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