T h e ca lcu la t ions for t hese

va lues of n a re s h o w n in a ) . T h u s for n = { 1 , 2 . 3} . nl is

less t han 2" .

18. e .g. , First, f igure ou t how m a n y w a y s D a d e n e a n d

Arno ld can be p laced next to e a c h o ther in the l ine.

Th in cnn ho f ound us ing a t i b l o

A r n o l d i

2

3

4

D a r l o n c

1

6

F rom the tab le I can see that there are 18 d i f ferent

w a y s for D a d e n e and Arno ld to be p laced next to

each o ther in the l ine. For every o n e of t hose

18 w a y s , there a re 8 o ther dance rs to be p laced in

8 d i f fe rent spo ts in the l ine. Let 1 represen t the

n u m b e r of pe rmuta t ions :

6 5 4 3 2

- 4

6)

3 - 2

1)

1)

1 - 1-3(8 - 7

I 18(8!)

1 18(8 7 6 5

L • 18(8 • 42 20

I, = 18(336 • 120)

1 - 13(40 320 )

/ - / 2 5 760

The re a re 725 7 6 0 poss ib le a r r a n g e m e n t s of the

dance rs for the R e d River J ig .

4-8 C h a p t e r 4: C o u n t i n g Methods

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 s

A r e D i s t i n g u i s h a b l e , p a g e 2 5 5

i a , P 5!

^ ^ ( 5 ^ 2 ) !

5 !

5 ^

31

5 4 3!

3!

, 1 ^ - 2 0

8!

)!

8 !

2!

c)

1 0 ^

10!

5!

, 1.) / •

P = 1 0 - 9 - 8 - 7 - 6

^Q.̂ g 3 b 2 ! 0

9!

° ( 9 ^ 0 ) 1

p . 9 1 ^ ° 9!

7!

( 7 ^ 7 ) 1

P . I ! ^ ^ 0 !

, P , = 7!

^P^ = 7 - 6 - 5 - 4 - 3 - 2 - 1

, P , = 5040

15!

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 0

2 ;/t <- i:

• Pt-rn i i i t i i f ic»r i

' J

i ^ ; 11}

' 12

P r e s i d e n t ^V ice P r e s i d e n t "

K,4j i

K.i tr i f ia

Knt r r vl

,lf">r

N e l / l f

_ N. i / i r

M o t u i m a d

M(>\um\ai:y

Mt.har f idd

N j / i i

Katrsn.'^

J t .

hat i i r t ; ,

_ Jess

b) „ p - ^

4 !

It a p res ident and v ice-

(4^2)1

4 !

2 !

4 ^ 3 ^

2!

, P , - 4 . 3

, P , = 1 2

T h e fo rmula for „Pr g ives an answe r of 12. Th is

ma tches my resul ts f r om part a)

3. a) ' ^ {n-r)l

P. = 6!

( 6 ^ 4 ) !

P h f. 4 3 2!

, P , 6 5 4 - 3

J'] 360

There are 360 diOerent w a y s the choco la te bars can

be d is t r ibu ted.

b ) P = r ^

« ^ ( 6 ^ 1 ) !

6!

5 !

6 - 5 !

5 !

T h e choco la te bars can be d is t r ibuted in 6 d i f ferent w a y s .

p =

F o u n d a t i o n s of Ma themat ics 12 S o l u t i o n s Manual 4-9

4 . loFs is larger, e .g . , I k n o w th is by look ing at the

fo rmu la for „Pr. T h e numera to r is the s a m e for both

va lues s ince n is the s a m e . T h e d e n o m i n a t o r wi l l be

smal le r for the f irst va lue s ince it has a g rea te r r. W h e n

you d iv ide a numera to r by t w o d i f ferent denom ina to r s ,

the f inal va lue is g rea te r for the o n e w i th the smal le r

denomina to r . B a s e d o n th is . I k n o w that WPB has the

larger va lue s ince its e x p a n s i o n has the smal le r

denomina to r .

9 IJ ci . As^-unimg that any o f t h e 10 d ig i ts can be put

in any i,\ tbp 'o rema in ing spo ts for the S I N s , let S

fct.ff.'M-(it t in- n u m b e r of soc ia l i nsu rance n u m b e r s :

S - 10 10 HJ !(i 10 1(1 10 10

- K l '

S 1f'(J DUO 000

ri»M-f. . i ro 100 0 0 0 000 d i f ferent S INs that can be

r(;fji.'.t';r<!<l in cMch of t hese g roups of p rov inces a n d

lerri ! f)nf); '

5. , / '

J'

P

9 ]

6!

fd

/ ' L 0 /

,P 'M\

The re are 504 d i f ferent w a y s the pos i t ions can be f i l led,

M.O 4) .

10. a) } '

P « - 11 !

1 5 ^ =

1 5 - 1 4 - 1 3 - 1 2 - 1 1 t

11!

^gP^ = 1 5 - 1 4 - 1 3 - 1 2

, , P , - 3 2 7 6 0

The re a re 32 760 poss ib le execu t i ve commi t t ees .

7. 3P3 8!

^ ( 8 - 8 ) 1

8!

^ 0 !

P 3 '

1

P tJ

P H 7 b

, P, - 40 520

There fo re , 4 0 320 d i f ferent s igna ls cou ld be c rea ted .

' ( 5 0 0 0 - 3 ) !

p ^ 5 0 0 0 ! 5000 3 4 g g 7 ,

5000^3 =

5 0 0 0 - 4 9 9 9 - 4 9 9 8 - 4 9 9 7 !

4 9 9 7 !

5 0 0 0 - 4 9 9 9 - 4 9 9 8

5Qj,j,P3 = 124 925 010 000

The re are abou t 124 9 2 5 010 000 d i f ferent w a y s the

t ickets can be d r a w n .

1 1 / f)H

12 '

/ •

12 11 10 q 7!

/ I

12 n Ui 0 f-'

0504(1

The re are 95 040 w a y s the coach can se lec t the

s tar t ing f ive p layers .

b l A

P^

( 1 1 ^ 4 ) 1

11!

7!

1 1 - 1 0 - 9 - 8 - 7 !

7!

„ F ^ = 1 T 1 0 - 9 - 8

, , P , - 7 9 2 0

T h e r e a re 7 9 2 0 w a y s the coach can se lec t the

s tar t ing f ive p layers , if the ta l lest s tuden t mus t star t at

the cent re pos i t ion .

p . 1 ^ 10^3 7,

p ^ 1 0 - 9 - 8 - 7 !

7!

^ j , P 3 = 1 0 - 9 - 8

, „ P 3 = 7 2 0

Mul t ip ly by 2, s ince S a n d y and Na tasha can p lay the

guard pos i t ions in e i ther order . (720) (2 ) = 1440

The re a re 1440 w a y s in w h i c h the c o a c h can se lec t

the star t ing f ive p layers , if S a n d y a n d Na tasha mus t

p lay the two guard pos i t ions.

1 1 . a ) n > 0 a n d

n - 1 > 0

n> 1

The re fo re , the express ion is de f ined for n > 1 ,

w h e r e n e I.

b ) n + 2 > 0

f i > - 2

The re fo re , the express ion is de f ined for n > - 2 ,

w h e r e 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

c ) ri + 1 > 0 A N D n >0

n> 1

There fo re , the exp ress ion is de f ined fc

w h e r e n e L

d) n + 5 > 0 A N D n + 3 > 0

n > - 5 n > - 3

The re fo re , the exp ress ion is de f ined for n > ^ 3 ,

w h e r e n e I.

12 . a) ^,P,

6 ^

6!

6 ^ =

( 6 ^ 4 ) 1

6 !

2!

6 - 5 - 4 - 3 - 2 !

2!

6 - 5 - 4 - 3

360

T h e r e are 360 w a y s to d r a w the four marb les if you do

not rep lace the marb le each t ime.

b | Let L represent the n u m b e r of w a y s :

L = 6 • 6 - 6 - 6

L = 6'

L = 1296

T h e r e are 1296 w a y s to d raw the four marb les if you

rep lace the marb le each t ime.

c ) e .g. , Y e s ; if you rep lace the marb le , there are more

possib i l i t ies for the next d raw.

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 6

20 Pg = 1 8 6 0 4 8 0

The re are 1 860 4 8 0 di f ferent w a y s to awa rd the

scho la rsh ips .

b) Let L represen t the n u m b e r of w a y s :

1 = 20 - 20 - 20 - 20 - 20

L = 20^

1 = 3 200 000

The re a re 3 200 0 0 0 d i f ferent w a y s to awa rd the

scho la rsh ips .

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 0

b) Sub t rac t the total poss ib le n u m b e r s by the a n s w e r

to part a ) .

104 = 10 000

10 000 - 5040 - 4960

The re a re 4 9 6 0 d i f ferent phone n u m b e r s .

15 u) • I eed to so lve . ^ = 20 ( n - 2 ) !

fi r ^tiO 1 1 ^ 2 > 0

r i > 2

nl There fo re ,

( n ^ 2 ) !

{n){n^i):n - ' j p ; , h h ) (2)(1

"i'^ \ ' ] ( ' ' ' ; ( t )

{ „ ) ( „ :

20 is de f ined for n > 2. w h e r e n e I.

T- = 20

20

^ = 20 (11^2) !

( n ) ( n ^ l ) = 20

0 ^ - 0 = 20

n " ^ f i ^ 2 0 = 0

( n + 4 ) ( n - 5 ) = 0

/? > 4 0 or n - 5 = 0

n -4 n = 5

T h e root n = -4 is not a so lu t ion to n > 2

C h e c k f l = 5 LS RS

5P2 20

5!

( 5 ^ 2 ) 1

5!

3!

5 - 4 - 3 !

3!

5 - 4

20

The re is one so lu t ion , n = = 5.

b) 1 need to so 'vo {n t

in 1 1

1'

2)1

n - f 1 > 0 A N D n '\-Z -? 0

n>-1 11 1 - 0

n > 1

There fo re ,

n e I.

( n + 1 ^ 2 ) ! 72 IS de f ined for n > 1 , w h e r e

T h e r e are 5040 d i f ferent phone numbers poss ib le .

F o u n d a t i o n s of Mathemati u t ions Manual 4 t <

: . . I - >)<

(r, + l ) !

C h e c k f = 2

72

72

72

= 72

(11^1).

( r i + 1 | ( / / l ( » 1 i | u - ••') t . J ) i 2 | i l )

2 ) " V ' ) ( 2 | ( 1 )

( i i + l ) ( n ) = 72

+ n = 72

n ^ + n - 7 2 = 0

(o + 9 ) ( f i - - 8 ) = 0

n + 9 = 0 o r f i - ^ 8 = 0

n = - 9 f l = 8

T h e root n = 9 is not a so lu t ion to n > 1.

Check n = 8

LS RS

8 +1P2 72

9P2

(9 2)1

9!

7!

9 - 8 - 7 !

JI

9 - 8

72 The re is o n e so lu t ion , n = 8.

1 e. a l T h e equa t i on I need to so lve = 30 .

^ 0 - r)l

6 - r > 0

r < 6

There fo re . = 30 is de f ined for 0 < r < 6, w h e r e ' ( 6 - - f ) l

re I.

'.e «

6 J ^ 4 3 2 1

16 r i !

720

M

30

= 30

( 6 ^ r ) . = I 2 0

30

( 6 - f ) ! = 24

= 4

r 2

LS RS

6P2 30

6!

( 6 ^ 2 ) !

6!

4 !

6 - 5 - 4 !

4 !

6 - 5

30 The re is o n e so lu t ion , r = 2.

b ) I h<- ....luation I need to so lve is 2

7 r 0

7! 420

Th<.i.-f.,rf; 2

w h e r e r e l .

7! 420 IS de f ined f o rO < r < 7.

2 -if

71

>'/ ,n

7 6 - 5 - 4 3 2 1

, 1 !

5040

( 7 ^ r r

( 7 ^ f ) i :

420

210

210

210

5040

210

( 7 - f ) ! = 24

7 r - 4

r 3

Check r = 3

LS R S

2(^P^ 4 2 0

( 7 - 3 )

7 - 6 - 5 - 4 !

4 !

2 ( 7 - 6 - 5 )

2 ( 2 1 0 )

4 2 0

The re is one so lu t ion , 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

1? I : RS

nfn nPn ^ 1

nl f l !

nl n!

0 ! [ f l + 1

nl n!

1 1! n! nl

1

n! LS = R S

18. a) e .g. , The fo rmu las for both „P„ and r,Pr have a

numera to r of nl. However , the fo rmu la for „ P „ has a

denom ina to r of 1 and the fo rmu la for „Pr has a

denomina to r of ( o ^ r ) L

b ) e .g . , A g roup of f r iends each o rder a d i f ferent

f lavour of ice c r e a m f rom a shop wi th 12 f lavours .

H o w m a n y possib i l i t ies are there if the g roup is 12

peop le? If the g roup is 7 peop le?

19. a) n = 52 and r = 5

P ^ 52 !

' ( 5 2 ^ 5 ) !

5 ^ 52 5

^.^:,^l: .50.,: '*.9,:48-47!

4 7 !

52P5 = 5 2 - 5 T 5 0 - 4 9 - 4 8

ggPg^ 3 1 1 8 7 5 2 0 0

The re a re 311 875 200 poss ib le a r rangemen ts ,

b) n = 26 and r = 5

26 !

( 2 6 ^ 5 ) 1

2 6 !

2« ' 2 1 !

As = 2 6 - 2 5 - 2 4 - 2 3 - 2 2 - 2 1 !

2 1 !

^ePg = 2 6 - 2 5 - 2 4 - 2 3 - 2 2

26P5 = 7 8 9 3 6 0 0

L ike l ihood = 7 893 600

•100% 311 875 200

L ike l ihood = 0 .025. . . -100%

L ike l ihood = 2 . 5 3 1 . . . %

There fo re , there is abou t a 2 . 5 3 % chance that an

a r rangemen t con ta ins b lack cards only.

1 -•

p - i ^ ' - : 11 l u y

P \ ' r U l l

U K e l , h o o d = J ^ ^ . . . 1 0 0 % 3 1 1 8 7 5 2 0 0

L ike l ihood = 0 , 0 0 0 , , . - 1 0 0 %

L ike l ihood = 0 . 0 4 9 . . . %

There fo re , there is abou t a 0 . 0 5 % chance that an

a r rangemen t conta ins 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 - 157

nl

(n7 I j i

nl ( n - r - l ) ! n-r

{n^rjnl

= ( r , ^ r ) „ P

M a t h i n A c t i o n , p a g e 2 5 7

a) e.g. , Janua ry 5, Apr i l 23 , Ju ly 24 , and Oc tobe r 15 w o u l d be 5, 113, 2 0 5 , and 2 8 8 .

b) 365 • 365 • 365 • 365 = 17 748 900 630

c) i) = 0.983.. . or about 9 8 . 4 % 365

i i) 1 ^ ^ ^ = 0.016,. , or about 1.6% 365

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 u t i o n s M a n u a l 4^13

d ) e .g . : i | ^ = 0 J 6 6 . . . or 9 6 . 7 % 30

i i ) 1 ^ ^ = 0.033.. . or 3 .3% ' 30

e) For e x a m p l e , they w e r e c lose 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 9

1 . T h e n u m b e r of subs to c h o o s e f r o m , S, is based on

the n u m b e r of buns (b). the n u m b e r of co ld cu ts (cc) ,

the n u m b e r of c h e e s e s (c) , the n u m b e r of topp ings (f) ,

and the n u m b e r of sauces (s) :

S = (# of b) • (# of cc) - (# of c) • (# of t) - (# of s)

S = 3 - 5 • 3 - 12 - 3

S = 1620 So , Mar io can c h o o s e f r o m 1620 d i f ferent subs .

2 «; (J V o u < ;ii< lis* on»- ot K W and C 1r)f the f irst

'Ji.ir.-K.tr-i o i l . ; n j VC ,j(>perooM lottr j rs lor the

s.'-cond .md third oh.mjcto 'c ; , nod one i d tho

yj] u() f« ' to. iso loUfMs Ol o b lank for t f io lost charac ter .

l u o m this I g.; l I t io l o l l owmg calc utat ion

ft o\ . lot ion f .omos - :< 2() 26 ?f

li i-l . tdt ion i .nmoh 04 / u b

I ho to fo ro , .station n a m e s are poissible.

3. t vent A Rol l ing a 2 O R h v e n l B- Rol l ing 10

1 1 i 2_ 3 4 5 6~

2 ') %j

4 5^ 6 7

[ 2 ' 3 6 7 ' 8

i 3 4 b " e 7^ 8 9

r 4 ^ ' " h i 7 8 " 9 "10

[ 5 6 f ^ 8 9 ' J O ^ ' 11 "

7 " 8 10^ U ~ 12 "

F r o m the tab le a b o v e , the re is o n e w a y to roll a s u m

o l I w i lh a pair f)f d i r e a n d th ree w a y s to roll a s u m of

10 wi th a pa i ! of d ice .

fi(A ' >B)- niA) < 0 ( 8 )

n(A . ' 8 ) ' 1 + 3

niA ' B) - 4

There a rc 4 w a y s that a s u m of 2 or a s u m of 10 can

be rol led w i th a pair of d ice .

4 , 1 0 - 9 - 8 = 720

The re are 720 w a y s to se lec t 3 ho rses to c o m e f irst,

s e c o n d , th i rd in a 10-horse race.

5. a) 8!

8!

b ) 6! • 3!

6! • 3!

6! • 3!

8 - 7 - 6 - 5 - 4 - 3 - 2 - 1

40 320

(6 • 5 • 4 • 3 • 2 • 1) • (3 • 2

( 7 2 0 ) • ( 6 )

4 3 2 0

1)

^ = 9 - 8 - 7 - 1

d )

9!

6!

9!

6!

9 !

6!

91

6!

9!_

6!

I l l

5

10

'E'

11 5-

l i

5

10

5-

10

5-

f: / h ) 4 3 2 1

fi f- 4 3 2 1

r. f. 4 3 2 1 = b il {

(i ') 4 3 2 1

504

10 i f ) 8 / fi f) 4 3 2 I I

U (8 / b 5 4 3 2 11

1 ^ 9 5 8 !

3 7 a

b / 6

8 !

4 3 z

4 3 2

^ 1 0

^ 5

2 - 9

18

9 1

6. The re are n ine p layers on the t e a m so the re are 9

d i f ferent pos i t ions. Let L represent the n u m b e r of

(>ormutations:

I - 9 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1

L -

L 9 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1

£ - / 2 - 42 • 20 • 6

L - 362 880

The re are 362 880 d i f fe rent l ineups tha t can be

f o rmed by n ine p layers on a sof tbal l t e a m .

7. a) ( f i + 5 ) ( n + 4 ) !

= (f, + 5 ) [ ( r i + 4 ) ( r ) + 3 ) ( r i + 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 + 12

4 -14 C h a p t e r 4 : C o u n t i n g M e t h o d s

1 (.

d )

I nl

{n + 2}i

nl

{n + 2]i

nl

nl

I ' j ' u ' / A ^ - j ' - > v 2 ) ( i )

2 ) ( r , + l )

= + n + 2n + 2

= i i " + 3 o + 2

8. 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 ) ! 72

n + 8 = 0 o r / i - 9 = 0

n = - 8 f l = 9

Check n = 8 I s RS

( Bj- 72

t 8 } ! r - , is unde f ined f 10)!

Check 17 = 9

LS R S

9! 72

( 9 ^ 2 ) !

9 !

7!

9 - 8 - 7 !

7!

9 8

72

I

111

;.< , -J - I -

L'c

1-4 1,1

1-4

{ Oj.

Check A? = 7

" '\\j> i ) h ' ' - -1) ( ' d ( 2 i i : i j

( f i - l ) ( n ^ 2 ) = 30

n ' - 2 f i ^ n + 2 = 30

n ^ - 3 / i + 2 = 30

f i ^ - 3 f i - 2 8 = 0

( f i + 4 ) { f i ^ 7 ) = 0

/. f _ 0

i R S

unde f ined

30

LS RS

( 7 - 1 ) ! 30

( 7 - 3 ) !

6!

4 !

6 - 5 - 4 I

4 !

6 5

30

The re is o n e so lu t ion , n = 7.

i . a , P . 9 !

9 ! 9 ^ 2 7!

9 - 8 - 7 !

7 !

, P , = 9 - 8

A = 72

4 !

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 0

The re is one so lu t ion , n = 9.

4^15

12!

( 1 2 - 1 0 ) !

p . 1 2 !

« 2!

IV 11 I f ) iJ a 7 6 0 4 .5 >'i

P !>' I 1 If) 9 p. / C, 0 1 i 12 10

^2P,o = 2 3 9 5 0 0 8 0 0

10. a) i: f. • 5 0 ANIJ / i 4 ('

I! f; _ I

rhs- »:i|jr*;'. lois I • fl«'fin(id lot n - 1 A f i e re n e I.

t.- ' / I 4 C' A N ! 1 / 1 2 0

/!_ -I /* 2

r i i f cxpt ;sMi-'!i I'. dufit ioO for n - w h e r e n e !.

r,; r» '1 - 0 A N [ ) tt 5 0

n > 4 n > 5

T h e exp ress ion is de f ined for n > 5, w h e r e n g I,

d : n + 2 > 0 A N D n s 0

f i > - 2

T h e exp ress ion is de f ined for n > 0, w h e r e ne L

b ) a : n > 0 A N D n ^ 2 > 0

n>2

T h e exp ress ion is de f ined for n > 2, w h e r e n g I.

b: r7 ---1 > 0 A N D n - 3 > 0

n > 1 n > 3

T h e exp ress ion is de f ined for n > 3. w h e r e n e I.

11. n = 20 and r = 6

20 !

( 2 0 - 6 ) !

p . 2 ^ 20 e 14!

2 0 ^ = 20 19 IB 17 16 15 14!

~ l ' 4 !

2oPg = 2 0 - 1 9 - 1 8 - 1 7 - 1 6 - 1 5

„,P ,̂ 2 7 9 0 7 2 0 0

Renn ie can load his C D p layer in 27 907 2 0 0 d i f ferent w a y s .

12. n = 14 and f = 2

14!

P - 13?

T h e n ; a i o 182 w a y s that M a n n y and 2 o ther p layers

(,;in line up to rece ive the c h a m p i o n s h i p t rophy .

13 . A. j t fH ' o g . T h e n u m b e r of w a y s to c h o o s e a

pr..'!>i!irnt . ir.d a v ice-pres iden t f r om a g r oup of f ive

5! s tuden ts is 20 . I cou ld a lso use the

F u n d a m e n t a l Coun t i ng Pnnc ip le b e c a u s e the re a re

f ive cho ices for p res ident and four cho ices rema in ing

for v ice-pres ident : 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 e

I d e n t i c a l , p a g e 2 6 6

l . a )

b)

.'! 7 6 5 4 .3 2 1

3 !2 ! (3 2 1) l 2 1)

71 7 3 2J 4Ji

3197 " ( 3 2 I I

7!

l ' 2 l

7!

3 !2 !

8! _ f w 6y 4J 2 '

2 2 V :

7 4

420

2 ! 2 ! 2 !

8!

2 ! 2 ! 2 !

8!

- ^ 8 M i fs <

5040

c)

2 !2 !2 !

10! 10 9 8 7 6 5 4 3 2 J

4T3T2"! 1 3 2 1 3 2 1 2 1

10!

4 ! 3 ! 2 !

10!

10 9 7 4

: 12600

d)

4 ! 3 ! 2 !

12! 1 2 - 1 1 - 1 0 - 9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1

2 ! 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 - 7

83160

4 16 C h a p t e r 4 : C o u n t i n g M e t h o d s

— * • ' ie a r r a n g e m e n t of 6 f lags : r r angemen ts :

7 4-7-

•i r iu ic i j i ^ UU Gi i lo icnt s ignals that can be m a d e f r o m the 6 f lags hung in a ver t ical l ine,

3. Let C represent the n u m b e r of w a y s :

6!

3 L 3 !

C = 20

The re are 20 d i f ferent w a y s th ree co ins land as h e a d s and th ree co ins land as tai ls,

4 . Let R represent the n u m b e r of w a y s :

18! R

R =

1 0 ! 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 - 8

R 2 450 448

The re are 2 4 5 0 4 4 8 w a y s that this record cou ld have occur red

5. Let C represen t the n u m b e r of w a y s :

2 ! 3 ! 4 !

'? : 3 •' 1 1 V '

f" 0 ( {, !

C 1260

The re are 1260 w a y s that No rm can d is tnbu te 1 cook ie to each g randch i ld ,

6. a) Let A represent the n u m b e r of a r r angemen ts : A = 5!

/ \ = 5 4 • 3 2 1

A = 120

The re are 120 di f ferent a r rangemen ts that can be

m a d e us ing all the let ters.

b) Let A represent the n u m b e r of a r rangemen ts :

2 !

^ 7 6 - 5 - 4 - 3 - 2 !

A ^ 7 6 - 5 4 - 3

A 2520

1 here a re 2 5 2 0 d i f ferent a r rangemen ts that can be m a d e us ing all the let ters.

8 - 7 - 6 - 5 - 4 - 3 - 2 !

/• ., ' i f

inu.f i f , ' , ,,i , . li men ts that can be

d ; i • 7i ' • • [ ) ! ' • <>nr m.• ..iifnU.-- • „ . r rangements :

2 -1 -3 2 1

A 3 9 9 1 6 8 0 0

The re are 39 916 800 d i f ferent a r r a n g e m e n t s that can be m a d e us ing all the let ters.

7. a) Let A represent the n u m b e r of a r r angemen ts :

5 !5 !5 !

, 11 n f 1 1 "f •. f. ^ ^ 'I . 2

-% .1 - • 4 '. V : c . ll o T

14 r . ! ! » / C

A 7 5 6 7 5 6

The re a re 756 756 d i f ferent w a y s he can a r range the books on the shelf . h i ^'.r.-iJi. l lu sets of 5 together .

A - ' 2 \

A = B

rays he can a r range the books ,

8. e g, , A sh ish kabob skewe r has 4 p ieces of beef,

2 p ieces of g reen pepper , and 1 p iece each of

m u s h r o o m and on ion . How m a n y d i f ferent

comb ina t i ons a re poss ib le?

R

R

9. a) Let R represent the n u m b e r of routes :

9f

5'4"!

9 HJ 6 - 5 4 3• 2• 1

5 4 3 2 '̂i 4 3 2 [

R - 7 6 4 3

R 126

The re are 126 routes t ravel l ing f r om point A to point B if you t ravel on ly sou th or east .

4-17

b) Let R rep resen t the n u m b e r of rou tes :

13!

7 !6 !

1 i 1 11 I f) u .". . (•) u 4 ; / 1 R ~ - - - - -

I f, 4 2 1 n 0 1 A ? 1 L i 1 ! 4

H I / ' l b

!h»,-ic. ;i ir- i r i r . l ou tes t ravel l ing f r o m point A to

|ic)ini H ll '/'HI i rave l on ly sou th or east .

10. Let R rep resen t the n u m b e r of routes :

„ 13 !

8 !5 !

I ' l IV 1 ' 1(8 u .3 / h 'J 4 :̂ 2 1

f, { \] '-. 4 / T ' 1 5 4 3 / 1

/< !:•• 11 t«

R = 1287

The re are 1287 routes t ravel l ing f r o m the house of

Jess to the house of her f r iend if she t rave ls on ly

nor th or wes t .

11 - a) o.g I dl .vw th« fo l lowing d i ag ram to show the

n u m b e r of w a y s to get to each in tersect ion

B , :^ . : : ;HJ.. !s: .

e4

^ 4--

1 ; t !i

T h e s u m of t he n u m b e r s on the top r ight and bo t tom

left co rners of e a c h b lock is equa l to t he n u m b e r of

routes to the top left co rner of each b lock. The re are

560 d i f ferent rou tes f r o m A to B. if you t rave l on ly

nor th or wes t .

b) I need to go nor th tw ice and wes t four t imes , for a

tota l of 6 m o v e s , to t rave l the f irst 2 by 4 b lock of the

route. I need to g o nor th once and w e s t o n c e , for a

tota l of 2 m o v e s , to t ravel the next 1 by 1 b lock of the

route. I need to g o nor th tw ice and wes t tw ice , for a

tota l of 4 m o v e s , to t rave l the last 2 by 2 b lock of the

route.

Let R represen t the n u m b e r 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 - 180

The re are 180 d i f ferent rou tes f r o m A to B. if you

t ravel on ly nor th or wes t .

12. Let P represen t the n u m b e r of pe rmuta t ions :

r 5 '3 i 8 ' V. r. 4 i ^ 1

^ r. 4 3 2 1 .3 2

p -i 2

P - 5 6

The re a re 56 d i f ferent pe rmuta t i ons of a n s w e r s that

the teache r can c rea te .

13. a) Let P rep resen t the n u m b e r of pe rmuta t ions : p = 7l

P = 7 - 6 - 5 - 4 - 3 - 2 - 1

P = 5040

The re a re 5040 d i f ferent a r r a n g e m e n t s poss ib le for

the new to tem po le .

b) Let P represen t the n u m b e r of pe rmuta t ions :

7!

2 !2 I

/ R .h • 4 3_ ^ 1

2 1 . ]

P - / 6 •

P 1260

f here are i2bO d i f te ien t a r r angemen ts poss ib le for

the new to tem po le .

14. e .g. , nPn wi l l be too h igh ; it g ives the n u m b e r of

a r r angemen ts of all n i tems, but s o m e of the

a r rangemen ts wil l be ident ica l because of the

a ident ica l i tems in the g roup .

15. a) e .g. , I a m a s s u m i n g that the co ins of the s a m e

denomina t i on a re cons ide red ident ical ob jec ts . Let A represent the n u m b e r of a r rangemen ts :

9!

P

P

4 ! 3 ! 2 !

9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1

4 - 3 - 2 - 3 - 2 - 2

A 1260

The re are 1260 w a y s the 9 co ins can be a r ranged in a

l ine.

i - i 3 C h a p t e r 4 : C o u n t i n g Methods

11 ' i g that the co ins of the s a m e

flsidered ident ica l ob jec ts . Let A if o f a r r a n g e m e n t s :

1 ' •.

A = 35

The re are 35 w a y s ns can be a r ranged in a l ine.

1S. T h e n u m b e r of w a y s to d iv ide the 8 rema in ing

f reez ies a m o n g s t the o ther 8 ch i ld ren is w h a t I want .

Let P represent the n u m b e r of pe rmuta t ions :

215!

^ 2 - 1 - 5 - 4 - 3 - 2 - 1

F = 7 - 6 - 4

P 168

The re are 168 w a y s to d is t r ibute the 10 f reez ies

1 / , t; ifW mber of pe rmuta t ions :

p ^ y f ' : •<

P 560

The re are 560 pe rmuta t ions poss ib le if you must star t w i th A and end w i th C.

b) e g. , If you start by put t ing the I's in the f irst and

second pos i t ions, and then in the second and th i rd

pos i t ions, and so on and so for th up unti l you put t h e m in

the ninth and tenth pos i t ions , there are 9 d i f ferent

a r rangemen ts of the I's just on thei r o w n . T h e n u m b e r of

d i f ferent a r rangemen ts of all the let ters in e a c h of t hese

9 a r rangemen ts is the n u m b e r of w a y s to o rgan ize the

other 8 let ters. S ince the o ther 8 let ters a re a lways the

s a m e , the n u m b e r of pe rmuta t ions of the let ters for each

a r rangemen t of the I's is the s a m e . Let P represen t the

number of pe rmuta t ions :

P 9 I 313!

P = 9 • 7 - 6 - 6 - 4 - 3 - 2 - 1

3 - 2 - 1 - 3 - 2 - 1

P 9 ( 8 7 5 - 4 )

P 9 (1120)

P 10080

The re are 10 080 permuta t ions poss ib le if the two I's mus t be together .

18. e .g . , B A N D I T S has 7 d i f ferent let ters, so the

n u m b e r of pe rmuta t ions is 7! B A N A N A S a lso has 7

let ters, but there are 3 A s and 2 Ns so you must

d iv ide 7! by 3! - 2 ! = 12,

19. T h e shor tes t poss ib le route con ta ins 3 m o v e s

d iagona l l y to tho r ight, 3 m o v e s d iagona l ly to the left,

a n d 3 m o v e s d o w n Let R represent the n u m b e r of

routes :

f: U U.'.f

U .'• ,' t . •. i:

3 ! .

P :. i ; / 1̂

1 •» ii ' d».''.u j . ri the top rear ver tex of the

• . t". Ml. '< >.-j •'. '••'•'ll of the cube ,

20. a) e .g. , Th is is the s a m e as a r rang ing the

20 p layers t hen d iv id ing by 2 ! ten t imes because the o rder of pa i rs d o e s not mat ter . Let T represen t the n u m b e r of pa i rs :

2 P

r = 2 ,375 . . , x10 ' ^

T h e r e a re abou t 2 .38 x 10^^ w a y s to ass ign 20

p layers to 10 doub le rooms .

b) e .g . , Th is is the s a m e as a r rang ing the 20 p layers

then d iv id ing by 4 ! f ive t imes because the o rder of

pai rs does not mat ter . Let T represent the n u m b e r of

pa i rs ;

- f

r = 3 . 0 5 5 . , . x l O "

The re are abou t 3.06 x 10^^ w a y s to ass ign 20

p layers to 5 guad rup le rooms

21. a) e .g. , I can m a k e a tab le to show all o f t h e a r r a n g e m e n t s that cou ld be m a d e . Pos i t ion 1 in the tab le be low is the le f tmost pos i t ion , and posi t ion 4 is the r igh tmost pos i t ion.

Pos i t ion

2

R

R

K

W

W

R

R

\N

W

w

Pos i t ion

3

' l ~.

W

w

R

Pos i t ion 4

W

R

R

w VV

R

R

W

W

R

R

"vv

w

w

w

R

W

R

w

w

ut ions Manual 4-19

From the tab le , I see that 14 d i f ferent a r r a n g e m e n t s

migh t be m a d e .

b) e .g . , F r o m the tab le a b o v e , 1 out of the

14 a r rangemen ts , f r o m left to right, w o u l d be red ,

wh i te , wh i te , red . The re fo re , there is a 1 in 14 c h a n c e

that the a r rangemen t , f r om left to right, w o u l d be red ,

wh i te , wh i te , red .

Apply ing Prob lem-So lY ing St ra teg ies , page 27©

A . 4 0 4 4 pa ths

b)

C a n n e d ' G o o d s I n u t s a n d

G o o d s V e g e t a b l e s

B n a n Rache l le 1 i:ili

B n a n 1 inh RachoHo

Rache l le Br ian ' l i n h

Rache l le Linh Hi

Linh R a c h f l i o B n a n

Linh Br ian j Rache l le

c ) S ince all 3 vo lun teers are be ing used to he lp

un load the veh ic les , there is only o n e w a y they can be

c h o s e n for th is j ob ,

d) Par t a ) a n d b) invo lve permuta t ions a n d part c)

invo lved comb ina t i ons . I k n o w because in par t a ) and

b) , the o rder in wh i ch the vo lun teers w e r e se lec ted for

the jobs ma t te red . In part c ) the o rder d id not s ince all

the vo lun tee rs w e r e be ing se lec ted to d o the s a m e

job .

2. e .g. , T h e ma in d i f fe rence is that for the permuta t ions ,

the o rder of the 4 ob jec ts mat te rs , a n d for the

comb ina t i ons , it does not. For the pe rmuta t i ons , you

cou ld have mul t ip le a r r a n g e m e n t s w i th the s a m e

ob jec ts s ince there is m o r e t han one w a y to o rder a

g roup of four d i f ferent ob jec ts . Th is is not poss ib le for

comb ina t i ons s ince you jus t need one a r r a n g e m e n t for

each g r oup of 4 , regard less of the order ,

3. Let C represen t the n u m b e r of d a n c e commi t t ees

poss ib le :

c - ; io The re are 210 w a y s that 4 of the m e m b e r s can be

c h o s e n to se rve on the d a n c e commi t t ee .

B. 2 (924) + 2 ( 2 5 0 8 ) + 2 ( 3 4 9 8 ) = 13 860 pa ths

C . Yes . The re a re 2 (3936 ) , or 7872 . pa ths tha t lead to

no m o n e y at a l l , but 17 904 pa ths that resul t in the

con tes tan t w inn ing s o m e t h i n g . T h e con tes tan t has a

bet ter chance of w inn ing some th ing than no th ing , so

it's a fair g a m e f r o m the con tes tan t ' s point of v iew.

L e s s o n 4 . 5 : E x p l o r i n g C o m b i n a t i o n s ,

p a g e 2 7 2

1. a) Let l/V represen t the number of w a y s :

W= 3!

W/= 3 2 1

W=6

There are 6 d i f ferent w a y s that B n a n , Rache l le , and

L inh can be c h o s e n for t hese j obs .

4. Let C represen t the n u m b e r of comb ina t i ons :

C = 12C3

C = 220

T h e r e are 2 2 0 w a y s 3 of the 12 dogs can be se lec ted

to appear .

L e s s o n 4 . 6 : C o m b i n a t i o n s , p a g e 2 8 0

1 -a)

F l a v o u r 1 F l a v o u r 2

vani l la s t rawber ry

vani l la choco la te

vani l la bu t te rsco tch

s t rawber ry vani l la

s t rawber ry choco la te

s t rawber ry bu t te rsco tch

choco la te vani l la

choco la te s t rawber ry

choco la te bu t te rsco tch

but tersco l vani l la

lu t terscotch s t rawber ry

' lu t terscotch hoco la te

4-20 C h a p t * ' » . -unt ing Methods

b)

I kfvom 1

•.Ml l.l! t

f l . i vcu f 2

• IV,. {.I.lf«.-

rKj t tcm. ot( h

l i ie huf t iUcI Ul l/vU-lldVOUi OUIllblhalluUS buUdUSe

each two- f lavour comb ina t i on can be wr i t ten in two

d i f ferent w a y s

2 « ;sent the n u m b e r of c o m m i t t e e s :

3. Let T rep resen t the n u m b e r of poss ib le t e a m s :

( , (11/

b ' I 1 1 . ' i:

i: •] i 2 1

*, 4 ;•- 2 I

J / 1 . c /

1 l i f .u - i!if H,^4 w a y . h peop le can be se lec ted f r o m a

y r u u p ot i z lo f o n n a dodge-ba l l t e a m .

4 . a) C = 5!

The re are 10 poss ib le commi t t ees .

) resent the n u m b e r of c o m m i t t e e s :

2 1 ( 5 ^ 2 ) !

2 «

L 0 2

C - 1 0 T h e r e are 10 poss ib le commi t t ees ,

c ) e .g. , My a n s w e r s for parts a ) and b) a re the s a m e .

Th is occur red because the s u m of 2 and 3 is 5.

' 3 ! ( 5 - 3 ) !

^ mil

C = l ± l i s " 312 1

c , 03 = 5 - 2

b) 9C3 9!

8 ! ( 9 - 8 ) !

9!

811!

9 8!

8!1

9

1

,.C, =

A =

A =

6!

4 ! ' h

6!

4 ! 2 !

6 - 5 - 4 !

4 ! 2 - 1

6^5

2 -1

e C , = 3 . 5

X , = 1 5

0 !10!

i o C o = 1

10!

01(10--0)1

10!

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 4

8!

1 ! (8 -1 ) !

8 !

1!7!

8 - 7 !

1-7!

1

a = B

F o u n d a t i o n s o f Ma themat i ^ i l y f i o n s M a n y a l 4 -21

5. Let C represen t the n u m b e r of comb ina t i ons :

i I -

10'

I C

u"4<

1IJ U / l . l

f . ' f 0 ? 1

10 U H /

1 ? i

I [, .'. -J ^

210

: li«>re are 2 1 0 w a y s 6 p layers can be c h o s e n to star t

,1 vol leybal l g a m e f r o m a t e a m of 10.

6. Let C represen t the n u m b e r of comb ina t i ons :

C = 55C5

C =

c

55!

55!

5 ! -50 !

5 1 5 i 52 C1 5 0 !

5 i 2 2 fo-O!

55 04 52 :.J ',}

5 4 2. 2 1

c: - 1 • 27 (.3 13 1 /

C 3 478 761

The re are 3 4 7 8 761 d i f ferent comb ina t i ons of h ip -hop

songs you can d o w n l o a d for f ree .

7. Let H represen t the n u m b e r of hands :

^ = 5 2 ^ 8

H

H

H

52!

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 5

8 - 7 - 6 - 5 - 4 - 3 - 2 - 1

H = 1 3 - 1 7 - 1 0 - 7 - 4 7 - 4 5 - 2 3

W = 752 5 3 8 1 5 0

The re are 752 538 150 d i f ferent 8~card hands that

can be deal t .

b) Let L rep resen t the n u m b e r of d i f ferent l ineups,

n = 14 and r = 8 because Conn ie mus t be the p i tcher

of the s tar t ing l ineup.

L

= . .C „

14!

8f(l4-8)! 141

HIU!

14 L i 12 11 10 9 8!

o n ; S 4 2 I

11 'H \7 .1 1 0 - 9

0 i> 4 3 2 1

i:-. I I .5

3003

T h e r e are 3 0 0 3 w a y s that the coach can c h o o s e his

s tar t ing l ineup of 9 p layers , if Conn ie mus t be the

pi tcher.

9. a) Yes , I do ag ree .

e.g. ,

L =

LS RS

6C2

6! 6!

2 1 ( 6 ^ 2 ) ! 4 ! l 6 4 l i

6! 01

2! 4 !

0 f; i '

2 1 / " 1 ' 2 1

6 5

2 1

3 5

15 15

LS - K S

b) e .g. , S o m e o ther cases w i th the s a m e re la t ionsh ip

as par t a ) are aCi = gC?, eCo = eCe, a n d 12C7 = 12C5.

I not ice that if you have two comb ina t i ons w i th the

s a m e n, and the s u m of the 2s for t hose comb ina t i ons

is equa l to n, then the va lue of the comb ina t i ons wi l l

be the s a m e .

c) e.g. , n

n - r

8. a) T h e p rob lem invo lves comb ina t i ons e.g. ,

because it does not s tate that the o rder of the star t ing

l ine mat ters .

4-22 C h a p i unt ing Methods

10.

Let T represen t the Let S represen t the

n u m b e r of comb ina t i ons n u m b e r of comb ina t i ons

' • . ' .M= hers . for the s tuden ts ;

• I

8!

3 !5 !

8 / '• '

•• i

8 / C

sent the n u m c

S - 4

S 56

commi t t ees ;

; - 10

3

I l.t,tu are 560 g radua t ion c o m m i t t e e s that the

pr inc ipal has to choose f r o m .

' ' a i Let C represen t the n u m b e r of commi t t ees ;

c

10!

5 ! ( l 0 - 5 ) !

10!

5 !5 !

1 0 - 9 - 8 - 7 - 6

/ ' ,

C 252

The re are 252 commi t t ees that can be f o rmed if there are no cond i t ions , b)

Let W represent the

n u m b e r of comb ina t i ons

for the w o m e n ;

Let M represen t the n u m b e

o f comb ina t i ons for the mer

M = 4 !

M

M

21(4-^2) !

4 !

2 ! - 2 !

4 - 3 - 2 !

M

2 - 1 - 2 !

i l l 2-1

M = 2 - 3

6 M =

Let C represen t the n u m b e r of commi t t ees :

C=W-M

C = 20 6

C= 120

The re are 120 commi t t ees that can be f o r m e d if there mus t be exact ly 3 w o m e n .

c ) Let C represent the n u m b e r of commi t t ees :

6!

1 ! ( 6 -1 ) !

6!_

1!-5!

6 5!

1-5!

6

1

C = 6

The re are 6 commi t t ees that can be f o r m e d if t he re mus t be exact ly 4 m e n .

d ) Let C represent the n u m b e r of commi t t ees ;

c = . a

c

c =

c

c = -

c = 6!

c =

c =

c

5 1 ( 6 - 5 ) !

6!

5 ! - 1 !

6 5!

5!-1

6

1

C 6

The re a re 6 commi t t ees that can be f o r m e d if t he re can be no m e n .

e) e .g. , C a s e 1 : 3 m e n and 2 w o m e n

4 ! 6!

3 ! - 1 ! ' 2 ! - 4 !

60 C,

C a s e 2 : 4 m e n and 1 w o m a n

' ' ' ' 4 ! - 0 ! I I 5!

, q - , ^ = 1-6

4 C , - e q = 6

N u m b e r of commi t t ees = 60 + 6 N u m b e r of commi t t ees = 66

66 5 -person commi t t ees can be f o rmed if there mus t be at least 3 m e n .

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

12 . e . g . Let 's say I w a n t to ass ign s tuden ts to the 0 ! 1!

r o o m wi th 5 beds f irsL Let A represent the n u m b e r of «l o H ^ ^ ^ j , " I i H o i ( l - ^0)«

w a y s to ass ign the 12 s tuden ts to the 5 beds : ' n ^'

5 ! M : ' .'dt 0 ^ 0 - ^ 1 ^ 0 - ^

12! C = 1 n = l /A 0 0

5 ! -7 !

I M I In ' I d ,C, =

' b A M

1!

1!

1L0 !

A-,> I I 2 2 4 ^C, = ;J C=1

N(»w H i c i f ,m-' 12 - s or 7 s tuden ts left to ass ign . Let 's ^

. i sMf jh s tuo ' j t ; ! ' . to 'hu r o o m wi th 4 beds n e x t Let B iCo, i C i = 1 , 1

-111 t l i r 1 IMI ll KM of w a y s to ass ign 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 ) !

c

J ' ; / - 4 H 1 ^ 2 J !

4 1 2 ' X = 1 ^ 2

41 3 2 i ,C, 2

2 I " 2 2 2 1 ( 2 - 2 ) 1

B - 35 2 2 2 ! -0 !

N o w there are 7 - 4 or 3 s tuden ts left to ass ign to the

room w i th 3 beds . S ince all of these s tuden ts wil l be A =

ass igned to the r o o m , there is on ly one comb ina t i on

for t h e m . Let C n o w represen t the n u m b e r of d i f ferent 2 M = '

ass ignmen ts : 2C0, 2C1, 2C2= 1 , 2, 1

C = 792 . 35 0 1 ( 3 ^ 0 ) ! ^"^^ ^ 1 ! ( 3 ^ l ) ! C = 27 720 01 31 T h e r e are 27 720 w a y s the 12 s tuden ts can be C = 3 = '

ass igned to t hese r o o m s . 0 ! -3 ! ' ' 1!-2!

13 . a | i ) 5 ob jec ts , 3 in each comb ina t i on s M ^ 3^1 , | ,2 i

i i ) 10 ob jec ts , 2 in e a c h comb ina t i on ^ ^ .| 3

i i i ) 5 ob jec ts , 3 in each comb ina t i on ^ ° a ^ M

b ) e.g. , i) H o w m a n y w a y s can you c h o o s e 3 co ins ^ ^ o

f r om a bag con ta in ing a penny , a n icke l , a d i m e , a 3 ^ ^ ^

guar ter , and a loon ie2

4-24 C h a p t e . 4 t . c u n t i n g M e t h o d s

3^2 3!

3C3 3!

3^2 2 ! ( 3 ^ 2 ) !

3C3

3^2 3!

2 ! - 1 !

3!

3 ! -0 !

^ 3 - 2 !

"""''mi 3^3 ^ 1

^ ' 1

^ 3 - ^

3^3 = 1

:p2 = 3

3C0, 3C1,3C2, 3C3= 1 . 3, 3, 1

4 ! „ 4 !

01(4^0)! ^^ •̂̂ ^11(4--1)1

C ^'

C

, q = 4

4! 4!

2 1

- ^ ^ ^ 2 . 1

, q = 2 - 3

4 C M 6

^ ^ ^ ^ 4 ! ( i r 4 ) ,

C " * 4 ! - 0 !

c ^ J L

" ' 3! 1

4Co, 4C1, 4C2, 4C3, 4C3 = 1 , 4 , 6 , 4 . 1

c ) e .g. , T h e n u m b e r s on the left and r ight s ides are all 1s; every o ther n u m b e r is the s u m of the two n u m b e r s above it.

d ) s ixth row: 1 , 5, 10, 10, 5,1 seven th row: 1 , 6, 15, 20 , 15, 6, 1

e) e.g. , T h e n u m b e r in each square of Pasca l ' s Tr iang le is equa l to the n u m b e r of pa thways to it f rom the top square .

« a l T h e equat ion I need to so lve is nl

2 ! ( o ^ 2 ) ! '

II • ' 4 j D n - 2 > 0

n > 2

15 IS de f ined for n > 2, w h o r e n e N.

15

2 ! ( n ^ 2 ) !

nl

V / - r;;

nl

; c - M ) l

nl

nl

= 15

= 1 5 - 2 !

= 1 5 ( 2 )

= 30 ( n ^ 2 ) !

n(n~^i){n^2,y J i j \ i 2 n i .

(n^2]{„A] n}(2){\f'

n ( i i - 1 ) = 30

n " - / i = 30

(o + 5 ) ( o - 6 ) = 0

11 + 5 = 0 O R n - 6 = 0

n = - 5 n = 6

Based on the rest r ic t ions, n = -5 canno t be a so lu t ion .

There fo re , n = 6 is the so lu t ion to the equa t i on .

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

b) T h e equa t ion I need to so lve is n !

4 ! i p 4)1 = 3 5 .

I) AN I ^ n 4 f

An,! .} 3:3 c, du f i i i ed for n > 4 . w h e r e n e N.

35

35 41

840

840

840

840

.,(.•,. mn ?){n i)(n-4i(n^5).„{3)(2)m

4\in -51 . ( i l p f l )

M n - l ) ( n - 2 ! i r / 3)

n* - 5n ' ' + 6n' - r f + bit - 6 n = 840

„ M + 1 1 l l " ^ 6 f i - 8 4 0 = 0

W n t e out all of the fac to rs of - 8 4 0 ; ± 1 , ± 2 , ±3 . ±4 . ±5 ,

±6 . ±7 , ±8 . ± 1 0 . ±12 , ± 1 4 , ±15 . ± 2 0 , ± 2 1 , ±24 . ±28 ,

± 3 0 , ± 3 5 , ± 4 0 , ±42 . ±56 , ± 6 0 , ± 7 0 , ± 8 4 , ± 1 0 5 , ± 1 2 0 ,

± 1 4 0 , ±168 . ±210 . ±280 . ± 4 2 0 . ±840 . T h e s e are all

the poss ib le roots of the equa t i on . Subst i tu te t h e m

into the equa t ion and if the equa t ion g o e s to 0, then I

have a root of the equa t ion . By tr ial a n d error, the

roots are -4 and 7. T h e o ther roots are not rea l . S ince

4 is out of the d o m a i n , the on ly real so lu t ion is n = 7.

c ) T h e egu. i t ion I need to so lve is

n! j 1 " ' >'|! _

2 \ { n - / f \ 3 ! i / < ' 2 3 ) ! '

n > 0 kU\) 2 0 A N l i / ; 2 (:

n _ ? // ^ -2

A N D r n 2 3 = i

0

1

( n + 2 ) !

r ^ -1

n

1

2 ! ( / 7 - 2 ) t

w h e r e n e N

3 ! (n i 2 3 ) ! is de f ined for n > 2,

_̂ ! "/J i

2 ! (n 2 ' i ' " 'AH,; :> ' iH

/2 i n ^ 2)1

2(f7 21 ' 6 ( » ^ 2 31 '

2 n ! «n . -21!

( n - 2 , ) ! " 6 ( u - r 2 - 3 ) !

2n ! I n 1 2)1

( n ~ 2 H " " 6 l n - l H

2n(M 1) (/; 1 2 H , 7 f l ) ( n )

6

12/11/' 1) 1-5 • 2 ) ( n + l ) ( n )

1 2 ( n - l ) - ( n + 2 ) i / ; +1) 0

1 2 n - 1 2 - ( n ^ + n + 2 u i 2) <'

1 2 n - 1 2 - n ^ - o - 2 ' i ^ 0

~ n ' + 9 n 11 0

An-2)iii 7 t 0

n 2 0 or n - 7 0

n 2 11 = 7

Both the roots are w i th in the d o m a i n , so there are two

so lu t ions , n = 2 and n = 7. 6!

d ) T h e equa t ion I need to so lve is i ( 6 - r ) !

15 .

r > 0 A N D 6

6! 15 is de f ined for 0 < r < 6, w h e r e r e I.

15

r ! ( 6 - r ) ! =

r ! ( 6 - ^ f ) !

r ! ( 6 ^ r ) !

6!

6!

15

720

15

r ! ( 6 - f ) ! = 4 8

By subst i tu t ing each of the in tegers r for 0 < r < 6, I

get r = 2 or r = 4.

16 . a) 1 , e .g . , the p layer can on ly w in if the six

n u m b e r s t hey c h o o s e are the s a m e and in the s a m e

order as the six n u m b e r s d r a w n .

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 61

6 - 5 - 4 - 3 - 2 - 1

ggCe = 1 1 - 1 3 - 1 6 - 2 1 - 3 1 - 6 1

9 0 8 5 8 7 6 8

The re are 90 8 5 8 7 6 8 d i f ferent w a y s the p layer can

w in .

4 -26 C h a p t e r 4 : C o u n t i n g M e t h o d s

c | e , g „ No , Even if e v e r y o n e in the c i t y p l a y s , it is

very u n l i k e l y that a n y o n e wil l w i n s i n c e each p l a y e r

•>-'v I !•! 'K, / 6 7 c h a n c e o f w inn ing .

' *' ' t} I .':«• ! i , in , in- : ;:t s ides in a p o l y g o n is e q u a l to

' • • • » ! : l -. f \\i t t ie n u m b e r o f 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 h i iq .> v e r t e x t h a t is not d i r e c t l y

s ic . i d . - t l i . t i ^ Uu u u m b e r o f 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 ry iug i - lm „ W >N\i\i the v a l u e s f r om t h e p o l y g o n s

on the s ide 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 /» -i

Rea r rang ing , d = „C2- n. T h u s , the n u m b e r of

d iagona ls for an n-s ided po lygon can be de te rm ined

us ing nC-i-n.

18. a) C a s e 1: 2 boys and 3 girts

C c - ^ ' ' " " ^ 2 !5 ! 3110!

7 ^ 2 - 1 3 ^ 3 = 2 1 - 2 8 6

, C , - „ C 3 = 6006

C 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 , = 2730

C a s e 3: 4 boys and 1 giri

c c = ^ i l L

' " ' 4 ! 3 ! 1!12!

, q - „ C , = 3 5 - 1 3

, C , - „ C , = 455

C a s e 4 : 5 boys and 0 gir is

C C " 5 ° 5 !2 ! 0113!

, C , . „ C „ = 2 1 . 1

, q . „ C „ = 2 1

N u m b e r of g roups = 6006 + 2 7 3 0 + 4 5 5 + 21

N u m b e r of g roups = 9212

The re are 9212 d i f ferent g roups of 5 s tuden ts wi th at

least 2 boys to choose f r om.

b) N u m b e r of g r o u p s wi th no cond i t ions ;

' ° 5 ! -15!

C a s e 1 : 1 boy and 4 g i r ls

1 1 6 ! ' 4 1 9 1

. X , - „ C , = 5005

C a s e 2 : 0 boys and 5 girts

7! J 3 ! ^

5 !8 ! . C . = • " 0 !7 !

, C „ - „ C , = 1287

N u m b e r of g roups w i th at least two boys ;

2 0 C 5 - 7 C 1 • 1 3 C 4 - 7Co - 1 3 C 5 = 9 2 1 2

The re a re 9 2 1 2 d i f ferent g roups of 5 s tuden ts w i th at

least 2 boys to choose f r o m .

c ) e .g . . I prefer indirect reason ing because fewer

ca lcu la t ions are n e e d e d .

19. a) e .g . . Comb ina t i ons and pe rmuta t i ons both

invo lve choos ing ob jec ts f r om a g roup . For

pe rmuta t ions , o rder mat te rs . For comb ina t i ons , o rder

d o e s not mat ter . For e x a m p l e , a be and bac a re

d i f ferent pe rmuta t ions , but the s a m e comb ina t i on .

b) D iv ide nPr by rf to get „Cr. For e x a m p l e . e C = i

and oP.i 360 ; 15

20. First, de te rm ine the tota l n u m b e r of o u t c o m e s

poss ib le 111 a s s u m e that o n c e a song is se lec ted , it

canno t be se lec ted aga in . T h e n u m b e r of o u t c o m e s , O.

is;

0 = ^ 5166!

O , 1 3 0 1 9 9 0 9

a) N u m b e r of t imes the even t cou ld occur ;

* ' 5 ! 2 1 !

3,,C, 6 5 7 8 0

Probabi l i ty (P) ;

p 6 5 7 8 0

1 3 0 1 9 9 0 9

P = 0.505. . .%

The re is abou t a 0 . 5 1 % chance that the f ive songs wi l

be f r om C D 2 and C D 4 .

b) N u m b e r of t imes the even t cou ld occur ;

12 14 15 12 18 5 4 4 3 2 0

Probabi l i ty (P) ;

„ 5 4 4 3 2 0

x 1 0 0 %

1 3 0 1 9 9 0 9 x 1 0 0 %

P = 4 .180 . . .%

The re is abou t a 4 . 1 8 % chance that one of the f ive

s o n g s wil l be f rom each C D .

F o u n d a t i o n s of Ma themat ics 12 S o l u t i o n s Manual 4 -27

c ) The re is on ly one t ime w h e r e your favour i te song

f r o m each of the 5 C D s wi l l be p layed .

Probabi l i ty (P) :

p = 1 x 1 0 0 % 1 3 0 1 9 9 0 9

P = 0 . 000008%

The re is abou t a 0 , 0 0 0 0 0 8 % or 1 in 13 019 909

c h a n c e that your favour i te s o n g f r o m each of the

5 C D s wi l l be p layed .

2 1 . A + rP2 + A

nl nl nl

3 ! | f i - ^ 2 ! { n - 2 ) ! 1.l(n-l)l

nl nl nl

6 ( 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)l

i ( n 3 ) '

n ( n - 1 | ( u 2 ) i 3 r t ( u i | , On

6 " ' ~ ^

n{f)2-^2u I M 2 t 3// 3 I 6)

n ( f r I 5 )

6

2 2 . e .g. ,

LS_^

pin 1 r ) !

RS

C^ + Ar

nl nl

r \ { n ^ r f { r ~ ^ l ) { n ^ ( r ^ i ) }

" n - ^ ( f - l ) ] f i l r{n\)

r\\t> (/ 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)l

LS - R S

There fo re . „ + i C r = „C, t .C,

L e s s o n 4 . 7 : Solwlng C o y n t i n g P r o b l e m s , page 2 8 8

1 . a) Th is s i tuat ion invo lves comb ina t i ons b e c a u s e the

order of the 3 topp ings on the p izza does not matter .

b ) Th is s i tuat ion invo lves pe rmuta t i ons b e c a u s e the

th ree spo ts for the cand ida tes w h o are se lec ted are all d i f ferent so order mat te rs .

c ) Th is s i tuat ion invo lves pe rmu ta t i ons b e c a u s e for a

g r o u p of 3 n u m b e r s , the re are d i f ferent w a y s to roll

t hose th ree n u m b e r s b e c a u s e of the d i f fe rent co lours

of the d ice ,

d ) Th is s i tuat ion invo lves comb ina t i ons b e c a u s e the

5 ch i ld ren w h o are se lec ted a re all in the s a m e

pos i t ion. No in fo rmat ion is s ta ted in the ques t ion

abou t pos i t ions the ch i ld ren m a y play, so I can on ly a s s u m e that they a re not p lay ing in spec i f ic pos i t ions,

2 , e .g. , S i tuat ion A invo lves comb ina t i ons a n d s i tuat ion B invo lves pe rmuta t ions . For s i tuat ion A ,

o rder does not mat ter s ince the 3 peop le w h o are

se lec ted wil l all be cons ide red equa ls . For s i tuat ion B,

th is is not the case . Each of the 3 peop le w h o are se lec ted wi l l have a d i f ferent pos i t ion w i th a d i f ferent

a m o u n t of power and d i f ferent ro les.

3. a) , C = 3!

3 ! -0 !

3 ^ 3 = 1

There is 1 w a y that M a d d y can bid on 3 i tems if she

bids on on ly her 3 favour i te i tems.

b ) A = ^ ^ « ' 3 ! -5 !

T h e r e are 56 w a y s that M a d d y can bid on 3 i tems if

she b ids on any 3 of the 8 i tems.

13!

( , 3 q f = 2 8 5 6 1

The re are 28 561 d i f ferent four -card hands w i th one

card f r om each suit .

200 ! 5 a) P

200 ^ 5

2 0 0 - 1 9 9 - 1 9 8 - 1 9 7 - 1 9 6 - 1 9 5 !

195!

2„Pg = 200-199-198-197-196

2ooPg = 304 278 004 800

The re a re 304 278 004 800 w a y s that the top f ive

cash pr izes can be a w a r d e d if each t icket is not

rep laced w h e n d r a w n .

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

b) ( 2 0 0 f = 320 0 0 0 000 000

T h e r e a re 320 000 000 000 w a y s that the top f ive cash pr izes can be a w a r d e d if each t icket is rep laced w h e n d r a w n .

6.

1'./- i)i i ' lO

= 180

The re are 180 w a y s that the 5 star t ing pos i t ions on the basketba l l t e a m can be f i l led.

10 ! 7.

2 ! - 2 ! - 2 ! - 2 ! - 2 !

10!

y 2 2 2 •

= 1 0 - 9 - 7 - 6 - 5 - 3 ' 2 - 1

113400

2 ! - 2 ! - 2 ! - 2 ! - 2 !

_ 10!

2 ! - 2 1 2 1 ^ 2 ! ^ !

T h e r e are 113 4 0 0 w a y s that the f ive d i f ferent pairs of ident ica l t eddy bears can be a r ranged ,

8. C a s e 1 : 3 f lags a re used

5!

^5 I h

5 ^

5 4 3 2 1

5 P 3 = 5 - 4 - 3

60

C a s e 2: 4 f lags a re used

^ ^ ^ ^ ^ ( 5 ^ 4 ) .

5 ^

A 120

5!

' 1!

5!

C a s e 3: 5 f lags are used

•P.', = 5!

5P5= 120

Let S represen t the n u m b e r of d i f ferent s igna ls that

can be sent us ing at least th ree of the f lags;

5 = 60 + 120 + 120

S = 300

The re are 300 d i f ferent s igna ls that can be sent us ing

at least th ree of the f lags.

9 e .g. , First m a k e a tab le to s h o w the n u m b e r of w a y s the two cab in c ru isers can be a r ranged next to each other.

CC 1 C C '• Af i , u \ y i M n c n t 1 ' '

i A r r a t i y e m e i i t 2

A 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 A

A r u m g c - m e u t 5 h

Arramnmnmt G i i i

! A r r . m j e m o n l / : p I J - y - '

Ar r ; i nc | . , i nen t 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 7i

For each of these a r r a n g e m e n t s , the n u m b e r of w a y s

the SIX boats can dock is the n u m b e r of w a y s that the

o ther four boats can dock . Let D rep resen t the tota l

m'Tb'jr of w a y s that the boats can dock ;

: 4 '

24 n ..p.

3 240 w a y s that the six boats can dock .

I i . e .g. . Each row of sea ts is d i f ferent , and wi th in a

row, the sea ts are a s s u m e d to be d i f ferent . There fo re , there are 10 d i f ferent peop le be ing sea ted in

10 d i f ferent spots . Let A represen t the n u m b e r of seat ing a r rangemen ts ; A = 10!

71 = 3 628 800

The re are 3 628 800 w a y s that the 10 p layers can sit in the van .

1 1 . ' 2 !

60

The re are 60 d i f ferent a r r a n g e m e n t s that are poss ib le for the letters if there are no cond i t ions . b) 3! = 6

There are 6 d i f ferent a r r a n g e m e n t s that are poss ib le for the let ters if each a r r a n g e m e n t mus t star t and end wi th an N.

12. e g . W h e n there is an even a m o u n t of n u m b e r s ,

hal f o f t h e m wil l be o d d . In th is case there a re

100 poss ib le n u m b e r s that each n u m b e r can be.

There fo re . ^ , or 50 of t h e m are o d d . S ince I w a n t

each n u m b e r to on ly be 1 of these 50 odd n u m b e r s ,

the n u m b e r of s e q u e n c e s S is;

S = 50 50 • 50

S= 125 000

The re a re 125 000 comp le te l y odd s e q u e n c e s .

13. 11!

5h&. = 462 Y o u can take 4 6 2 d i f ferent routes .

F o u n d a t i o n s of Mathemati dut ions Manual 4-29

14. e.g. , Let 's ass ign people to ttie 5-person car first.

Let J represent the nunnber of w a y s to ass ign the

people to this car:

5!11!

J = 4368

Now there are 16 - 5 or 11 people left to ass ign to the

remaining two vehicles. Let 's ass ign people to the

4-person car next. Let K represent the number of

w a y s to ass ign the people to this car:

17. e.g. ,

K 11!

4!7!

K = 330

Now there are 11 - 4 or 7 people left to ass ign to the

remaining vehicle. There is only 1 way to ass ign these

people to the 7 -passenger van b e c a u s e all of them

are going to be ass igned to it. Now let T represent the

total number of assignments:

r = J K-1 7 = 4 3 6 8 • 330

7 = 1 441 440

There are 1 441 440 w a y s the 16 people can be

ass igned to the 3 vehic les. Top of Board

2 6 10 6 15.

Start

Number of Paths = 2 + 6 + 10 + 6 Number of Paths = 24 There are 24 paths that the red checker can follow.

16. C a s e 1: 0 hearts and 5 non-hearts: 13C0 • 39C5

C a s e 2: 1 heart and 4 non-hearts: 13C1 39C4

C a s e 3: 2 hearts and 3 non-hearts: 13C2 39C3

C a s e 4: 3 hearts and 2 non-hearts: 13C3 • 39C2

Let H represent the number of hands with at most 3

hearts:

H = 13C0 • 39C5 + 13C1 • 39C4 + 13C2 • 39C3 + 13C3 • 39C2

H = 1 575 757 + 13 82 251 + 78 9139 + 286 741

H = 2 569 788

There are 2 569 788 different five-card hands that

contain at most three hearts that can be dealt.

ords-r in.il'ii l r

yes

'U-.,i' peiuiiit-ition'v, .,r, Civ !)!nl>iii.ni')i

KlfiUlCfll:'

yes

toi inHiti'.Hi

iJ iv ic i , by 1!. -.vf.r-i'. t h e n u n i l H ' i 1

ir.lenUCtll itt.TTl

AND OR

iJbu l-UMdiinu;ntdl CtHuUinq Princ iple. multiply the number

of ways ejir.h tHsk (.rin ocnir

of way; ej< h task f.an occuf

,C„ =

18. Number of Total Outcomes:

13!

6!-7!

i3Ce = 1716

Number of Outcomes Where 3 B o y s and 3 G ids C a n G o :

e 3 T ^ 31.31 31-4!

^ ^ 3 ^ = 700

Probability (P):

700 P = x 1 0 0 %

1716

P = 40.792. . .%

There is about a 4 0 . 8 % c h a n c e that there will be three

boys and three giris on the trip.

19. e.g. . If I have an A a s the first letter, there are 4

possibilities 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 K

E a c h one of these h a s 3 possibilities for the fourth

letter. 4 ( 3 ) = 12

If the second letter is L, S , or K: 3 possibilities for the third letter: A and 2 of L, S , and

K (depending on which letter is second) T h e A ' s have 3 possibilities for the fourth letter, and

the other two letters have 2 possibilities for the fourth

letter. 3 + 2(2) = 7

Total for all three second letters that are L, S , or K:

7 ( 3 ) = 21

4-30 C h a p t e r 4: C o u n t i n g Methods

Tota l if A i s t t ie f irst letter;

2 33

There fo re , if the f irst let ter is A, there are 33 poss ib le

a r r a n g e m e n t s .

If the f irst let ter is L, S. or K, there are th ree possib i l i t ies

for the s e c o n d letter; A, and 2 of L, S, and K ( the ones

that are not the f irst let ter) . If A is the second letter;

3 possib i l i t ies for the th i rd letter: A, and 2 of L. S. and K T h e A has 3 possib i l i t ies for the four th let ter and the

o ther two let ters have 2 . 3 + 2(2) = 7

If A is not the s e c o n d letter;

/ .sibi i i t ies for t he third letter

T h e A has 2 possib i l i t ies for the four th letter and the

o ther letter has 1. 2 + 1 = 3

Tota l for both s e c o n d let ters that are not A ;

3 (2) = 6

Tota l for one of th ree t imes w h e r e first letter is L, S. or K:

7 + 6 = 1 3

Tota l w h e n f irst let ter is L, S, or K; 3 (13) = 39

Tota l a r r a n g e m e n t s : 39 + 33 = 72

There fo re . 72 four- le t ter a r r a n g e m e n t s can be m a d e us ing all o f the let ters in the w o r d A L A S K A .

20. If I have an O as the f irst letter, there are

4 possib i l i t ies for the second letter, each of wh i ch has 3 possib i l i t ies for the third letter. 4 (3 ) = 12

The re fo re , there ar^; 1 / uo^..sible a r r a n g e m e n t s w h e n

O is the f irst letter.

If the f irst letter is B, K, or S;

The re are 3 possib i l i t ies for the second letter; O, and t w o of B, K. and S (the ones that are not the f irst

let ter) . O has 3 possib i l i t ies for the third let ter wh i le

the o ther 2 have 2. 3 + 2 (2 ) = 7 Tota l if the f irst let ter is B K, or S:

3 ( 7 ) = 21

Tota l a r r a n g e m e n t s :

21 + 12 = 33

There fo re , 33 th reedet te r a r r angemen ts can be m a d e

us ing all of the let ters in the w o r d 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 0

A . Y e s . Each n u m b e r f rom 0 to 127 is ass igned a

d i f ferent charac te r or s ymbo l on the keyboard . S ince

the n u m b e r s a l ready have an es tab l i shed order , the

charac te rs and s y m b o l s ass igned to these n u m b e r s do , as wel l .

B. Y e s . Each n u m b e r in ASCI I (p ronounced "askey")

mus t be conver ted into a s tnng of Os and I s to c reate the b inary code , so o rder mat te rs . Each 0 or 1 is

assoc ia ted w i th a pos i t ion in the s tnng . A d i f ferent

pe rmuta t ion of Os and I s represents a d i f ferent

n u m b e r in the A S C I I code sys tem.

C. The re a re 128 n u m b e r s in ASCI I that mus t be

rep resen ted by a st r ing of Os and I s . Y o u need to

de te rm ine the length of the s tnng needed to c reate 128 d i f ferent a r rangemen ts of Os and I s . Y o u can

beg in by th ink ing abou t a s tnng of length of 5.

A b o x d i a g r a r ' j . , f ;an help you de te rm ine the . , i . . : it)fr < - ' ' A S ( , i | n u m b e r s you can represent .

Wi th in each box you can p lace a 0 or a 1. The re a re

two cho ices for each box. s ince repet i t ion of Os and

I s is a l l owed . So for a s tnng length of 5, there a re

2 - 2 - 2 • 2 - 2 = 2^ or 32 A S C I I n u m b e r s that can be

rep resen ted . Obv ious ly , the s tnng mus t be longer for 128 n u m b e r s . If n represen ts the st r ing leng th , and

128 n u m b e r s mus t be rep resen ted , then 2" = 128. By

tr ial a n d error , n = 7.

A b inary s tnng of length 7 is needed to represen t

each ASCI I code .

C h a p t e r S e l f - T e s t , p a g e 291

1 n l Let N represen t the n u m b e r of d i f fe rent ser ia l

numbers :

IV = 26 - 26 • 10 - 10 - 10 - 3

IV = 2 028 000

The re are 2 028 000 d i f ferent ser ia l n u m b e r s

poss ib le , if repet i t ion of charac te rs is a l l owed ,

b) Let N represent the n u m b e r of d i f fe rent sena l

numbers :

IV = 2 5 - 2 4 - 1 0 - 9 - 8 - 3

IV = 1 296 000

T h e r e are 1 2 9 6 000 d i f ferent sena l n u m b e r s

poss ib le , if no repet i t ion is a l l owed .

2. Event A: D raw ing a s p a d e

Event B: D raw ing a d i a m o n d

n(A ' ' m = iiiA) + n(B)

niA H) - 13 + 13

ni'A •• li) ^- ?u

There fo re , there are 26 w a y s to d raw 1 card that is a s p a d e or a d i a m o n d .

3. a) n + 9 > 0

/ 7 > - 9

(n + 10)(n + 9) ! is de f ined for n > - 9 . w h e r e 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 > 0

n > 2

(n^2)\ ^—A is de f ined for n > 2, w h e r e 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 /•?'' - n

F o u n d a t i o n s of Ma themat ics 12 S o l u t i o n s Manual 4-31

4 a) hi - 1/0

r hnn : f o ro , there a re 120 d i f ferent w a y s that the 5 cars

t.ai i Ix ; p a i k o d s ide by s ide.

b ) I el B rofJHjsent the n u m b e r of a r r angemen ts ;

B . / f . / ' ,

P ~ 2» 4 '

f i - 2 1 4 3 2 1

[i - 48

1 h(;r(4<)fu, there are 4 8 d i f ferent w a y s the cars can be

pa rked so the 2 b lack ca rs a re next to e a c h other .

The re are 126 d i f ferent four -book se lec t ions that can

be m a d e

( 9 T 4 ) !

0 '

fd

9 8 f o b\

• - V

j \ -M a / h

The re are 3024 d i f ferent four -book se lec t ions can be

a r ranged in o rder of p re fe rence .

c ) e .g. , T h e o rder mat te rs in part b) . The re are still

126 w a y s to choose the four books f rom the n ine

op t ions , but there are a lso 4 ! = 24 w a y s to a r range the

books , (126 • 24 = 3 0 2 4 )

m P ,^ '^pr

P

6. „ P , = 8 4 ( „ q )

( " - 4 ) 1 ( n - 2 j

f ) ( n - l ) ( f ] - - 2 ) ( n - 3 ) = 4 2 n ( n ^ 1 )

( n - ^ 2 ) ( r , - 3 ) = 42

n-' - 3A7 - 2r? t 6 42

fP 5n 36 -:- 0

{ n - r 4 ) ( n - ^ 9 ) = 0

17 + 4 = 0 or n - 9 0

n = - 4 n = 9

Check n = --4

l A

" A

i 4)1

( 4 4)1

is unde f ined

841

84 2 ! ( - 4 2 ) !

C h e c k n = 9 LS

9!

5!

u y, ( h 5!

5!

9 8 7 6

3024

R S

8 4 ( , q

84

84

84

84

9!

21(9 2)1

9!

^2L7!^

g 8 71

2 1 P

g a '

2 1

84i^< 4

3024

The re is one so lu t ion , n = 9.

6 ! 8!

, C , - , C 3 = 8 4 0

The re a re 8 4 0 d i f ferent w a y s that a 5 -person

commi t t ee can be se lec ted if there mus t be 2 boys

and 3 gir is .

b ) C a s e 1 : 2 boys a n d 3 gir is ;

C (• - ^ ' 6 2 n '3 2 !4 I 3 !5 !

ApAS-^Q

C a s e 2 : 3 boys and 2 gir is :

« « 2 3,3, 216!

, C 3 . , C , = 560

C a s e 3 : 4 boys and 1 gir i ;

IL 6 4 - 8 i " ^ 4 ! 2 ! ' i ! 7 !

120

C a s e 4 : 5 boys and 0 girts:

C . c ^ ^ . ^ 6 5 8 0 5 , ^ , Q,8|

As-sCo = Q

Let C represen t the n u m b e r of 5 -person commi t t ees

w i th at least 2 boys :

C = 840 + 560 + 120 + 6

C= 1526

The re a re 1526 d i f ferent w a y s that a 5 -person commi t t ee

can be se lec ted if there mus t be at least 2 boys .

^UA-'''

A, 3 !9 !

220

4-32 Chapter 4 . . r^un t i ng M e t h o d s

The re a re 220 d i f ferent w a y s that a 5 -person • .i.'i< ' .11. t . fed if Dav id and S u s a n mus t

d ) C a ^ e 1- i l . V '\ . girts

ni H! { : f :

" P

X „ ' X , = 8 4 0

f ; . :sc 2

f r.

C. = 420

C a s e 3 : 0 boys and 5 gir is

" ° « '"̂ 0 ! 6 ! 5 !3 !

56

Let C represen t the n u m b e r of 5 -person commi t t ees w i th m o r e gir ls t han boys:

C = 840 + 4 2 0 + 56

C = 1316

The re are 1316 d i f ferent w a y s that a 5 -person commi t t ee can be se lec ted if there mus t be more gir is

t han boys,

8. - ^ = 30 2 ! 2 !

T h e r e a re 30 d i f ferent a r r angemen ts of the letters in the w o r d 1 r f 111

9. 5! • 4 ! = 2 8 8 0

The re a re 2880 d i f ferent a r r a n g e m e n t s poss ib le .

C h a p t e r R e v i e w , p a g e 2 9 3

1 . e .g. , T h e Fundamen ta l Coun t i ng Pnnc ip le is used

w h e n a coun t ing p rob lem has d i f ferent tasks re la ted by the wo rd A N D , For e x a m p l e , you can use it to

f igure out how m a n y w a y s you can roll a 3 w i th a d ie

and d raw a red card f rom a deck of cards .

Quarter Toonie

heads

tails

heads

tails

heads

ta.L.

Loonie

Iv.-ads

tails

heads

tails

heads

tails

heads

rails

3 I r-* A : rr ': the n u m b e r of se ts of answe rs :

, 4 - 4 • 4 • 4 • 4 A = 4^° /•. i ' . ; ! . , c / c

" i - • . In. ' . , : I ' . • ) . . l i ve 1 048 576 di f ferent sets of answers .

4 a | /. .• u A N D n > 0

, '- y'f ! • ' ' u f i ned for n > 0, w h e r e n c I.

in^ 2)1 ^ = 20

( f i ) ( i i -^4 ) . . , ( . i i l2) ( l ,

l ) = 20

n^+n I I / 20 = 0

ll . 3// 18 = 0

l) = 0

n + 6 = 0 or n ^ 3 = 0

n = ~6 n 3

T h e root n =• 6 is ou ts ide the res tnc t ions on the

var iab le in the equa t i on , so it canno t be a so lu t ion . The re is one so lu t ion , n = 3. b) T h e s impl i f ied ve rs ion of the egua t ion is

n + 1>0 A N D n - 1 > 0

n > - 1 n > 1

In , 1)1 132 IS de f ined for n > 1, w h e r e r? e I.

// 1 !

132

132

(11-^1)1

{n l ) f u 2 , . (3) (2) (1)

(o + l ) ( n ) = 132

n ' + n = 132

rf * n - 132 0

(r7 + 1 2 ) ( n - 1 l ) = 0

0 + 12 = 0 or n - 11 0

,0 - 1 2 n = 11

T h e root n = - 1 2 is ou ts ide the res tnc t ions on the

vanab le in the egua t i on , so it canno t be a so lu t ion .

The re is one so lu t ion , n = 1 1 , T h e t ree d i a g r a m s h o w s there are 8 poss ib le w a y s

that the th ree co ins can land .

F o u n d a t i o n s of Mathematic • - ' . o iu t ions Manual 4-33

5. e .g . , 6 ^ 6 has a larger va lue , e .g . , I k n o w b e c a u s e

^ is the factonat exp ress ion for the permuta t ion 6!

exp ress ion 8 ^ 2 . Here , I have m o r e ob jec ts than for

ePe, but I a m not us ing all o f t h e m . Th is leads to f ewe r

poss ib le a r rangemen ts , or in o ther w o r d s , a lower

8! va lue for

6!

6. Let O represen t the n u m b e r of o rde rs :

0 = 12!

O = 4 7 9 001 600

The re are 4 7 9 001 600 d i f fe rent o rde rs in wh i ch the

s ingers cou ld pe r fo rm the 12 songs .

7. , , P , = 25 ! ^5 3 22 !

25P3 = 1 3 8 0 0

T h e r e a re 13 800 d i f ferent w a y s a d i rec tor of

educa t i on , a super in tenden t of cu r r i cu lum, and a

supenn tenden t of f i nance can be se lec ted .

b) Let A mnrf^cppf ^hn n u m b e r of a r r angemen ts :

10! i n 'I 8 / h > 1 4 2 1

2 ! 2 ! 2 !

10!

/ 1

Vi <' ! i\ 4 ^

= 4 5 3 6 0 0

2 ! 2 ' 2 !

10!

2 ! 2 ! 2 !

The re are 4 5 3 6 0 0 d i f fe rent a r r a n g e m e n t s that a re

poss ib le if al l the let ters a re u s e d , but each

a r r a n g e m e n t mus t beg in w i th the C.

1 1 . a) 14 '

2 5 2 2 5 2 0 2 ! 3 ! 4 ! 5 !

I IKJHJ a re 2 522 520 d i f fe rent w a y s T ina can s tack the

b locks in a s ing le t o w e r if t he re a re no cond i t ions .

b) 2 7 7 2 0 3 ! 4 ! 5 !

The re a re 27 720 d i f ferent w a y s T ina can s tack the

b locks in a s ing le t owe r if there mus t be a ye l low

b lock at the bo t tom of the t owe r and a ye l low b lock at

the top .

25P^o=11 861 676 288 000

2 5 P „ - 1 . 1 8 r K . . x 1 0 "

The re are 11 861 676 2 8 8 0 0 0 or a b o u t 1.2 x 1 0 "

d i f ferent w a y s the test can be c rea ted if there are no

cond i t ions .

23Pg = 19 769 4 6 0 4 8 0

23Pg=1 .976 . . . x10 ' °

T h e r e are 19 769 4 6 0 4 8 0 or abou t 2 .0 x 10^°

d i f ferent w a y s the test can be c rea ted if the eas ies t

gues t ion of the 25 is a lways first and the most di f f icul t

gues t ion is a lways last.

9 p . 5 2 !

.^P^ 3 1 1 8 7 5 2 0 0

The re are 311 875 200 d i f ferent f i ve-card

a r rangemen ts poss ib le .

10. a) Let A represent the n u m b e r of a r r angemen ts :

11! 1 1 - 1 0 - 9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1

2 ! 2 ! 2 !

11 !

2 I 2 I 2 !

11!

2 - 1 - 2 - 1 - 2 - 1

1 1 - 1 0 - 9 - 7 - 6 - 5 - 4 - 3 - 2 - 1

4 9 8 9 6 0 0 21212!

The re are 4 9 8 9 600 d i f ferent a r r a n g e m e n t s that are

poss ib le if all the letters are used .

12. ™ 5 5 ! 5 !

ioCs = 252

C

• ^ ^ 7 ! 4 !

= 3 3 0

^5 2 2113!

A 105

The re fo re , nd resu l ts in the g rea tes t va lue .

13. C, =• 20 !

4116!

20 = 4 8 4 5

The re a re 4 8 4 5 d i f ferent se lec t ions of 4 books that

Ruth can c h o o s e .

14. No . e g. , Each comb ina t i on can be a r ranged in

m a n y d i f ferent w a y s to m a k e a pe rmuta t i on , so there

are more pe rmuta t i ons t han comb ina t i ons

15. a) „ C = 19!

4 !15 !

, ,C^ 3876

The re are 3 8 7 6 d i f ferent w a y s that a commi t t ee of

4 peop le can be c h o s e n if there are no cond i t ions .

9! 10 !

2 ! 7 ! ' 2 ! 8 !

. C =36-45

b) A ,C =

c . 1620

The re a re 1620 d i f ferent w a y s that a commi t t ee of

4 peop le can be chosen if there mus t be an equa l

n u m b e r of m e n and w o m e n on the commi t t ee .

' '° ' 4 ! 6 !

. „ C , 210

The re are 210 d i f ferent w a y s tha t a commi t t ee of

4 peop le can be chosen if no m e n can be on the

commi t tee .

4-34 Chapte r 4- C o u n t i n g Methods

1S. e .g. , Let A represent the n u m b e r of w a y s to

ass ign teachers to the f irst g roup of 5:

5 !10 !

A = 3003

N o w there are 15 - 5 or 10 teachers left to ass ign .

Lc i f ; i r -p tesent the n u m b e r of w a y s to ass ign the

rema in ing teache rs to the second g roup of 5;

B

• . rs left to ass ign to

1 w a y that th is can

I n u m b e r of w a y s to

10!

5 !5 !

8 252

N o w there ' i-

the last grot f i '- ' h. i-.-

be d o n e . I • ' - i - -i !

ass ign the t

T=A-B

T= 3003 252

T = 756 756

The re are 756 756 d i f ferent w a y s 15 teachers can be

d iv ided into 3 g roups of 5.

17 . c q. . T h e f irst point can be jo ined wi th 11 more

[soinO- to form st ra ight l ines. T h e second point can

then be jo ined wi th 10 more points to f o rm stra ight

l ines (s ince it w a s a l ready jo ined w i th the f irst po int ) .

T h e th i rd point can be jo ined w i th 9 more points to

f o rm st ra ight l ines (s ince it w a s a l ready jo ined w i th the

f irst two po in ts) . Th is pat tern con t inues on unti l I get to

the seconddas t point that can on ly be j o i ned w i th the

last point (s ince it w a s a l ready j o i ned wi th the o ther

10 po in ts) . T h e last point canno t be jo ined any fu r ther

s ince It IS a l ready jo ined to every o ther point in the

c i rc le. Us ing this obse rved pat te rn . I can ca lcu la te the

n u m b e r of s t ra ight l ines (L) :

/. = 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1

1 = 66

The re are 66 d i f ferent w a y s the points can be j o i ned

to f o rm st ra ight l ines.

18. a) S ince there is one more boy than there are

gir is, the l ine mus t a lways fo l low this pa t te rn :

B G B G B G B G B G B G B . T h u s the boys are a r ranged in

7 pos i t ions, and the gir is in 6 pos i t ions.

7! 6! = 3 628 800

The re are 3 628 8 0 0 w a y s in wh i ch the ch i ld ren can

be a r ranged in one row if the boys and gir is mus t

a l te rnate pos i t ions .

b) G r o u p the tnp le ts as one . The re are 3! w a y s in

wh i ch the tnp le ts can a r range t hemse l ves . Let B

represen t the n u m b e r of d i f ferent a r rangemen ts :

e = 1 1 ! 3!

6 = 39 916 800 6

B = 2 3 9 500 800

The re are 2 3 9 500 800 w a y s in wh i ch the ch i ld ren can

be a r ranged in one row if the t r ip lets mus t s tand next

to each other.

19. C a s f 1 face ca rds and 3 non- face cards : 12C2 • 40

C a s e 2 : 3 face ca rds and 2 non face ca rds : 12C3 • 40C2

C a s e 3: 4 face ca rds and 1 non- face ca rd : 12C4 - 40C1

C a s e 4: 5 face ca rds and 0 non- face cards : 12C5 • 40Co

Let H represen t the n u m b e r of hands w i th at least 2

t .̂ ca rds :

I'lC'z • 40C3 + 12C3 • 40C2 + 12C4 • 4oCi + 12C5 - 40Co

/ ' 36 • 9 8 8 0 + 2 2 0 • 780 + 4 9 5 • 40 + 792 1

544 272

! L i - .e are 844 2 7 2 d i f ferent f i ve-card hands wi th at

least two face ca rds .

C h a p t e r T a s k , p a g e 2 9 5

A . Comb ina t i ons . T h e order in wh i ch the d ice a re

tossed d o e s not mat ter (note that p layers toss all

8 d ice s imu l taneous ly ) nor does the w a y the d ice a re

a r ranged w h e n they land mat ter . W h a t is impor tan t is

the o u t c o m e of each toss a comb ina t i on of n u m b e r

of d ice that land w i th the s a m e s ide up A N D n u m b e r

of d ice that land w i th a d i f ferent s ide up, for e x a m p l e ,

7 d ice land w i th the s a m e face up A N D 1 d ie w i th the

oppos i te face up .

B . Each o u t c o m e can h a p p e n two w a y s . For examp le .

8 wi th the s a m e s ide up cou ld occu r as 8 of the

u n m a r k e d s ides face up or 8 of the marked s ides face

up. Tha t is w h y each ca lcu la t ion is the s u m of two

comb ina t i on va lues :

8 d ice land w i th the s a m e s ide up

^ f + or 1 + 1 or 2

7 d ice land w i th the s a m e s ide up

i f f

, 1

6 d ice land wi th the s a m e s ide up

r i . ^

1 i ^ l 7 or 8 + 8 or 16

or 28 + 28 or 56 8V21^ 8 ¥ 2

^6)l2j''"i6ji2^ 5 d ice land w i th the s a m e s ide up

or 56 + 56 or 112

4 d ice land w i th the s a m e s ide up

8 ¥ 4 '

3 d ice land wi th the s a m e s ide up

+ I 4 J I

or 70 + 70 or 140

or 56 + 56 or 112

F o u n d a t i o n s of Ma themat ics 12 S o l u t i o n s Manual 4-35

C . Y e s , i th ink it is fair.

O u t c o m e P o i n t s N u m b e r of W a y s O u t c o m e C a n

O c c u r

8 d ice land s a m e

s ide up

10 2

7 d ice land s a m e

s ide up

4 16

6 d ice land s a m e

s ide up

2 56

5, 4 , or 3 d ice

land s a m e s ide up

0 364

T h e h ighest n u m b e r of b e a n s (10) is a w a r d e d for the

o u t c o m e that can h a p p e n in the least n u m b e r of

w a y s , 8 d ice land ing s a m e s ide up ; 4 b e a n s a re

a w a r d e d for the o u t c o m e that can h a p p e n in the

second fewes t n u m b e r of w a y s , 7 d ice land ing s a m e

s ide up; 2 beans a re a w a r d e d for the o u t c o m e that

can h a p p e n in the th i rd f ewes t n u m b e r of w a y s , 6 d ice

land ing s a m e s ide up. T h e mos t l ikely o u t c o m e s of 5,

4 , a n d 3 d ice land ing s a m e s ide up ail rece ive the

lowest n u m b e r of beans (0 ) . So the point s ys tem

rewards the least l ikely o u t c o m e s wi th t he mos t beans

and the mos t l ikely o u t c o m e s wi th the f ewes t 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 9

1 . a )

Coin

heads

t a i l s

30 i*o>u^b«« OutoEMnet

heads and red

heads and o range

heads and pu rp le

heads and yeUa<H heads and green

ta i ls irdi red ta i ls and orange

ta i ls and pu rp le

ta i ls andyellOK^f

ta i ls and green gf een

b) b) e .g. . By look ing at the t ree d i a g r a m , there is one

w a y he cou ld f l ip a head and sp in g r e e n , and ten

poss ib le o u t c o m e s .

P (heads and g reen ) = ^ 10

or 1 0 % .

2. a)

C h i l d 1 C h i l d 2 C h i l d 3

B B B

B G B

B B G

G B B

G G B

G B G

B G G

G G G

b) The re a re \wo w a y s all t h ree ch i ld ren wi l l be the

s a m e gender , e i ther all boys or all g ids , and there are

e ight poss ib le o u t c o m e s .

2 P(al l boys or all g ids) = - or 2 5 % c h a n c e , a s s u m i n g

8

that hav ing a boy or g id is egua l ly l ikely.

3. a) B ' = { the set of e l emen ts not in S}

B ' = {a , b, c, d, e, i, o, u} b) Au B = { the set of e l emen ts in A and B}

Au B = {a, b, c, d, e, i, o, u, x, y, z} c) Ar\B = { the set of e l emen ts 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

5 . e .g. , Let x be the n u m b e r of g o o d s ingers w i thout

danc ing ski l ls. Let A be the set of s ingers and B the

set of dance rs .

n{A) = X + 6

n{B) = 10 + 6 or 16

niAnB) =6

niAuB) = 2 4

niA KJB) =niA) + niB) - niA n B)

24 = x + 6 + 1 6 - 6

24 = x + 16

8 = x

Use a V e n n d i a g r a m to help so lve the p rob lem.

c ) ii) the in tersect ion of sets A and S

atuditioning gsr Is

s f e r i c m ' ^ ^ singers \ ^

10 + 6 + x = 2 4

1 6 + x = 2 4

X = 8

The re we re 8 g ids w h o w e r e g o o d s ingers but not g o o d 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 2

1. a) v ) d is jo int sets A and B

d) i) t ree d i a g r a m

first coin Mcofid eofci

e) iv) o u t c o m e tab le

1 2 3 4 5 6 t 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12

F o u n d a t i o n s of Ma themat ics 12 S o l u t i o n s Manual 4-37

7. e .g. , / is the un iversa l set of in tegers . E is the

subse t of e v e n in tegers . O is the subse t of odd

in tegers .

3. a) n{A uB) = n{A) + n{B) - niA n B)

niA)=^2, niB) = 9, niAnB) = 5

r)(/\u 8 ) = 12 + 9 - 5 = 16

b) niA u 8 ) = n ( ^ ) + n (B) - niA n B)

niA) = 23 , n ( 8 ) = 16, n(>\ n 8 ) = 1

niA uB) = 23+ 1 6 - 1 = 3 8

4. a) Let S represen t t he un iversa l se t of al l s tuden ts .

Let A rep resen t the s tuden ts w h o a t tended the f irst

schoo l d a n c e , a n d let 8 represen t s tuden ts w h o

a t tended the s e c o n d schoo l d a n c e . T h e n niA u 8 ) is

the n u m b e r of s tuden ts w h o wen t to o n e of the f irst

t w o schoo l d a n c e s .

niA uB) = niA) + niB) - niA n 8 )

niA) = 4 2 0 , niB) = 4 8 0 , niA nB) = 285

niA u 8 ) = 4 2 0 + 4 8 0 - 2 8 5 = 615

6 1 5 s tuden ts w e n t to o n e of the f irst two schoo l

d a n c e s of the year .

b) niA u 8 ) ' is the n u m b e r of s tuden ts w h o d id not

a t tend e i ther d a n c e .

niA u 8 ) ' = S - niA u 8 )

S = 1200 , n { A u 8 ) = 6 1 5

n ( / \ u 8 ) ' = 1 2 0 0 - 6 1 5 = 585

585 s tuden ts d id not a t tend e i ther d a n c e .

5. a) B = {set of b lack face ca rds in a s tandard deck of

p lay ing cards } = { J * , Q * , K * , J * , Q A , K * }

b) D = {set of d i f fe rent three-d ig i t n u m b e r s us ing the

dig i ts 1 , 3, and 5} = {135 , 153, 315 , 3 5 1 , 513 , 531}

c ) S = {set of al l poss ib le s u m s w h e n a pair of d ice is

ro l led} = {2 , 3 , 4 , 5, 6, 7, 8, 9, 10, 1 1 , 12}

d) T = {set of al l t he days of the w e e k w i th n a m e s that

beg in w i th T} = {Tuesday , Thu rsday }

6. a) { } , { red} , {b lue} , { red , b lue}

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 ) { } , {Apn l } , {May } , {June} , {AphI , May} , {Apn l , June} ,

{May , June } , {Apn l , May , June }

d) { } , {100}

8. S = { l , N , T , E, R, S, C O }

a) A = {set of vowe l s in S}

A = { I , E, 0 }

b) 8 = {set of let ters in S E C T I O N }

8 = {S , E, C, T , I, O, N}

c ) A u 8

= {set of vowe l s in S a n d set of let ters in S E C T I O N }

= {S , E, C, T , I, O, N}

d) A 8 = {e lemen ts in bo th A a n d 8 }

A n 8 = {t, E, 0 }

e) A ' u 8 ' = {e lemen ts not in A and e l e m e n t s not in 8 }

A'uB'= { N , T, R, S, C, E, I, 0 }

f) A ' n 8 ' = {e lemen ts in ne i ther 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 e l emen ts of S not in A}

A ' = { 1 , 3, 5, 7, 9, 1 1 , 12, 13, 14}

8 ' = { the e lemen ts of S not in 8 } = {2 , 4 , 6, 8, 10, 12, 14}

S

4-38 C h a p t e r 4: C o u n t i n g Methods

A u B = { the e lemen ts of A and B}

= { 1 , 2 , 3 , 4 , 5, 6, 7, 8, 9, 10, 1 1 , 13}

A n B = { the e lemen ts 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 e lemen ts 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 * }

A'

e ' = { the e lemen ts 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 * }

S

A u B = { the e l emen ts of A a n d 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 e l emen ts in bo th A and S}

A n B = { 6 A , 6 A , 6 ¥ , 6 * }

s

/ I \

A i \ { \ B )

\ /

10. e .g. ,

a) S is the un iversa l set of all s tuden ts in my ma th c lass.

A is the subse t o f s tuden ts w h o a re tal ler t han 6 ft .

6 is the subse t of all s tuden ts w i th b lack hair.

A n B ( shaded) is the subse t of s tuden ts w h o are

tal ler than 6 ft and have b lack hair.

b) S is the un iversa l se t of all ca rds in a s tandard

deck of p lay ing ca rds .

A is the subse t of all f ace ca rds .

B is the subse t of all red ca rds .

Au B ( shaded) is the subse t of all f ace ca rds and all

red cards .

11. Let S represent the un iversa l se t of all f i rs t -year s tuden ts .

Let C represen t the s tuden ts w h o take ca lcu lus .

Let A represen t the s tuden ts w h o take a lgebra . T h e n

n ( C u A)' wil l be the n u m b e r of f i rs t -year s tuden ts

w h o take ne i ther ca lcu lus nor a lgebra .

n{S) = 2 0 0

n (C) = 110

n{A) = 75

n ( C r ^ A ) = 60

n{C uA) = n{C) + n{A) - n{C n A)

n{CuA) = 110 + 7 5 - 6 0

n{CuA)= 125

n{CuAy = S-n{CuA)

r 7 ( C u A ) ' = 2 0 0 - 1 2 5

n{CuAY = 75

The re are 75 f i rs t -year s tuden ts w h o take nei ther ca lcu lus nor a lgebra .

F o u n d a t i o n s of Ma themat ics 12 S o l u t i o n s Manual 4-39

Chapter 4 Test, TR page 282 1. e .g . , T h e tab le s h o w s the poss ib le w ins a n d losses

of o n e of the t e a m s . S h a d e d cel ls ind icate g a m e s that

w o u l d not ac tua l ly be p layed , s ince o n e of the t e a m s

wi l l have a l ready w o n or lost two g a m e s . Tha t m e a n s

there are on ly six d i f fe rent o u t c o m e s .

G a m e 1 G a m e 2 G a m e 3 O u t c o m e s

W W W 1 o u t c o m e

W W

w L w 1 o u t c o m e

w L L 1 o u t c o m e

L W W 1 o u t c o m e

L W L 1 o u t c o m e

L L W 1 o u t c o m e

L L

b) T h e r e a re bNO d i f fe rent w a y s a t e a m can lose

exac t l y one g a m e but stil l w i n the c h a m p i o n s h i p :

W L W or L W W .

2. a) T h e s a n d w i c h has fou r e l emen ts : e g g sa lad or

ch i cken sa lad , le t tuce or t oma to , but ter or

m a y o n n a i s e , and w h o l e w h e a t bun or w h o l e gra in

bage l or s e s a m e seed bun . Us ing the F u n d a m e n t a l

Coun t i ng Pnnc ip le :

2 2 2 3 = 24 w a y s to m a k e a s a n d w i c h . I a s s u m e d

that the s a n d w i c h has exac t l y one i tem f r o m e a c h of

set of cho ices .

b) For each of the four d ig i ts , there are six op t ions

a n d repet i t ion is a l l owed . Us ing the F u n d a m e n t a l

Coun t i ng Pnnc ip le :

6 • 6 • 6 6 = 1296 w a y s to m a k e a p a s s w o r d . I

a s s u m e d star t ing the p a s s w o r d w i th a 0 w a s a l l owed .

c ) The re are 13 hear ts and 13 c lubs in a s tandard

deck of 52 ca rds , and the se ts of hear ts and c lubs are

d is jo int . The re are 13 + 13 = 26 w a y s to d raw a hear t

or c lub. I m a d e no assump t i ons .

d) The re are f ive let ters in T E E T H w i th bo th T and E

repeat ing tw ice . The re a re = 30 a r r a n g e m e n t s of

the let ters. I m a d e no assump t i ons .

e) The re are 25 d i f ferent t opp ings , and J im mus t

c h o o s e 3. O rde r d o e s not mat ter . J im can o rder

25 2300 d i f ferent p izzas . I a s s u m e d he w o u l d

c h o o s e exac t ly th ree topp ings and each w o u l d be

d i f ferent .

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 2 4 0

120

= 252

3. a) S ince o rder d o e s not mat ter :

' 1 0 1 10!

. 5

10

5

noi 5

10

5

10

5^

The re a re 252 se lec t ions tha t can be m a d e .

b) S ince o rder mat te rs :

'° ' ( 1 0 - 5 ) !

1 0 - 9 - 8 - 7 6 5 !

1 0 ^ 5 = 1 0 . 9 . 8 - 7 . 6

1 0 ^ - 3 0 240

The re a re 30 2 4 0 se lec t ions if the se lec t ions a re

o rde red by p re fe rence .

c ) e.g. , O rde r does not mat te r in part a ) but it d o e s

mat te r in part b) , so part a ) invo lves comb ina t i ons ,

w h e r e a s part b) invo lves pe rmuta t i ons .

d) e .g. . T h e a n s w e r to par t a ) is 5! or 120 t imes

smal le r than the a n s w e r to par t b) . Th is is because

the 30 240 f i ve-nove l se lec t ions f rom part b) mus t be

d iv ided by 5! to e l im ina te comb ina t i ons that a re the

s a m e , because o rder d o e s not mat ter .

4. T o so lve the p rob lem, look at the wa lk in th ree

sec t ions . In the f irst sec t ion , M a h a has to wa lk th ree

b locks eas t and th ree b locks sou th for a tota l o f six

b locks ; in the second sec t ion she has to wa lk one

block eas t and one b lock sou th for a total of t w o

b locks; in the th i rd sec t ion she has to wa lk t w o b locks

east a n d hwo b locks sou th for a total o f four b locks .

First sec t ion :

6 ! _ 6 - 5 4 3!

3 ! 3 ! " " 3 ! 3 - 2 1

6 ! _ 6 - 5 4

3 ! 3 ! ~

6 !

3 !3 !

6 !

3 !3 !

S e c o n d sec t ion :

1 ! 1 ! ^ 1-1

_2!^

1!1!

3 2 1

5 4

20

= 2

4-40 C h a p t e r 4: C o u n t i n g Methods

2 3

21

2!

2 ! 2 !

2 ! 2 !

Us ing the Fundamen ta l Coun t ing Pr inc ip le :

20 • 2 • 6 = 240

S h e can wa lk 240 d i f ferent v /ays.

Th is p rob lem can a lso be so lved us ing a pa thway d i ag ram;

5. a) Order does not matter , so this is a comb ina t i on

p rob lem. The re are (5 2 ) w a y s to choose two boys

f rom f ive, and (6 2) w a y s to choose two gir is f r om six.

2 i! / I

f

f, ,

f ,

• 4 - 3 ! 6 - 5 - -

'••2 ' t h '

4 D O

- 1 0 0 ,

= 150

There are 150 ways to choose a committee with two boys and two giris.

b) At least two giris means two, three, or four giris:

6!

2 ' l i

6!

[ 6 ^ 4 ) ! 4 !

AW

CJ 1; .Ji;«: ,1 i c f: t. 1 •;•>.!• commi t t ee , there c i c n ine fl. f . p l - ioi t t j i f i d i im ig two pos i t ions :

l i f l

1111

1

\3 \

^roi,,^ 1 / 2 712!

1 l l j l 2 j • • 7 !2 !

1 ¥ 1

1 1 5

i ¥ 9 i

12 = 36

The re are 36 committees that can be made if Jim and

Nanci must be on the committee.

d) More boys than giris means three or four boys:

5! 6 ! 5!

1 ; U

5

1 ; u

5

1 / 14

5

17 V4.

6 l fs' + i i u

6^ rs 1 J"U 6 l (5

+ 1 ; U

1

( 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 5 2 1 1

10-6 + 5̂ 1

-1

60 + 5

There are 65 committees that can be made with more boys than giris.

F o u n d a t i o n s of Mathemat ics 12 S o l u t i o n s Manual 4-41

B. a) 0 0 1 , C . l

60 . " ' :

' •« ! / } 1 ) : / ) ( / ! - 2 H / ' 3 ) 60 , -

1,1 ? | ( u 3 ) 30

u 5f} - 24 0

(// B i ( f i i 3 ) 0

n 8 or /I 3

-3 is f x i r r i n o f j u s

n = 8 b | /? - 8. .'/ i- 4 „P_, ^ f>0(.,c;,3 5j<j / i r- 4 and n > 2 H U n > 4

7. i f you pluf.o t h r f a n d K as requ i red . Ihrs uan

li.;jp[)on only O O P w a y for oaf:h pos i t ion . Fhu

roma in inp jsovon le t ters can then t)o a r ranged in

b e t w e e n , keep ing in m ind that there are repea led

IH tom- two 72s, two S's. and two O's-

1.1 " - iO ' ' h lV"

. 2 I2 '2 ! I 212-2

l ^ 2 ' 2 ' 2 ' ; ^ 2 2

U!2!2!j i 4

1,1^1 = 630 l 2 ! 2 ! 2 ! j

The re a re 6 3 0 d i f fe rent w a y s to a r range the let ters

w i th the g i ven cond i t i ons

4 -42 C h a p t e r 4 C o u n t i n g M e t h o d s