Counting methods solutions

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 black
red red/khaki red/black
blue biue/khaki blue/black
greer green/khaki green/black
Each X represents a different combination. There are
6 x's; therefore, there are six different vanations of the
outfit to choose from,
b) The number of outfit variations, O, is related to the
'-"r of shirts and the number of shorts;
O .number of shirts) • (number of shorts)
2
Ml I' are six different variations of the outfit to
choose from. This matches the part a) result.
2. a)
Upholstery
leather
cloth
Colour
red
•.il.-. r
red
1.1,1' It
•.V-xt.-
8 uThereto".• thf ro 0 Mpholstery-colour choices that
are available.
b) The number of upholstery-colour choices, U, is
related to the number of colours and the number of
kinds of upholstery;
U - (Ml mber of colours) • (number of upholstery)
iJ 1 2
8
There are 8 upholstery-colour choices that are
available. This matches the part a) result.
3. a) The Fundamental Counting Pnnciple does not
apply because tasks in this situation are related by
the word OR.
b) The Fundamental Counting Pnnciple does apply
because tasks in this situation are related by the word
A N D .
c) The Fundamental Counting Pnnciple does not
apply because tasks in this situation are related by
the word OR.
d) The Fundamental Counting Pnnciple does apply
because tasks in this situation are related by the word
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 looking at the tree diagram. I can see there are
2 ways in which Kim's team can win the series
despite losing one g a m e .
5. The number of colour-size vanations, C, is related
to the number of colours and the number of sizes;
C = (number of colours) • (number of sizes)
C = 5 - 4
C - 20
There are 20 colour-size variations that are available.
6. The number of computer systems, S, the
employees can build for their customers is related to
the number of desktop computers (dc), the number of
monitors (m), the number of printers (p), and the
number of software packages (sp);
- (# of y ) (# of m) • (# of p) • (# of sp)
^ S 4 f. 3
.5 -• .}uO
rhfciufo;u, Ihe employees can build 360 different
computer systems for their customers.
7. The number of possible meals. M, is related to the
number of soups (s), the number of sandwiches (sw), the
number of drinks (dr), and the number of desserts (P):
M = (# of s) - (# of sw) • (# of dr) • (# of d)
/W = 3 5 • 4 2
M= 120
Therefore, there are 120 different meal possibilities.
8. Event A: Selecting a rap C D
OR
Event B: Selecting a classic rock C D
n{A u B) = n{A) + n{B)
n{A ;„„ B) = 8 + 10
n{A u B) = 18
Therefore, Chadene can select from 18 C D s to play in
her car stereo that will match T o m ' s musical tastes.
9. a) The number of different PIN combinations, C. is
related to the number of digits from which to select for
each digit of the PIN. F:
C = P l • P2--P3- P4- Ps
C = 9 9 9 9 9
C = 59 049
There are 59 049 different five-digit PIN combinations.
F o u n d a t i o n s of Mathemati'. s «/ solutions Manual 4-1

b) llv: niirnlHjro! <iilferont PIN cfmibinations, N. )s
n.'lali'fl to llu' iiiinibcr ol digits Itorn w h i r j i to st;lt)rj for
each diyil ol llir; PIN. D'
N ' /)i n, / ) ; Dt Di
N M p. / 0 5
W 15 17f)
IhtjK.- or<: only 1f» 120 difforoni livc-ffigit PIN
f;urnl)inolif)iis in w h i d i the digits cannot repeal.
10. f ho n(jrnt)t;r of different t)ytes that can bo created,
N. IS rolatfHl lo Ifie number of digits from which to
f.hoosf! for each ditji! of the byte. B:
N - th H • / i . Bi Bs • Bf, B, By.
N - 7 2 2 2 2 2 2 - 2
N • 256
Thf.Tutonj. 2b6 different byles can bo creaUid
1 1 . a) 1 he fiurntwf of different upper-~caso letter
f.ossibilitips. H. IS related to the number of upper-case
letters ftom which to choose for each odd position of
the country's postal code, P:
W - Pi P ; P .
H 26 • 26 26
A/-- 1 / 57t>
Tlie number of different digit possibilities. D, is related
to the number of digits from which to choose for each
even position of the country's postal code, P;
D'^P'- Pi Pf.
D---^10-10 10
O 1000
The number of different postal codes that are possible
in this country. C. is related to the number of upper-
case letter possibilities, N, and the n u m b e r o f digit
possibilities, D:
C W • D
C = 17 576 • 1000
C ^ 17 576 000
Tticrefore, 1 7 576 000 postal codes are possible,
b) Tlie number of different upper-case letter
possibilities, N. is related to the number of upper-case
letters from which to choose for each odd position of
the country's postal code, P;
Al = P, Po • Pf,
A / - 2 1 - 2 1 - 2 1
N = 9261
The number of different digit possibilities, D, remains
the same since all digits can be used. The number of
different postal codes that are possible in Canada, C, is
related io the number of uppercase letter possibilities,
Ay. and the number of digit fK)SSibi!ities, D;
C = N D
0 = 9261 1000
C = 9 261 000
Therefore. 9 261 000 postal codes are possible in
C a n a d a .
12. T o answer this question, I need U) determine how
many digit r-.oml.nnations ttioro arc for the last four
digits of one ot these two phone n u m l j e i s . and then
multiply it by 2,
The number ot digit combinations. C, is rolnted to the
number of possible digits for each of the last four
digits of one of tho phono numbers, P'
c p , . p... p.,. f>,
C = 10 - 10 - 10 - 10
C-^ 10 000
The number of phone numtjers is 2C since there an)
two given tt.nnplates for the phone numbers in tho
question.
2 C 2(10 000)
2 C =^ 20 000
Therefore. 20 000 different phone numbers are
possible for this town,
13. The number of different codes, C. is related to
number of positions from which to select for each
switch of the garage door opener. G:
C^Gi- G2- Ga • G,, • Gr, • Gr, • G, - Gr; G:.
C = 3 • 3 - 3 - 3 ^ 3 • 3 • 3 - 3 3
C - 19 683
Therefore, 19 683 different codes are possible.
14. Event A; Selecting a pickup truck O R
Event B: Selecting a passenger van O R
Event C: Selecting a car O R
Event D: Selecting a sports utility vehicle
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
Therefore, a customer has 65 choices w h e n renting
just one vehicle.
15. a) Multiply the number of sizes of the crust, by the
number of types of the crust, by the number of types
of cheese, by the number of types of tomato sauce.
2 • 2 - 2 • 2 = 16
Multiply this number by the number of different
toppings.
1 6 - 2 0 = 320
Therefore, there are 320 different pizzas that can be
made with any crust, cheese, tomato sauce, and
1 topping.
b) Multiply the number of types of cheese by the
number of types of tomato sauce.
2 - 2 = 4
Therefore, there are 4 different pizzas that can be
made with a thin whole-wheat crust, t o m a t o sauce,
cheese, and no toppings.
4-2

1 i . a | The number of different ypper-^case letter
possibilities, N, is related to the number of upper^case
letters from which to choose for each of the first three
positions of the Alberta licence plate, P:
W = 24 - 24 • 24
N= 13 824
fl... number of different digit possibilities, O, is related
so In. - number of digits from which to choose for each of
H..- I . r i three positions of the Alberta licence plate. P
D - P.-P^^Ps
- ••0 - 10 - 10
P' m o o
The number of different possible Alberta licence plates,
I = -I J ! the number of upper^case letter
, / ; and the number of digit possibilities, O;
^ / i O
< ' , i ' ' ' , I f . 0 0
1 ; i.j'i CO.' Alberta licence plates are possible,
b) I he number of different upper-case letter
possibilities, N, remains the s a m e since the number of
letters in the plates and the number of letters that can
be used is the s a m e as in a).
The number of different digit possibilities, 0 , is related
to the number of digits from which to choose for each of
the last four positions of the Alberta licence plate. P;
D = P4 • Ps • Fe • P?
D = 10 • 10 • 10 - 10
D = 10 000
The number of different possible Alberta licence plates,
C, is related to the number of upper-case letter
possibilities. N, and the number of digit possibilities, D:
C = N- D
C = 13 824 - 10 000
C = 138 240 000
138 240 000 13 824 000 = 124 416 000
So, 124 416 000 more licence plates are possible,
17. e g,, If multiple tasks are related by A N D , it means
the Fundamental Counting Pnnciple can be used and
the total number of solutions is the product of the
solutions to each task. For example, A 4-digit PIN
involves choosing the 1st digit A N D the 2nd digit A N D
the 3rd digit A N D the 4th digit. So the number of
solutions IS 10 10 10 10 = 10 000. O R means the
solution must meet at least one condition so you must
add the number of solutions to each condition, and
then subtract the number of solutions that meet all
conditions. For example: Calculating the number of
4-digit PINs that start with 3 O R end with 3. The
solution IS the number of PINs that start with 3, plus
the number of PINs that end with 3. minus the number
of PINs that both start and end with 3:
1000 + 1 0 0 0 - 100 = 1900.
18. a) i) Event A Drawing a king O R
Event B; Drawing a queen
rif/. = n(A) + niB)
n{A u e ) = 4 + 4
n{A u e ) = 8
Likelihood = ^
52
Likelihood = ^
13
Therefore, there is a 2 in 13 chance that a king or a
queen will be drawn.
ii) Event A: Drawing a diamond
OR
Event B: Drawing a club
I >'•'"• P) -•.'/• I I''p.)
n{A u S) = 26
Likelihood = ^
2
Therefore, there is a 1 m 2 chance that a diamond or
a club will be drawn,
iii) Event A: Drawing an Ace O R
Event B: Drawing a spade
n(A u B) = n{A) + n(B) - n{A n B)
n(Au B) = 4 +13^1
n(A u e ) = 16
Likelihood = ^
52
Likelihood =
13
Therefore, there is a 4 in 13 chance that an ace or a
spade will be drawn.
b) No. e.g.. because the Fundamental Counting
Principle only applies w h e n tasks are related by the
word A N D .
19. e.g.. T o begin, there are 90 two-digit numbers.
There are 10 with a 1 in the tens column, 10 with a 2
in the tens column, and this pattern continues until I
reach the 10 with a 9 in the tens column. Next. I must
determine the numbers that are divisible by either 2 or
5. I know that every other number is even and thus
divisible by 2. This means that or 45 of the two-
2
digit numbers are divisible by 2. The two-digit
numbers that are divisible by 5 can be found by
starting at the first two-digit number, 10, and then
counting by 5 until I get to a three-digit number.
By doing this. I can determine that the two-digit
numbers that are divisible by 5 are: 10, 15, 20, 25. 30.
35, 40, 45. 50. 55, 60. 65, 70. 75, 80, 85. 90. 95.
There are 18 of them. I see that half of them'or 9 are
even and thus divisible by 2.
F o u n d a t i o n s of Mathemati' ^ V -..lutions Manual
4-3

Therefore, there are 9 numbers that are divisible by 5
and not by 2. If I add this together with the number of
two-digit numbers that are divisible by two (45). I see
that there are 54 two-^digit numbers divisible by 2 or 5,
Whatever is leftover from the two digit numbers are
the ones that are not divisible by either 2 or 5, This
amount is:
90 54 = 36. Thus, there are 36 two-digit numbers
that are not divisible by either 2 or 5.
20. The number of different outcomes for a student's
test. N, is related to the number of possible answers
for each question 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 perfect score is only 1 out of these 1024 outcomes;
therefore, there is a 1 in 1024 chance that the student
will get a perfect score.
2 1 . This question is solved by constant application of
the Fundamental Counting Pnnciple.
If an item from each category is selected:
0 = 3 5 4 2
0 = 120
If no soup is selected:
0 = 5 4 2
0 = 40
If no sandwich is selected:
0 = 3 4 2
0 = 24
If no drink is selected:
0 = 3 5 2
0 = 30
If no dessert is selected:
0 = 3 5 - 4
O = 60
If no soup or sandwich is selected:
0 = 4 2
0 = 8
If no soup or drink selected:
0 = 5 2
0 = 10
If no soup or dessert is selected:
0 = 5 4
0 = 20
If no sandwich or dnnk is selected:
0 = 3 2
0 = 6
If no sandwich or dessert is selected:
0 = 3 4
0 = 1 2
If no dnnk or dessert is selected:
0 = 3 5
0 = 1 5
If only n <^nur s a n d w i r h drink or dessert is selected:
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 herefure, 359 meals are possible if you do not have
to choose an item from a category.
L e s s o n 4 . 2 : I n t r o d u c i n g P e r m u t a t i o n s a n 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! 362880
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
-20
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

PosstioJ! Ptji^ilmn *^(jsitifm
I'firmii.itKtti
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
Permutattor*
4
Permutat,ir,n
5
Permutation
6
Raj Sarah Ken
b) Let L represent the total number of permutations:
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!
98!
4. Expressions a), c), and d) are undefined because
factorial notation is only defined for natural numbers,
5. a) 8-7-»^d S / ^1 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(180)
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 Mathematics V/ Solutions IVIanual

b) , •• + 2|!
1,1 41b; . ,.({n + 2){n + i)(n)in^i}...(3)i2}{t
in-. 41'
( o . 1)1 ( / n n i n p 1)(» ? | (3)(2)(1)
m in)(n ~1](n ?] l 3 ) ( 2 ) ( l )
p i i 1!'
c)
/: ^ 1
' ' ^ l i y ^ P 4 | ( n ^ 5 ) . . . ( 3 ) ( 2 ) ( l )
i|! (/) 3)(r, 4 H r r : i ^ 3 ) ( 2 ) ( i r
,1 ^ ^ ( n ) ( r i ^ l ) ( / i 2)(,/ 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.'
ffl 5)1
f n .4)!
^ = (ri + 5)(/i + 4)
= f)2 + 9o + 20
[n I f . { V i ^ 2 ) ( n - ^ 3 ) ( 0 ^ 4 ) . . . { 3 ) ( 2 ) ( 1 )
^> ( . r i , r ( , r i ) ( ; ^ r 2 ) ( ^ ^
(rt 1)! ( n - l ) { n - - 2 ) !
( » ^ 2 ) ! ^ 1
(/? 1 ) ! ^ n - - 1
7. There are nine students in the lineup, so there are
nine possible positions. Let L represent the total
number of permutations:
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
There are 362 880 permutations for the nine students
at the Calgary Stampede.
8. There are five students in the club and there are
five possible positions. Let L represent the total
L = 5 4 3 2 1
/
/ - -b 4 3 • 2 • 1
- 2 0 - 6
/ - 120
There are 120 different w a y s to select m e m b e r s for
the five positions.
9. There are six activities to do and there are six days.
Let L represent the total number of permutations:
hi
.' H 5 - 4 • 3 - 2 • 1
I • : i . 4 6
•' 1 / 0 - 6
/ - 7-'0
1 ht;ru are 720 different w a y s they can sequence
ihos(< activities over the six days.
10. There are 28 movies, so there are 28 possible
spots for the movies to go. Let L represent the total
L = 28!
L = 3.048... X 10^®
There are about 3.05 x 10^® possible permutations of
the movie list.
fl!
(n + l ) ( n ) ( i i ^ l ) . . . ( 3 ) ( 2 ) ( l )
(Vr+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
There is one solution. 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)(ii + l)(ii!)
( n + :
+ n + 2n + 2 = 6
n'' + 3n + 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 undefined
Check n = 1
LS RS
< i 1 2)1 (.
1!
3!
11
3 - 2 - 1 !
1!
3 2
6
There is one solution, 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
There is one solution, 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
fi + 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 undefined
C h e c k n = 6
LS RS
3 ( 6 + 1)1 126
( 6 ^ 1 ) !
3(7!)
5!
3 - ( 7 - 6 - 5 ! )
5!
3 - 7 - 6
126
There is one solution, n = 6.
F o u n d a t i o n s o f M a t h e m a t i c s 12 S o l y t i o n s M a n u a l 4-7

L • r p i e s u iil till ni
I a / (, L 4
I
1 - 8 / h H 1
L 8 A? 20
12. Ihcu: .Ko '»nht more players left to organize so
It (OH- c>ifih! iriore spots left in the batting order. Let
mber of permutations:
3 2 1
3 - 2 - 1
L - :VAb 120
L = 40 320
There are 40 320 possible batting orders.
13. There are 7 possible digits to use and there are 7
digits in each serial number. Let L represent the
1 = 7 - 6 - 5 - 4 - 3 - 2 - 1
1 = 7!
There are 7! possible serial numbers. This makes sense
because, e.g., the integer in the factorial (7 in this case)
for the number of permutations is normally equal to the
number of spots in which there are things to place. There
are seven spots in the serial number so this means that
the number of permutations should be 7! which matches
the answer that w a s found.
14. There are 5 cars to be arranged between the
engine and the caboose so there are 5 spots in which
the cars can be lined up. Let L represent the number
of permutations:
1 = 5 4 3 2 1
1 = 5!
1 = 5 4 - 3 2 1
1 = 20 6
1 = 120
There are 120 w a y s for the cars to be arranged
between the engine and the caboose.
15. There would be 7 c h u c k w a g o n s behind Brant's so
there are 7 spots where the other dnvers could finish.
Let 1 represent the number of permutations:
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 wins, there are 5040 different orders
in which the eight chuckwagons can finish.
16. a) e.g.. Y K O N U , Y U K N O , Y K N O U
b) There are five letters so there are five spots to put the
letters. Let 1 represent the number of permutations:
1 = 5 - 4 3 2 1
1 = 5!
There are 5! possible permutations. This makes
sense because e.g.. the integer in the factohal (5 in
this case) for the number of permutations is normally
equal to the number of spots in which there are things
to place. There are five spots to place the letters so
this means that the number of permutations should be
5! which matches the answer that w a s found.
17. a) e.g.. Using tnal and error. I have the following
calculations:
1! = 1,2^ = 2 ; 2! = 2, 2" = 4;
3! = 6. 2 ' = 8; 4! = 24. 2'' = 16
I notice that for n = 4, nl is greater than 2". This
continues for n > 4 because 2** will keep getting
multiplied by 2. while 4! will keep getting multiplied by
numbers greater than 2 to obtain the higher factonals.
b) e.g.. Using what I have in a), I know that for n < 4,
nl IS not greater than 2". The calculations for these
values of n are s h o w n in a). Thus for n = { 1 , 2 . 3}. nl is
less than 2".
18. e.g., First, figure out how m a n y w a y s Dadene and
Arnold can be placed next to each other in the line.
Thin cnn ho found using 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
From the table I can see that there are 18 different
w a y s for Dadene and Arnold to be placed next to
each other in the line. For every one of those
18 ways, there are 8 other dancers to be placed in
8 different spots in the line. Let 1 represent the
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
There are 725 760 possible arrangements of the
dancers for the Red River Jig.
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-rniiitiific»ri
' J
i ^
; 11}
' 12
P r e s i d e n t ^Vice P r e s i d e n t "
K,4ji
K.itrifia
Kntrr vl
,lf">r
Nel/lf
_ N.i/ir
M o t u i m a d
M(>umai:y
Mt.harfidd
N j / i i
Katrsn.'^
Jt.
hatiirt;,
_ Jess
b) „ p - ^
4!
It a president and vice-
(4^2)1
4!
2!
4 ^ 3 ^
2!
, P , - 4 . 3
, P , = 1 2
The formula for „Pr gives an answer of 12. This
matches my results from 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 chocolate bars can
be distributed.
b) P = r ^
« ^ ( 6 ^ 1 ) !
6!
5!
6 - 5 !
5!
The chocolate bars can be distributed in 6 different ways.
p =
F o u n d a t i o n s of Mathematics 12 S o l u t i o n s Manual 4-9

4. loFs is larger, e.g., I know this by looking at the
formula for „Pr. The numerator is the s a m e for both
values since n is the s a m e . The denominator will be
smaller for the first value since it has a greater r. W h e n
you divide a numerator by t w o different denominators,
the final value is greater for the one with the smaller
denominator. Based on this. I know that WPB has the
larger value since its expansion has the smaller
denominator.
9 IJ ci . As^-unimg that any o f t h e 10 digits can be put
in any i, tbp 'o remaining spots for the SINs, let S
fct.ff.'M-(it tin- number of social insurance numbers:
S - 10 10 HJ !(i 10 1(1 10 10
- Kl'
S 1f'(J DUO 000
ri»M-f. .iro 100 000 000 different SINs that can be
r(;fji.'.t';r<!<l in cMch of these groups of provinces and
lerri!f)nf);'
5. ,/'
J'
P
9]
6!
fd
/ ' L 0 /
,P 'M
There are 504 different w a y s the positions can be filled,
M.O 4).
10. a) }'
P
« - 11!
1 5 ^ =
15-14-13-12-11t
11!
^gP^ = 1 5 - 1 4 - 1 3 - 1 2
, , P , - 3 2 7 6 0
There are 32 760 possible executive committees.
7. 3P3
8!
^ ( 8 - 8 ) 1
8!
^ 0 !
P
3'
1
P tJ
P H 7 b
, P, - 40 520
Therefore, 40 320 different signals could be created.
' ( 5 0 0 0 - 3 ) !
p ^ 5000!
5000 3 4gg7,
5000^3 =
5 0 0 0 - 4 9 9 9 - 4 9 9 8 - 4 9 9 7 !
4997!
5 0 0 0 - 4 9 9 9 - 4 9 9 8
5Qj,j,P3 = 124 925 010 000
There are about 124 925 010 000 different w a y s the
tickets can be drawn.
1 1 / f)H
12'
/•
12 11 10 q 7!
/ I
12 n Ui 0 f-'
0504(1
There are 95 040 w a y s the coach can select the
starting five players.
b l A
P^
(11^4)1
11!
7!
1 1 - 1 0 - 9 - 8 - 7 !
7!
„ F ^ = 1 T 1 0 - 9 - 8
, , P , - 7 9 2 0
There are 7920 w a y s the coach can select the
starting five players, if the tallest student must start at
the centre position.
p . 1 ^
10^3 7,
p ^
1 0 - 9 - 8 - 7 !
7!
^ j , P 3 = 1 0 - 9 - 8
, „ P 3 = 7 2 0
Multiply by 2, since Sandy and Natasha can play the
guard positions in either order. (720)(2) = 1440
There are 1440 w a y s in which the coach can select
the starting five players, if Sandy and Natasha must
play the two guard positions.
1 1 . a ) n > 0 a n d
n - 1 > 0
n> 1
Therefore, the expression is defined for n > 1,
where n e I.
b) n + 2 > 0
f i > - 2
Therefore, the expression is defined for n > - 2 ,
where n e I,
4-10 C h a p t e r 4 I o u n t i n g M e t h o d s

c) ri + 1 > 0 A N D n >0
n> 1
Therefore, the expression is defined fc
w h e r e n e L
d) n + 5 > 0 A N D n + 3 > 0
n > - 5 n > - 3
Therefore, the expression is defined 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
There are 360 w a y s to draw the four marbles if you do
not replace the marble each time.
b | Let L represent the number of ways:
L = 6 • 6 - 6 - 6
L = 6'
L = 1296
There are 1296 w a y s to draw the four marbles if you
replace the marble each time.
c) e.g., Yes; if you replace the marble, there are more
possibilities for the next draw.
13. a)
2 0 - 1 9 - 1 8 - 1 7 - 1 6 - 1 5 !
15!
2„P5 = 2 0 - 1 9 - 1 8 - 1 7 - 1 6
20 Pg = 1 8 6 0 4 8 0
There are 1 860 480 different w a y s to award the
scholarships.
b) Let L represent the number of ways:
1 = 20 - 20 - 20 - 20 - 20
L = 20^
1 = 3 200 000
There are 3 200 000 different ways to award the
scholarships.
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) Subtract the total possible n u m b e r s by the answer
to part a).
104 = 10 000
10 000 - 5040 - 4960
There are 4 9 6 0 different phone numbers.
15 u) • I eed to solve . ^ = 20
( n - 2 ) !
fi r ^tiO 1 1 ^ 2 > 0
r i > 2
nl
Therefore,
( n ^ 2 ) !
{n){n^i):n - ' j p ; ,h h )(2)(1
"i'^ '](''';(t)
{ „ ) ( „ :
20 is defined for n > 2. where n e I.
T- = 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 solution to n > 2
Check fl = 5
LS RS
5P2 20
5!
( 5 ^ 2 ) 1
5!
3!
5 - 4 - 3 !
3!
5 - 4
20
There is one solution, 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
Therefore,
n e I.
(n + 1 ^ 2 ) !
72 IS defined for n > 1, where
There are 5040 different phone numbers possible.
F o u n d a t i o n s of Mathemati utions Manual 4 t <

:.. I- >)<
(r, + l)!
C h e c k f = 2
72
72
72
= 72
(11^1).
( r i + 1 | ( / / l ( » 1i|u- ••') t . J ) i 2 | i l )
2 ) " V ' ) ( 2 | ( 1 )
(ii + 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 fl = 8
The root n = 9 is not a solution to n > 1.
Check n = 8
LS RS
8 +1P2 72
9P2
(9 2)1
9!
7!
9 - 8 - 7 !
JI
9 - 8
72
1 e. a l The equation I need to solve = 30 .
^ 0 - r)l
6 - r > 0
r < 6
Therefore. = 30 is defined for 0 < r < 6, where
' ( 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
There is one solution, r = 2.
b) I h<- ....luation I need to solve is 2
7 r 0
7!
420
Th<.i.-f.,rf; 2
where r e l .
7!
420 IS defined forO < 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 RS
2(^P^ 4 2 0
( 7 - 3 )
7 - 6 - 5 - 4 !
4!
2 ( 7 - 6 - 5 )
2 ( 2 1 0 )
420
There is one solution, r = 3.
4-12 C h a p t e r 4 : C o u n t i n g M e t h o d s

1?
I : RS
nfn nPn ^ 1
nl fl!
nl n!
0! [fl + 1
nl n!
1 1!
n! nl
1
n!
LS = RS
18. a) e.g., The formulas for both „P„ and r,Pr have a
numerator of nl. However, the formula for „ P „ has a
denominator of 1 and the formula for „Pr has a
denominator of ( o ^ r ) L
b) e.g., A group of friends each order a different
flavour of ice cream from a shop with 12 flavours.
How many possibilities are there if the group is 12
people? If the group is 7 people?
19. a) n = 52 and r = 5
P ^ 52!
' ( 5 2 ^ 5 ) !
5 ^
52 5
^.^:,^l:.50.,:'*.9,:48-47!
47!
52P5 = 5 2 - 5 T 5 0 - 4 9 - 4 8
ggPg^ 311875200
There are 311 875 200 possible arrangements,
b) n = 26 and r = 5
26!
( 2 6 ^ 5 ) 1
26!
2« ' 2 1 !
As =
2 6 - 2 5 - 2 4 - 2 3 - 2 2 - 2 1 !
21!
^ePg = 2 6 - 2 5 - 2 4 - 2 3 - 2 2
26P5 = 7 8 9 3 6 0 0
Likelihood =
7 893 600
•100%
311 875 200
Likelihood = 0.025...-100%
Likelihood = 2.531...%
Therefore, there is about a 2.53% chance that an
arrangement contains black cards only.
1 -•
p - i ^ ' - : 11 l u y
P ' r U l l
UKel,hood= J ^ ^ . . . 1 0 0 %
311875200
Likelihood = 0 , 0 0 0 , , . - 1 0 0 %
Likelihood = 0.049...%
Therefore, there is about a 0.05% chance that an
arrangement contains h o n d ^ only.
20. e.g.. „ . , P , ^ ^ /
n - 1 - n
^ n - o - - 1 - n - ^ 2 . ( f f ; 1 -11
= n{n)
= n^
2 1 . e.g., „ P . , =
(7- 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., January 5, April 23, July 24, and October 15
would be 5, 113, 205, and 288.
b) 365 • 365 • 365 • 365 = 17 748 900 630
c) i) = 0.983... or about 98.4%
365
ii) 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
ii) 1 ^ ^ = 0.033... or 3.3%
' 30
e) For example, they w e r e close but not the s a m e .
M i d - C h a p t e r R e v i e w , p a g e 2 5 9
1. The number of subs to choose from, S, is based on
the number of buns (b). the number of cold cuts (cc),
the number of cheeses (c), the number of toppings (f),
and the number of sauces (s):
S = (# of b) • (# of cc) - (# of c) • (# of t) - (# of s)
S = 3 - 5 • 3 - 12 - 3
S = 1620
So, Mario can choose from 1620 different subs.
2 «; (J Vou < ;ii< lis* on»- ot K W and C 1r)f the first
'Ji.ir.-K.tr-i oil.; nj VC ,j(>perooM lottrjrs lor the
s.'-cond .md third oh.mjcto'c;, nod one id tho
yj] u()f«'to.iso loUfMs Ol o blank for tfio lost character.
l u o m this I g.;l Itio lollowmg calc utation
ft o .lotion f.omos - :< 2() 26 ?f
li i-l .tdtion i.nmoh 04 / u b
I hotoforo, .station names are poissible.
3. t vent A Rolling a 2 O R
hvenl B- Rolling 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 "
From the table above, there is one w a y to roll a s u m
ol I wilh a pair f)f d i r e a n d three w a y s to roll a s u m of
10 with a pai! of dice.
fi(A ' >B)- niA) < 0(8)
n(A . ' 8 ) ' 1 + 3
niA ' B) - 4
There arc 4 ways that a s u m of 2 or a s u m of 10 can
be rolled with a pair of dice.
4, 1 0 - 9 - 8 = 720
There are 720 w a y s to select 3 horses to c o m e first,
second, third in a 10-horse race.
5. a) 8!
8!
b) 6! • 3!
6! • 3!
6! • 3!
8 - 7 - 6 - 5 - 4 - 3 - 2 - 1
40 320
(6 • 5 • 4 • 3 • 2 • 1) • (3 • 2
( 7 2 0 ) • ( 6 )
4320
1)
^ = 9 - 8 - 7 - 1
d)
9!
6!
9!
6!
9!
6!
91
6!
9!_
6!
Ill
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 if) 8 / fi f) 4 3 2 II
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. There are nine players on the team so there are 9
different positions. Let L represent the number 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
There are 362 880 different lineups that can be
formed by nine players on a softball t e a m .
7. a) (fi + 5 ) ( n + 4 ) !
= (f, + 5 ) [ ( r i + 4)(r) + 3)(ri + 2 ) . . . ( 3 ) ( 2 ) { l f
= (n + 5)(r, + 4 ) { i i + 3 ) ( n + 2)...(3)(2)(l)
= (n + 5)i
( n i 4 ) ( i i + 3 ) ( n + 2 ) ( o - M ) ( i i ) . . . ( 3 ) ( 2 ) ( l
- ~ , n . 2 ) ( n ^ ( n f i 3 ) m
= (n + 4 ) ( n + 3)
n' r 4n I 3n r 1 2
= n" + 7 n + 12

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 fl = 9
Check n = 8
I s RS
( Bj- 72
t 8}!
r - , is undefined
f 10)!
Check 17 = 9
LS RS
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
(fi + 4 ) { f i ^ 7 ) = 0
/. f _ 0
i RS
undefined
30
LS RS
( 7 - 1 ) ! 30
( 7 - 3 ) !
6!
4!
6 - 5 - 4 I
4!
6 5
30
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
4^15

12!
( 1 2 - 1 0 ) !
p . 1 2 !
« 2!
IV 11 If) 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 Afiere n e I.
t.- ' / I 4 C' A N ! 1 / 1 2 0
/!_ -I /* 2
r i i f cxpt ;sMi-'!i I'. dufitioO for n - where n e !.
r,; r» '1 - 0 A N [ ) tt 5 0
n > 4 n > 5
The expression is defined for n > 5, where n g I,
d: n + 2 > 0 A N D n s 0
f i > - 2
The expression is defined for n > 0, w h e r e ne L
b ) a : n > 0 A N D n ^ 2 > 0
n>2
The expression is defined for n > 2, where n g I.
b: r7 ---1 > 0 A N D n - 3 > 0
n > 1 n > 3
The expression is defined 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
Rennie can load his C D player in 27 907 200 different ways.
12. n = 14 and f = 2
14!
P - 13?
T h e n ; aio 182 ways that Manny and 2 other players
(,;in line up to receive the championship trophy.
13. A.jtfH' o g . T h e number of w a y s to choose a
pr..'!>i!irnt .ir.d a vice-president from a group of five
5!
students is 20 . I could also use the
Fundamental Counting Pnnciple because there are
five choices for president and four choices remaining
for vice-president: 5 • 4 = 20.
L e s s o n 4 . 4 : P e r m u t a t i o n s W h e n O b j e c t s A r 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 Methods

— * •'ie arrangement of 6 flags: rrangements:
7 4-7-
•i riuic i j i ^ UU Giiloicnt signals that can be m a d e from
the 6 flags hung in a vertical line,
3. Let C represent the number of ways:
6!
3 L 3 !
C = 20
There are 20 different w a y s three coins land as heads
and three coins land as tails,
4. Let R represent the number of ways:
18!
R
R =
10!5!3!
1 8 - 1 7 - 1 6 - 1 5 - 1 4 - 1 3 - 1 2 - 1 1 - 1 0 - 9 - 8 - 7 - 6 - 5 - 4 - 3 !
1 0 - 9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1 - 5 - 4 - 3 - 2 - 1 - 3 !
l? = 1 7 - 1 4 - 1 3 - 1 T 9 - 8
R 2 450 448
There are 2 450 4 4 8 ways that this record could have
occurred
5. Let C represent the number of ways:
2!3!4!
'? : 3 •' 1 1 V '
f" 0 ( {, !
C 1260
There are 1260 ways that Norm can distnbute
1 cookie to each grandchild,
6. a) Let A represent the number of arrangements:
A = 5!
/ = 5 4 • 3 2 1
A = 120
There are 120 different arrangements that can be
made using all the letters.
b) Let A represent the number of arrangements:
2!
^ 7 6 - 5 - 4 - 3 - 2 !
A ^ 7 6 - 5 4 - 3
A 2520
1 here are 2520 different arrangements that can be
made using all the letters.
8 - 7 - 6 - 5 - 4 - 3 - 2 !
/• ., ' i f
inu.fif,', ,,i , . li ments that can be
d ; i • 7i '••[)!' • <>nr m.• ..iifnU.-- •„ .rrangements:
2-1-3 2 1
A 3 9 9 1 6 8 0 0
There are 39 916 800 different arrangements that can
be made using all the letters.
5!5!5!
, 11 n f 1 1 "f •. f. ^ ^ 'I . 2
-% .1 - • 4 '. V : c . ll o T
14 r. ! ! » / C
A 756756
There are 756 756 different ways he can arrange the
books on the shelf.
h i ^'.r.-iJi. llu sets of 5 together.
A - ' 2
A = B
rays he can arrange the books,
8. e g,, A shish kabob skewer has 4 pieces of beef,
2 pieces of green pepper, and 1 piece each of
m u s h r o o m and onion. How many different
combinations are possible?
R
R
9. a) Let R represent the number 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
There are 126 routes travelling from point A to point B
if you travel only south or east.
4-17

b) Let R represent the number of routes:
13!
7!6!
1 i 1 11 If) 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. ;iir- i r i r . loutes travelling from point A to
|ic)ini H ll '/'HI iravel only south or east.
10. Let R represent the number of routes:
„ 13!
8!5!
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
There are 1287 routes travelling from the house of
Jess to the house of her friend if she travels only
north or west.
11 - a) o.g I dl .vw th« following diagram to show the
number of ways to get to each intersection
B,:^.::;HJ..!s:.
e4
^ 4--
1 ; t !i
T h e s u m of the numbers on the top right and bottom
left corners of each block is equal to the number of
routes to the top left corner of each block. There are
560 different routes from A to B. if you travel only
north or west.
b) I need to go north twice and west four times, for a
total of 6 moves, to travel the first 2 by 4 block of the
route. I need to go north once and west once, for a
total of 2 moves, to travel the next 1 by 1 block of the
route. I need to go north twice and west twice, for a
total of 4 moves, to travel the last 2 by 2 block of the
route.
Let R represent the number of routes:
R 61 „ 4!
4 ' 2 ' 2 0 "
4 5 V I 2 I 2 1 2 1 '
R -15 | 2 i i l ) !
P - 180
There are 180 different routes from A to B. if you
travel only north or west.
12. Let P represent the number of permutations:
r
5'3i
8 ' V. r. 4 i ^ 1
^ r. 4 3 2 1 .3 2
p -i 2
P - 5 6
There are 56 different permutations of answers that
the teacher can create.
13. a) Let P represent the number of permutations:
p = 7l
P = 7 - 6 - 5 - 4 - 3 - 2 - 1
P = 5040
There are 5040 different arrangements possible for
the new totem pole.
b) Let P represent the number of permutations:
7!
2!2I
/ R .h • 4 3_ ^ 1
2 1 . ]
P - / 6 •
P 1260
f here are i2bO difteient arrangements possible for
the new totem pole.
14. e.g., nPn will be too high; it gives the number of
arrangements of all n items, but some of the
arrangements will be identical because of the
a identical items in the group.
15. a) e.g., I a m assuming that the coins of the same
denomination are considered identical objects. Let A
represent the number of arrangements:
9!
P
P
4 ! 3 ! 2 !
9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1
4 - 3 - 2 - 3 - 2 - 2
A 1260
There are 1260 ways the 9 coins can be arranged in a
line.
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 coins of the same
flsidered identical objects. Let A
if of arrangements:
1 ' •.
A = 35
There are 35 w a y s ns can be arranged in a
line.
1S. The number of ways to divide the 8 remaining
freezies amongst the other 8 children is what I want.
Let P represent the number of permutations:
215!
^ 2 - 1 - 5 - 4 - 3 - 2 - 1
F = 7 - 6 - 4
P 168
There are 168 w a y s to distribute the 10 freezies
1 /, t; ifW mber of permutations:
p ^ y f ' : •<
P 560
There are 560 permutations possible if you must start
with A and end with C.
b) e g., If you start by putting the I's in the first and
second positions, and then in the second and third
positions, and so on and so forth up until you put t h e m in
the ninth and tenth positions, there are 9 different
arrangements of the I's just on their o w n . The number of
different arrangements of all the letters in each of these
9 arrangements is the number of ways to organize the
other 8 letters. Since the other 8 letters are always the
same, the number of permutations of the letters for each
arrangement of the I's is the same. Let P represent the
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
There are 10 080 permutations possible if the two I's
must be together.
18. e.g., B A N D I T S has 7 different letters, so the
number of permutations is 7! B A N A N A S also has 7
letters, but there are 3 As and 2 Ns so you must
divide 7! by 3! - 2! = 12,
19. The shortest possible route contains 3 moves
diagonally to tho right, 3 moves diagonally to the left,
and 3 moves down Let R represent the number of
routes:
f:
U U.'.f
U .'• ,' t. •.
i:
3 ! .
P :. i ; /
^1
1 •» ii ' d».''.u j . ri the top rear vertex of the
• . t". Ml. '< >.-j •'. '••'•'ll of the cube,
20. a) e.g., This is the same as arranging the
20 players then dividing by 2! ten times because the
order of pairs does not matter. Let T represent the
number of pairs:
2 P
r = 2,375..,x10'^
There are about 2.38 x 10^^ w a y s to assign 20
players to 10 double rooms.
b) e.g., This is the same as arranging the 20 players
then dividing by 4! five times because the order of
pairs does not matter. Let T represent the number of
pairs;
- f
r = 3 . 0 5 5 . , . x l O "
There are about 3.06 x 10^^ w a y s to assign 20
players to 5 guadruple rooms
21. a) e.g., I can make a table to show all o f t h e
arrangements that could be made. Position 1 in the
table below is the leftmost position, and position 4 is
the rightmost position.
Position
2
R
R
K
W
W
R
R
N
W
w
Position
3
' l~.
W
w
R
Position
4
W
R
R
w
VV
R
R
W
W
R
R
"vv
w
w
w
R
W
R
w
w
utions Manual 4-19

From the table, I see that 14 different arrangements
might be made.
b) e.g., From the table above, 1 out of the
14 arrangements, from left to right, would be red,
white, white, red. Therefore, there is a 1 in 14 chance
that the arrangement, from left to right, would be red,
white, white, red.
Applying Problem-SolYing Strategies,
page 27©
A. 4044 paths
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
Bnan Rachelle 1 i:ili
Bnan 1 inh RachoHo
Rachelle Brian ' l i n h
Rachelle Linh Hi
Linh R a c h f l i o Bnan
Linh Brian j Rachelle
c) Since all 3 volunteers are being used to help
unload the vehicles, there is only one w a y they can be
chosen for this job,
d) Part a) and b) involve permutations and part c)
involved combinations. I know because in part a) and
b), the order in which the volunteers were selected for
the jobs mattered. In part c) the order did not since all
the volunteers were being selected to do the same
job.
2. e.g., The main difference is that for the permutations,
the order of the 4 objects matters, and for the
combinations, it does not. For the permutations, you
could have multiple arrangements with the same
objects since there is more than one w a y to order a
group of four different objects. This is not possible for
combinations since you just need one arrangement for
each group of 4, regardless of the order,
3. Let C represent the number of dance committees
possible:
c - ;io
There are 210 ways that 4 of the m e m b e r s can be
chosen to serve on the dance committee.
B. 2(924) + 2(2508) + 2(3498) = 13 860 paths
C . Yes. There are 2(3936), or 7872. paths that lead to
no money at all, but 17 904 paths that result in the
contestant winning something. The contestant has a
better chance of winning something than nothing, so
it's a fair g a m e from the contestant's point of view.
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 represent the number of ways:
W= 3!
W/= 3 2 1
W=6
There are 6 different w a y s that Bnan, Rachelle, and
Linh can be chosen for these jobs.
4. Let C represent the number of combinations:
C = 12C3
C = 220
There are 220 ways 3 of the 12 dogs can be selected
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
vanilla strawberry
vanilla chocolate
vanilla butterscotch
strawberry vanilla
strawberry chocolate
strawberry butterscotch
chocolate vanilla
chocolate strawberry
chocolate butterscotch
butterscol vanilla
lutterscotch strawberry
' lutterscotch hocolate
4-20 C h a p t * ' » .-unting Methods

b)
I kfvom 1
•.Ml l.l! t
f l.ivcuf 2
• IV,. {.I.lf«.-
rKjttcm. ot( h
liie huftiUcI Ul l/vU-lldVOUi OUIllblhalluUS buUdUSe
each two-flavour combination can be written in two
different ways
2 « ;sent the number of committees:
3. Let T represent the number of possible teams:
(,(11/
b' I 1 1 . ' i:
i: •] i 2 1
*, 4 ;•- 2 I
J / 1. c /
1 lif.u- i!if H,^4 w a y . h people can be selected from a
yruup ot i z lo fonn a dodge-ball team.
4. a) C =
5!
There are 10 possible committees.
)resent the number of committees:
2 1 ( 5 ^ 2 ) !
2 «
L 0 2
C - 1 0 There are 10 possible committees,
c) e.g., My answers for parts a) and b) are the s a m e .
This occurred because the s u m of 2 and 3 is 5.
' 3 ! ( 5 - 3 ) !
^ 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 Mathemati^ i l y f i o n s M a n y a l 4-21

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 210 w a y s 6 players can be chosen to start
,1 volleyball g a m e from a team of 10.
C = 55C5
C =
c
55!
55!
5!-50!
5 1 5 i 52 C1 50!
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
There are 3 478 761 different combinations of hip-hop
songs you can download for free.
7. Let H represent the number 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 538150
There are 752 538 150 different 8~card hands that
can be dealt.
b) Let L represent the number of different lineups,
n = 14 and r = 8 because Connie must be the pitcher
of the starting lineup.
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 10-9
0 i> 4 3 2 1
i:-. I I .5
3003
There are 3003 w a y s that the coach can choose his
starting lineup of 9 players, if Connie must be the
pitcher.
9. a) Yes, I do agree.
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 other cases with the same relationship
as part a) are aCi = gC?, eCo = eCe, and 12C7 = 12C5.
I notice that if you have two combinations with the
s a m e n, and the s u m of the 2s for those combinations
is equal to n, then the value of the combinations will
be the s a m e .
c) e.g.,
n
n - r
8. a) The problem involves combinations e.g.,
because it does not state that the order of the starting
line matters.
4-22 Chapi unting Methods

10.
Let T represent the Let S represent the
number of combinations number of combinations
' • .'.M= hers. for the students;
• I
8!
3!5!
8 / '•'
•• i
8 / C
sent the numc
S - 4
S 56
committees;
; - 10
3
I l.t,tu are 560 graduation committees that the
principal has to choose from.
' ' ai Let C represent the number of committees;
c
10!
5 ! ( l 0 - 5 ) !
10!
5!5!
1 0 - 9 - 8 - 7 - 6
/ ' ,
C 252
There are 252 committees that can be formed if there
are no conditions,
b)
Let W represent the
number of combinations
for the w o m e n ;
Let M represent the numbe
of combinations 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
6M =
Let C represent the number of committees:
C=W-M
C = 20 6
C= 120
must be exactly 3 w o m e n .
c) Let C represent the number of committees:
6!
1!(6-1)!
6!_
1!-5!
6 5!
1-5!
6
1
C = 6
must be exactly 4 men.
d) Let C represent the number of committees;
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
can be no men.
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 !
60C,
C a s e 2: 4 men and 1 w o m a n
' ' ' ' 4!-0! II 5!
, q - , ^ = 1-6
4 C , - e q = 6
Number of committees = 60 + 6
Number of committees = 66
66 5-person committees can be formed if there must
be at least 3 men.
F o u n d a t i o n s o f M a t h e m a t i c s =2 S o l u t i o n s M a n u a l 4-23

12. e . g . Let's say I want to assign students to the 0! 1!
room with 5 beds firsL Let A represent the number of «l o H ^ ^^j, " I i H oi(l-^0)«
w a y s to assign the 12 students to the 5 beds: ' n ^'
5!M:' .'dt 0 ^ 0 - ^ 1 ^ 0 - ^
12! C = 1 n = l
/A 0 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 students left to assign. Let's ^
.isMfjh stuo'jt;!'. to 'hu room with 4 beds n e x t Let B iCo, i C i = 1, 1
-111 tlir 1 IMI ll KM of ways to assign the ., ^ -^^ n ..... ...^AL^
7 s t H d e n 1 s h - i ! H ; 4 h r d s : '"^ ^ ° 0 ! ( 2 ^ 0 ) ! ' ' " " " t i l - i ) !
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!
Now there are 7 - 4 or 3 students left to assign to the
room with 3 beds. Since all of these students will be A =
assigned to the room, there is only one combination
for them. Let C now represent the number of different 2 M = '
assignments: 2C0, 2C1, 2C2= 1, 2, 1
C = 792 . 35 0 1 ( 3 ^ 0 ) ! ^"^^ ^ 1 ! ( 3 ^ l ) !
C = 27 720 01 31
There are 27 720 w a y s the 12 students can be C = 3 = '
assigned to these rooms.
0!-3! ' ' 1!-2!
13. a | i ) 5 objects, 3 in each combination s M ^ 3^1 ,|,2i
ii) 10 objects, 2 in each combination ^ ^ .| 3
iii) 5 objects, 3 in each combination ^ ° a ^ M
b) e.g., i) How many w a y s can you choose 3 coins ^ ^ o
from a bag containing a penny, a nickel, a dime, a 3 ^ ^ ^
guarter, and a loonie2
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., The numbers on the left and right sides are all
1s; every other number is the s u m of the two numbers
above it.
d) sixth row: 1, 5, 10, 10, 5,1
seventh row: 1, 6, 15, 20, 15, 6, 1
e) e.g., The number in each square of Pascal's
Triangle is equal to the number of pathways to it from
the top square.
« a l The equation I need to solve is
nl
2 ! ( o ^ 2 ) ! '
II • ' 4 j D n - 2 > 0
n > 2
15 IS defined for n > 2, whore n e N.
15
2 ! ( n ^ 2 ) !
nl
V / - r;;
nl
; c - M ) l
nl
nl
= 15
= 1 5 - 2 !
= 15(2)
= 30
( n ^ 2 ) !
n(n~^i){n^2,y J i j i 2 n i .
(n^2]{„A] n}(2){f'
n ( i i - 1 ) = 30
n " - / i = 30
(o + 5 ) ( o - 6 ) = 0
11 + 5 = 0 OR n - 6 = 0
n = - 5 n = 6
Based on the restrictions, n = -5 cannot be a solution.
Therefore, n = 6 is the solution to the equation.
F o y n d a t i o n s o f M a t h e m a t i c s 12 S o l u t i o n s M a n y a !
4^25

b) T h e equation I need to solve is
n!
4 ! i p 4)1
= 3 5 .
I) ANI ^ n 4 f
An,! .}
3:3 c, dufiiied for n > 4. where n e N.
35
35 41
840
840
840
840
.,(.•,. mn ?){n i)(n-4i(n^5).„{3)(2)m
4in -51 . ( i l p f l )
M n - l ) ( n - 2 ! i r / 3)
n* - 5n'' + 6n' - r f + bit - 6 n = 840
„ M +11ll" ^ 6 f i - 8 4 0 = 0
W n t e out all of the factors of - 8 4 0 ; ± 1 , ±2, ±3. ±4. ±5,
±6. ±7, ±8. ±10. ±12, ±14, ±15. ±20, ± 2 1 , ±24. ±28,
±30, ±35, ±40, ±42. ±56, ±60, ±70, ±84, ±105, ±120,
±140, ±168. ±210. ±280. ±420. ±840. These are all
the possible roots of the equation. Substitute them
into the equation and if the equation goes to 0, then I
have a root of the equation. By trial and error, the
roots are -4 and 7. T h e other roots are not real. Since
4 is out of the domain, the only real solution is n = 7.
c) The egu.ition I need to solve 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
where n e N
3!(n i 2 3)!
is defined for n > 2,
_^ !
"/J i
2!(n 2'i ' " 'AH,; :> 'iH
/2 i n ^ 2)1
2(f7 21' 6 ( » ^ 2 31'
2n! «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 7t 0
n 2 0 or n - 7 0
n 2 11 = 7
Both the roots are within the domain, so there are two
solutions, n = 2 and n = 7.
6!
d) The equation I need to solve is
i ( 6 - r ) !
15 .
r > 0 A N D 6
6!
15 is defined for 0 < r < 6, where r e I.
15
r ! ( 6 - r ) ! =
r ! ( 6 - ^ f ) !
r ! ( 6 ^ r ) !
6!
6!
15
720
15
r ! ( 6 - f ) ! = 48
By substituting each of the integers r for 0 < r < 6, I
get r = 2 or r = 4.
16. a) 1, e.g., the player can only win if the six
numbers they choose are the same and in the same
order as the six numbers drawn.
66!
b)
6 1 ( 6 6 - 6 ) !
66!
6160!
6 6 - 6 5 - 6 4 - 6 3 - 6 2 - 6 1 - 6 0 !
6 - 5 - 4 - 3 - 2 - 1 - 6 0 !
66 65 64 63 62 61
6 - 5 - 4 - 3 - 2 - 1
ggCe = 11-13-16-21-31-61
9 0 8 5 8 7 6 8
There are 90 858 768 different w a y s the player can
win.
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 „ N o , E v e n if e v e r y o n e in t h e c i t y p l a y s , it is
very u n l i k e l y that anyone will win s i n c e each p l a y e r
•>-'v I !•! 'K, / 6 7 c h a n c e of w i n n i n g .
' *' ' t} I .':«• !i,in,in-: ;:t s i d e s in a p o l y g o n is e q u a l t o
' • • • » ! : l -. f i ttie n u m b e r of v e r t i c e s = rt
' • " . , (,!• n •..( 'u / •.' . . d i a g o n a l is f o r m e d b y e l i n e
'-ni:i';r.: nu hiiq .> v e r t e x t h a t is not d i r e c t l y
sic.id.-t li.ti^ Uu u u m b e r of v e r t i c e s t h a t w i l l m a k e
. H j o t , , j i vvifr . II. in ( < - s i d e d p o l y g o n is n - 2.
t ryiug i - lm „ W >Nii the v a l u e s from t h e p o l y g o n s
o n t h e s i d e of t h e t e x t b o o k p a g e , t h e r e is a p a t t e r n ;
(d = n u m b e r of d i a g o n a l s )
. a =
I 6 I 2 I 15 I 9
• U4H < 15 fl /» -i
Rearranging, d = „C2- n. Thus, the n u m b e r of
diagonals for an n-sided polygon can be determined
using 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 giris
C C
" 5 ° 5!2! 0113!
, C , . „ C „ = 2 1 . 1
, q . „ C „ = 2 1
N u m b e r of groups = 6006 + 2730 + 455 + 2 1
N u m b e r of groups = 9212
There are 9212 different groups of 5 students with at
least 2 boys to choose from.
b) N u m b e r of groups with no conditions;
'° 5!-15!
C a s e 1 : 1 boy and 4 girls
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 groups with at least two boys;
2 0 C 5 - 7 C 1 • 1 3 C 4 - 7Co - 1 3 C 5 = 9 2 1 2
There are 9 2 1 2 different groups of 5 students with at
least 2 boys to choose from.
c) e.g.. I prefer indirect reasoning because fewer
calculations are needed.
19. a ) e.g.. Combinations and permutations both
involve choosing objects from a group. For
permutations, order matters. For combinations, order
does not matter. For example, a be and bac are
different permutations, but the s a m e combination.
b) Divide nPr by rf to get „Cr. For example. e C = i
and oP.i 3 6 0 ; 15
20. First, determine the total n u m b e r of o u t c o m e s
possible 111 a s s u m e that once a song is selected, it
cannot be selected again. T h e n u m b e r of outcomes, O.
is;
0 = ^
5166!
O , 1 3 0 1 9 9 0 9
a) N u m b e r of times the event could occur;
* ' 5 ! 2 1 !
3,,C, 6 5 7 8 0
Probability (P);
p 6 5 7 8 0
1 3 0 1 9 9 0 9
P = 0.505...%
There is about a 0 . 5 1 % chance that the five songs wil
be from C D 2 and C D 4 .
b) N u m b e r of times the event could occur;
12 14 15 12 18 5 4 4 3 2 0
Probability (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...%
There is about a 4 . 1 8 % chance that one of the five
songs will be from each C D .

c) There is only o n e time where your favourite song
from each of the 5 C D s will be played.
Probability (P):
p = 1 x 1 0 0 %
1 3 0 1 9 9 0 9
P = 0.000008%
There is about a 0 , 0 0 0 0 0 8 % or 1 in 13 0 1 9 9 0 9
chance that your favourite song from each of the
5 C D s will be played.
2 1 . A + rP2 + A
nl nl nl
3!|fi- ^ 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
22. 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)
rt> (/ 1)]' rl[n Jf
(r) + M f ) f i ! + r ( r i ! )
nl{n i 1 t I r )
f)l(n + l )
H i n f l r)l
LS - R S
Therefore. „ + i C r = „C, t .C,
L e s s o n 4 . 7 : Solwlng Coynting Problems,
page 2 8 8
1. a) This situation involves combinations because the
order of the 3 toppings o n the pizza does not matter.
b) This situation involves permutations because the
three spots for the candidates w h o are selected are
all different so order matters.
c) This situation involves permutations because for a
group of 3 numbers, there are different w a y s to roll
those three numbers because of the different colours
of the dice,
d) This situation involves combinations because the
5 children w h o are selected are all in the s a m e
position. N o information is stated in the question
about positions the children m a y play, s o I c a n only
assume that they are not playing in specific positions,
2, e.g., Situation A involves combinations a n d
situation B involves permutations. For situation A,
order does not matter since the 3 people w h o are
selected will all be considered equals. For situation B,
this is not the case. Each of the 3 people w h o are
selected will have a different position with a different
amount of power and different roles.
3. a) , C =
3!
3!-0!
3 ^ 3 = 1
There is 1 w a y that Maddy can bid on 3 items if she
bids o n only her 3 favourite items.
b) A = ^
^ « ' 3!-5!
There are 56 ways that Maddy c a n bid o n 3 items if
she bids o n any 3 of the 8 items.
13!
( , 3 q f = 2 8 5 6 1
There are 28 561 different four-card hands with o n e
card from 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
There are 304 278 004 800 w a y s that the top five
cash prizes can be awarded if each ticket is not
replaced w h e n drawn.

b) ( 2 0 0 f = 320 000 000 000
There are 320 000 000 000 w a y s that the top five
cash prizes can be awarded if each ticket is replaced
w h e n drawn.
6.
1'./- i)i i'lO
= 180
There are 180 w a y s that the 5 starting positions on
the basketball team can be filled.
10!
7.
2!-2!-2!-2!-2!
10!
y 2 2 2 •
= 1 0 - 9 - 7 - 6 - 5 - 3 ' 2 - 1
113400
2!-2!-2!-2!-2!
_ 10!
2 ! - 2 1 2 1 ^ 2 ! ^ !
There are 113 400 ways that the five different pairs of
identical teddy bears can be arranged,
8. C a s e 1 : 3 flags are used
5!
^5 I h
5 ^
5 4 3 2 1
5 P 3 = 5 - 4 - 3
60
C a s e 2: 4 flags are used
^ ^ ^ ^ ^ ( 5 ^ 4 ) .
5 ^
A 120
5!
' 1!
5!
C a s e 3: 5 flags are used
•P.', = 5!
5P5= 120
Let S represent the number of different signals that
can be sent using at least three of the flags;
5 = 60 + 120 + 120
S = 300
There are 300 different signals that can be sent using
at least three of the flags.
9 e.g., First make a table to show the number of
ways the two cabin cruisers can be arranged next to
each other.
CC 1 CC
'• Af i,uyiMncnt 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 - '
Arr;inc|.,inent 8 ' _ i ; V. " i
: A r r . i n g o m e r i t 9 f, ' " 4
; A r i a n g e m e i i t 10 f. i 7i
For each of these arrangements, the number of w a y s
the SIX boats can dock is the number of ways that the
other four boats can dock. Let D represent the total
m'Tb'jr of ways that the boats can dock;
: 4 '
24
n ..p.
3 240 ways that the six boats can dock.
I i . e.g.. Each row of seats is different, and within a
row, the seats are a s s u m e d to be different. Therefore,
there are 10 different people being seated in
10 different spots. Let A represent the number of
seating arrangements;
A = 10!
71 = 3 628 800
There are 3 628 800 ways that the 10 players can sit
in the van.
1 1 .
' 2!
60
There are 60 different arrangements that are possible
for the letters if there are no conditions.
b) 3! = 6
There are 6 different arrangements that are possible
for the letters if each arrangement must start and end
with an N.
12. e g . W h e n there is an even a m o u n t of numbers,
half of them will be o d d . In this case there are
100 possible numbers that each number can be.
Therefore. ^ , or 50 of them are o d d . Since I want
each number to only be 1 of these 50 odd numbers,
the number of sequences S is;
S = 50 50 • 50
S= 125 000
There are 125 000 completely odd sequences.
13.
11!
5h&.
= 462 You can take 462 different routes.
F o u n d a t i o n s of Mathemati dutions Manual
4-29

14. e.g., Let's assign people to ttie 5-person car first.
Let J represent the nunnber of ways to assign the
people to this car:
5!11!
J = 4368
Now there are 16 - 5 or 11 people left to assign to the
remaining two vehicles. Let's assign people to the
4-person car next. Let K represent the number of
ways to assign the people to this car:
17. e.g.,
K
11!
4!7!
K = 330
Now there are 11 - 4 or 7 people left to assign to the
remaining vehicle. There is only 1 way to assign these
people to the 7-passenger van because all of them
are going to be assigned 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
assigned to the 3 vehicles.
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
iJivici, 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 Boys and 3 Gids C a n Go:
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 40.8% chance 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 has 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)
The 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

Total if A i s ttie first letter;
2 33
Therefore, if the first letter is A, there are 33 possible
arrangements.
If the first letter is L, S. or K, there are three possibilities
for the second letter; A, and 2 of L, S, and K (the ones
that are not the first letter).
If A is the second letter;
3 possibilities for the third letter: A, and 2 of L. S. and K
The A has 3 possibilities for the fourth letter and the
other two letters have 2 . 3 + 2(2) = 7
If A is not the second letter;
/ .sibiiities for the third letter
The A has 2 possibilities for the fourth letter and the
other letter has 1. 2 + 1 = 3
Total for both second letters that are not A;
3(2) = 6
Total for one of three times where first letter is L, S. or
K:
7 + 6 = 1 3
Total w h e n first letter is L, S, or K;
3(13) = 39
Total arrangements:
39 + 33 = 72
Therefore. 72 four-letter arrangements can be made
using all of the letters in the word A L A S K A .
20. If I have an O as the first letter, there are
4 possibilities for the second letter, each of which has
3 possibilities for the third letter. 4(3) = 12
Therefore, there ar^; 1 / uo^..sible arrangements when
O is the first letter.
If the first letter is B, K, or S;
There are 3 possibilities for the second letter; O, and
two of B, K. and S (the ones that are not the first
letter). O has 3 possibilities for the third letter while
the other 2 have 2. 3 + 2(2) = 7
Total if the first letter is B K, or S:
3 ( 7 ) = 21
Total arrangements:
21 + 12 = 33
Therefore, 33 threedetter arrangements can be made
using all of the letters in the word B O O K S .
H i s t o r y C o n n e c t i o n , p a g e 2 9 0
A. Yes. Each number from 0 to 127 is assigned a
different character or symbol on the keyboard. Since
the numbers already have an established order, the
characters and symbols assigned to these numbers
do, as well.
B. Yes. Each number in ASCII (pronounced "askey")
must be converted into a stnng of Os and I s to create
the binary code, so order matters. Each 0 or 1 is
associated with a position in the stnng. A different
permutation of Os and I s represents a different
number in the ASCII code system.
C. There are 128 numbers in ASCII that must be
represented by a string of Os and I s . You need to
determine the length of the stnng needed to create
128 different arrangements of Os and I s . You can
begin by thinking about a stnng of length of 5.
A b o x d i a g r a r ' j . , f ;an help you
determine the .,i..:it)fr <-''AS(,i| numbers you can
represent.
Within each box you can place a 0 or a 1. There are
two choices for each box. since repetition of Os and
I s is allowed. So for a stnng length of 5, there are
2 - 2 - 2 • 2 - 2 = 2^ or 32 ASCII numbers that can be
represented. Obviously, the stnng must be longer for
128 numbers. If n represents the string length, and
128 numbers must be represented, then 2" = 128. By
trial and error, n = 7.
A binary stnng of length 7 is needed to represent
each ASCII code.
C h a p t e r S e l f - T e s t , p a g e 291
1 nl Let N represent the number of different serial
numbers:
IV = 26 - 26 • 10 - 10 - 10 - 3
IV = 2 028 000
There are 2 028 000 different serial numbers
possible, if repetition of characters is allowed,
b) Let N represent the number of different senal
numbers:
IV = 2 5 - 2 4 - 1 0 - 9 - 8 - 3
IV = 1 296 000
There are 1 296 000 different senal numbers
possible, if no repetition is allowed.
2. Event A: Drawing a spade
Event B: Drawing a diamond
n(A ' ' m = iiiA) + n(B)
niA H) - 13 + 13
ni'A •• li) ^- ?u
Therefore, there are 26 ways to draw 1 card that is a
spade or a diamond.
3. a) n + 9 > 0
/ 7 > - 9
(n + 10)(n + 9)! is defined for n > - 9 . where n e I.
(n + 10)(n + 9)! = (n + 10)[(n + 9)(n + 8)...(3)(2)(1)]
(n + 10)(n + 9)1 = (n + 10)(n + 9)(n + 8)...(3)(2)(1)
( n + 10)(n + 9)! = ( n + 10)!
b) n - - 2 > 0 A N D n > 0
n > 2
(n^2)
^—A is defined for n > 2, where n c I.
nl
( n - 2 ) ! ( n - 2 ) ( „ ~ 3 ) . . . ( 3 ) ( 2 ) ( l )
n' ^ ; H „ - 1 ) ( „ - 2 ) ( „ - 3 ) . . . ( 3 ) ( 2 ) ( I )
nl n ( n - l )
nl /•?'' - n
F o u n d a t i o n s of Mathematics 12 Solutions Manual 4-31

4 a) hi - 1/0
rhnn:foro, there are 120 different ways that the 5 cars
t.aii Ix; paikod side by side.
b) I el B rofJHjsent the number of arrangements;
B ./f ./',
P ~ 2» 4'
f i - 2 1 4 3 2 1
[i - 48
1 h(;r(4<)fu, there are 48 different ways the cars can be
parked so the 2 black cars are next to each other.
There are 126 different four-book selections that can
be m a d e
( 9 T 4 ) !
0'
fd
9 8 f o b
• - V
j -M a / h
There are 3024 different four-book selections can be
arranged in order of preference.
c) e.g., The order matters in part b). There are still
126 ways to choose the four books from the nine
options, but there are also 4! = 24 ways to arrange the
books,(126 • 24 = 3024)
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 undefined
841
84
2 ! ( - 4 2)!
Check n = 9
LS
9!
5!
u y, ( h 5!
5!
9 8 7 6
3024
RS
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
6! 8!
, C , - , C 3 = 8 4 0
There are 840 different ways that a 5-person
committee can be selected if there must be 2 boys
and 3 giris.
b) C a s e 1: 2 boys and 3 giris;
C (• - ^ '
6 2 n '3 2!4I 3!5!
ApAS-^Q
C a s e 2: 3 boys and 2 giris:
« « 2 3,3, 216!
, C 3 . , C , = 560
C a s e 3: 4 boys and 1 giri;
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 represent the number of 5-person committees
with at least 2 boys:
C = 840 + 560 + 120 + 6
C= 1526
There are 1526 different ways that a 5-person committee
can be selected if there must be at least 2 boys.
^UA-'''
A,
3!9!
220
4-32 Chapter 4 ..r^unting M e t h o d s

There are 220 different ways that a 5-person
• .i.'i< ' .11. t . fed if David and Susan must
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 giris
" ° « '"^ 0!6! 5!3!
56
Let C represent the number of 5-person committees
with more girls than boys:
C = 840 + 420 + 56
C = 1316
There are 1316 different w a y s that a 5-person
committee can be selected if there must be more giris
than boys,
8. - ^ = 30
2!2!
There are 30 different arrangements of the letters in
the word 1 r f 111
9. 5! • 4! = 2880
There are 2880 different arrangements possible.
C h a p t e r R e v i e w , p a g e 2 9 3
1. e.g., The Fundamental Counting Pnnciple is used
w h e n a counting problem has different tasks related
by the word A N D , For example, you can use it to
figure out how many ways you can roll a 3 with a die
and draw a red card from 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 number of sets of answers:
, 4 - 4 • 4 • 4 • 4
A = 4^°
/•. i '.;!., c / c
" i - •.In.'.,: I '.•)..live 1 048 576 different sets of
answers.
4 a| /. .• u A N D n > 0
, '- y'f ! • ''ufined for n > 0, where n c I.
in^ 2)1
^ = 20
(fi)(ii-^4)..,(.iil2)(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
The root n =• 6 is outside the restnctions on the
variable in the equation, so it cannot be a solution.
b) The simplified version of the eguation is
n + 1>0 A N D n - 1 > 0
n > - 1 n > 1
In , 1)1
132 IS defined for n > 1, where 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
The root n = - 1 2 is outside the restnctions on the
vanable in the eguation, so it cannot be a solution.
There is one solution, n = 11,
The tree diagram shows there are 8 possible ways
that the three coins can land.
F o u n d a t i o n s of Mathematic •-'. oiutions Manual 4-33

5. e.g., 6^6 has a larger value, e.g., I know because
^ is the factonat expression for the permutation
6!
expression 8^2. Here, I have more objects than for
ePe, but I a m not using all of them. This leads to fewer
possible arrangements, or in other words, a lower
8!
value for
6!
6. Let O represent the number of orders:
0 = 12!
O = 4 7 9 001 600
There are 4 7 9 001 600 different orders in which the
singers could perform the 12 songs.
7. , , P , = 25!
^5 3 22!
25P3 = 13800
There are 13 800 different w a y s a director of
education, a superintendent of curriculum, and a
supenntendent of finance can be selected.
b) Let A mnrf^cppf ^hn number of arrangements:
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!
There are 453 600 different arrangements that are
possible if all the letters are used, but each
arrangement must begin with the C.
1 1 . a)
14'
2 5 2 2 5 2 0
2 ! 3 ! 4 ! 5 !
I IKJHJ are 2 522 520 different w a y s Tina can stack the
blocks in a single tower if there are no conditions.
b) 2 7 7 2 0
3!4!5!
There are 27 720 different w a y s Tina can stack the
blocks in a single tower if there must be a yellow
block at the bottom of the tower and a yellow block at
the top.
25P^o=11 861 676 288 000
2 5 P „ - 1.18rK..x10"
There are 11 861 676 288 000 or about 1.2 x 1 0 "
different w a y s the test can be created if there are no
conditions.
23Pg = 19 769 460 480
23Pg=1.976...x10'°
There are 19 769 4 6 0 480 or about 2.0 x 10^°
different w a y s the test can be created if the easiest
guestion of the 25 is always first and the most difficult
guestion is always last.
9 p . 5 2 !
.^P^ 311875200
There are 311 875 200 different five-card
arrangements possible.
11! 1 1 - 1 0 - 9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1
2!2!2!
11!
2I2I2!
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!
There are 4 989 600 different arrangements that are
possible if all the letters are used.
12.
™ 5 5!5!
ioCs = 252
C
• ^ ^ 7 ! 4 !
= 3 3 0
^5 2 2113!
A 105
Therefore, nd results in the greatest value.
13. C, =•
20!
4116!
20 = 4 8 4 5
There are 4 8 4 5 different selections of 4 books that
Ruth can choose.
14. No. e g., Each combination can be arranged in
many different w a y s to m a k e a permutation, so there
are more permutations than combinations
15. a) „ C =
19!
4!15!
,,C^ 3876
There are 3876 different w a y s that a committee of
4 people can be chosen if there are no conditions.
9! 10!
2 ! 7 ! ' 2 ! 8 !
.C =36-45
b) A ,C =
c. 1620
4 people can be chosen if there must be an equal
number of m e n and w o m e n on the committee.
' '° ' 4 ! 6 !
. „ C , 210
4 people can be chosen if no m e n can be on the
committee.
4-34 Chapter 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
assign teachers to the first group of 5:
5!10!
A = 3003
N o w there are 15 - 5 or 10 teachers left to assign.
Lci f; ir-ptesent the number of w a y s to assign the
remaining teachers to the second group of 5;
B
•. rs left to assign to
1 w a y that this can
I number of w a y s to
10!
5!5!
8 252
N o w there ' i-
the last grot f i '- ' h. i-.-
be done. I • ' -i- -i !
assign the t
T=A-B
T= 3003 252
T = 756 756
There are 756 756 different w a y s 15 teachers can be
divided into 3 groups of 5.
17. c q.. The first point can be joined with 11 more
[soinO- to form straight lines. T h e second point can
then be joined with 10 more points to form straight
lines (since it w a s already joined with the first point).
T h e third point can be joined with 9 more points to
form straight lines (since it w a s already joined with the
first two points). This pattern continues on until I get to
the seconddast point that can only be joined with the
last point (since it w a s already joined with the other
10 points). T h e last point cannot be joined any further
since It IS already joined to every other point in the
circle. Using this observed pattern. I can calculate the
n u m b e r of straight lines (L):
/. = 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1
1 = 66
There are 66 different w a y s the points can be joined
to form straight lines.
18. a) Since there is one more boy than there are
giris, the line must always follow this pattern:
B G B G B G B G B G B G B . T h u s the boys are arranged in
7 positions, and the giris in 6 positions.
7! 6! = 3 628 800
There are 3 628 8 0 0 w a y s in which the children can
be arranged in one row if the boys and giris must
alternate positions.
b) Group the tnplets as one. There are 3! w a y s in
which the tnplets can arrange themselves. Let B
represent the number of different arrangements:
e = 1 1 ! 3!
6 = 39 916 800 6
B = 239 500 800
There are 239 500 800 w a y s in which the children can
be arranged in one row if the triplets must stand next
to each other.
19. C a s f 1 face cards and 3 non-face cards: 12C2 • 40
C a s e 2: 3 face cards and 2 non face cards: 12C3 • 40C2
C a s e 3: 4 face cards and 1 non-face card: 12C4 - 40C1
C a s e 4: 5 face cards and 0 non-face cards: 12C5 • 40Co
Let H represent the number of hands with at least 2
t ^. cards:
I'lC'z • 40C3 + 12C3 • 40C2 + 12C4 • 4oCi + 12C5 - 40Co
/ ' 36 • 9880 + 2 2 0 • 780 + 4 9 5 • 40 + 792 1
544 272
! Li-.e are 844 2 7 2 different five-card hands with at
least two face cards.
C h a p t e r T a s k , p a g e 2 9 5
A. Combinations. T h e order in which the dice are
tossed does not matter (note that players toss all
8 dice simultaneously) nor does the w a y the dice are
arranged w h e n they land matter. W h a t is important is
the o u t c o m e of each toss a combination of number
of dice that land with the s a m e side up A N D n u m b e r
of dice that land with a different side up, for example,
7 dice land with the s a m e face up A N D 1 die with the
opposite face up.
B. Each o u t c o m e can happen two w a y s . For example.
8 with the s a m e side up could occur as 8 of the
unmarked sides face up or 8 of the marked sides face
up. That is w h y each calculation is the s u m of two
combination values:
8 dice land with the s a m e side up
^ f
+ or 1 + 1 or 2
i f f
, 1
r i . ^
1 i ^ l 7
or 8 + 8 or 16
or 28 + 28 or 56
8V21^ 8 ¥ 2
^6)l2j''"i6ji2^
or 56 + 56 or 112
8 ¥ 4 '
+
I 4 J I
or 70 + 70 or 140
or 56 + 56 or 112

C . Yes, i think it is fair.
O u t c o m e Points Number of W a y s
O u t c o m e C a n
O c c u r
8 dice land s a m e
side up
10 2
7 dice land s a m e
side up
4 16
6 dice land same
side up
2 56
5, 4, or 3 dice
land same side up
0 364
T h e highest number of beans (10) is awarded for the
outcome that can happen in the least number of
ways, 8 dice landing s a m e side up; 4 beans are
awarded for the o u t c o m e that can happen in the
second fewest number of ways, 7 dice landing s a m e
side up; 2 beans are a w a r d e d for the outcome that
can happen in the third fewest number of ways, 6 dice
landing same side up. The most likely outcomes of 5,
4, and 3 dice landing s a m e side up ail receive the
lowest number of beans (0). So the point system
rewards the least likely outcomes with the most beans
and the most likely outcomes with the fewest beans.
C h a p t e r 4 D i a g n o s t i c T e s t , T R p a g e 2 6 9
1.a)
Coin
heads
tails
30 i*o>u^b«« OutoEMnet
heads and red
heads and orange
heads and purple
heads and yeUa<H
heads and green
tails irdi red
tails and orange
tails and purple
tails andyellOK^f
tails and greengf een
b) b) e.g.. By looking at the tree diagram, there is one
way he could flip a head and spin green, and ten
possible outcomes.
P(heads and green) = ^
10
or 10%.
2. a)
Child 1 Child 2 Child 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) There are wo w a y s all three children will be the
same gender, either all boys or all gids, and there are
eight possible outcomes.
2
P(all boys or all gids) = - or 2 5 % chance, assuming
8
that having a boy or gid is egually likely.
3. a) B' = {the set of elements not in S}
B' = {a, b, c, d, e, i, o, u}
b) Au B = {the set of elements in A and B}
Au B = {a, b, c, d, e, i, o, u, x, y, z}
c) ArB = {the set of elements in both A and B}
AnB = {y}

5. e.g., Let x be the number of good singers without
dancing skills. Let A be the set of singers and B the
set of dancers.
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 diagram to help solve the problem.
c) ii) the intersection of sets A and S
atuditioning gsr Is
sfericm'^^ singers ^
10 + 6 + x = 2 4
1 6 + x = 2 4
X = 8
There were 8 gids who were good singers but not
good 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) disjoint sets A and B
d) i) tree diagram
first coin Mcofid eofci
e) iv) outcome table
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

7. e.g., / is the universal set of integers. E is the
subset of even integers. O is the subset of odd
integers.
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 represent the universal set of all students.
Let A represent the students w h o attended the first
school dance, and let 8 represent students w h o
attended the second school dance. T h e n niA u 8 ) is
the number of students w h o went to one of the first
two school dances.
niA uB) = niA) + niB) - niA n 8 )
niA) = 420, niB) = 480, niA nB) = 285
niA u 8 ) = 4 2 0 + 4 8 0 - 285 = 615
615 students went to one of the first two school
dances of the year.
b) niA u 8 ) ' is the number of students w h o did not
attend either dance.
niA u 8 ) ' = S - niA u 8 )
S = 1200, n { A u 8 ) = 615
n ( / u 8 ) ' = 1 2 0 0 - 6 1 5 = 585
585 students did not attend either dance.
5. a) B = {set of black face cards in a standard deck of
playing cards} = { J * , Q * , K * , J * , Q A , K * }
b) D = {set of different three-digit numbers using the
digits 1, 3, and 5} = {135, 153, 315, 3 5 1 , 513, 531}
c) S = {set of all possible sums w h e n a pair of dice is
rolled} = {2, 3 , 4 , 5, 6, 7, 8, 9, 10, 1 1 , 12}
d) T = {set of all the days of the w e e k with names that
begin with T} = {Tuesday, Thursday}
6. a) { } , {red}, {blue}, {red, blue}
b) { } . {2}, {4}, {6}, {8}, {2, 4}, {2, 6}, {2, 8}, {4, 6},
{4, 8}, {6, 8}, {2, 4, 6}, {2, 4, 8}, {4, 6, 8}, {2, 4, 6, 8}
c) { } , {Apnl}, {May}, {June}, {AphI, May}, {Apnl, June},
{May, June}, {Apnl, May, June}
d) { } , {100}
8. S = {l, N , T , E, R, S, C O }
a) A = {set of vowels in S}
A = {I, E, 0 }
b) 8 = {set of letters in S E C T I O N }
8 = {S, E, C, T, I, O, N}
c) A u 8
= {set of vowels in S and set of letters in S E C T I O N }
= {S, E, C, T, I, O, N}
d) A 8 = {elements in both A and 8 }
A n 8 = {t, E, 0 }
e) A' u 8 ' = {elements not in A and elements not in 8 }
A'uB'= { N , T, R, S, C, E, I, 0 }
f) A ' n 8 ' = {elements in neither A nor 8 } = { N , T, R}
9. a ) S = { 1 , 2 , 3, 4, 5, 6, 7, 8, 9, 10, 1 1 , 12, 13, 14}
A = {2, 4, 6, 8, 10}
8 = { 1 , 3 , 5, 7, 9, 1 1 , 13}
A' = {the elements of S not in A}
A ' = { 1 , 3, 5, 7, 9, 1 1 , 12, 13, 14}
8 ' = {the elements of S not in 8 } = {2, 4, 6, 8, 10, 12, 14}
S

A u B = {the elements of A and B}
= { 1 , 2 , 3 , 4 , 5, 6, 7, 8, 9, 10, 1 1 , 13}
A n B = {the elements in both A and 6} = { }
b) S = { A * , 2 * , 3 * , 4 A , 5 A , 6 A , 7 A , 8 A , 9 A , 1 0 A ,
A * , 2 * , 3 * , 4 A , 5 * , 6 * , 7 * , 8 * , 9 * , 1 0 * ,
A v , 2 ¥ , 3 ¥ , 4 ¥ , 5 ¥ , 6 ¥ , 7 ¥ , 8 ¥ , 9 ¥ , 1 0 ¥ ,
A * , 2 * . 3 4 , 4 * , 5 * , 6 » , 7 * , 8 * , 9 * , 1 0 * }
A = { 3 A , 6 A , 9 A , 3 A , 6 * , 9 * , 3 ¥ , 6 ¥ , 9 ¥ , 3 * , 6 * , 9 * }
e = { 2 A , 4 A , 6 A , 8 A , 1 0 A , 2 * , 4 * , 6 * , 8 * , 1 0 * ,
2 ¥ , 4 ¥ , 6 ¥ , 8 ¥ , 1 0 ¥ , 2 * , 4 » , 6 » , 8 » , 1 0 * }
A ' = {the elements of S not in A}
A' = { A A , 2 A , 4 A , 5 A , 7 A , 8 A , 1 0 A , A A , 2 * . 4 * ,
5 * , 7 * , 8 * , 1 0 A , A ¥ , 2 ¥ , 4 ¥ , 5 ¥ , 7 ¥ , 8 ¥ ,
1 0 ¥ , A * , 2 * , 4 * , 5 * , 7 * , 8 » , 1 0 * }
A'
e ' = {the elements of S not in B} = { A A , 3 A , 5 A , 7 A ,
9 A , A A , 3 A , 5 A , 7 A , 9 A , A ¥ , 3 ¥ , 5 ¥ , 7 ¥ , 9 ¥ ,
A * , 3 * , 54, 7 * , 9 * }
S
A u B = {the elements of A and S} = { 2 A , 3 A , 4 A , 6 A ,
8 A , 9 A , 1 0 A , 2 A , 3 A , 4 A , 6 A , 8 A , 9 A , 1 0 A ,
2 ¥ , 3 ¥ , 4 ¥ , 6 ¥ , 8 ¥ , 9 ¥ , 1 0 ¥ . 2 * , 3 * , 4 * ,
6 * , 8 * , 9 * , 1 0 * }
A n B = {the elements in both A and S}
A n B = { 6 A , 6 A , 6 ¥ , 6 * }
s
/
I
A
i
{ B )
/
10. e.g.,
a) S is the universal set of all students in my math class.
A is the subset of students w h o are taller than 6 ft.
6 is the subset of all students with black hair.
A n B (shaded) is the subset of students w h o are
taller than 6 ft and have black hair.
b) S is the universal set of all cards in a standard
deck of playing cards.
A is the subset of all face cards.
B is the subset of all red cards.
Au B (shaded) is the subset of all face cards and all
red cards.
11. Let S represent the universal set of all first-year
students.
Let C represent the students w h o take calculus.
Let A represent the students w h o take algebra. Then
n(C u A)' will be the number of first-year students
w h o take neither calculus nor algebra.
n{S) = 200
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
There are 75 first-year students w h o take neither
calculus nor algebra.

Chapter 4 Test, TR page 282
1. e.g., T h e table s h o w s the possible wins and losses
of one of the teams. S h a d e d cells indicate g a m e s that
would not actually be played, since o n e of the t e a m s
will have already w o n or lost two g a m e s . That m e a n s
there are only six different outcomes.
G a m e 1 G a m e 2 G a m e 3 O u t c o m e 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) There are bNO different w a y s a t e a m can lose
exactly one g a m e but still win the championship:
W L W or L W W .
2. a) T h e sandwich has four elements: egg salad or
chicken salad, lettuce or tomato, butter or
mayonnaise, and whole w h e a t bun or w h o l e grain
bagel or s e s a m e seed bun. Using the Fundamental
Counting Pnnciple:
2 2 2 3 = 24 w a y s to m a k e a sandwich. I a s s u m e d
that the sandwich has exactly one item from each of
set of choices.
b) For each of the four digits, there are six options
and repetition is allowed. Using the Fundamental
Counting Pnnciple:
6 • 6 • 6 6 = 1296 w a y s to m a k e a password. I
a s s u m e d starting the password with a 0 w a s allowed.
c) There are 13 hearts and 13 clubs in a standard
deck of 52 cards, and the sets of hearts and clubs are
disjoint. There are 13 + 13 = 26 w a y s to draw a heart
or club. I m a d e no assumptions.
d) There are five letters in T E E T H with both T and E
repeating twice. There are = 30 arrangements of
the letters. I m a d e no assumptions.
e) There are 25 different toppings, and Jim must
choose 3. Order does not matter. Jim can order
25
2300 different pizzas. I a s s u m e d he would
choose exactly three toppings and each would be
different.
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
^ 30 240
120
= 252
3. a) Since order does not matter:
' 1 0 1 10!
. 5
10
5
noi
5
10
5
10
5^
There are 252 selections that can be made.
b) Since order matters:
'° ' ( 1 0 - 5 ) !
1 0 - 9 - 8 - 7 6 5 !
10^5 = 1 0 . 9 . 8 - 7 . 6
1 0 ^ - 3 0 240
There are 30 240 selections if the selections are
ordered by preference.
c) e.g., Order does not matter in part a) but it does
matter in part b), so part a) involves combinations,
whereas part b) involves permutations.
d) e.g.. The answer to part a) is 5! or 120 times
smaller than the answer to part b). This is because
the 30 240 five-novel selections from part b) must be
divided by 5! to eliminate combinations that are the
same, because order does not matter.
4. To solve the problem, look at the walk in three
sections. In the first section, M a h a has to walk three
blocks east and three blocks south for a total of six
blocks; in the second section she has to walk one
block east and one block south for a total of two
blocks; in the third section she has to walk two blocks
east and hwo blocks south for a total of four blocks.
First section:
6! _ 6 - 5 4 3!
3 ! 3 ! " " 3 ! 3 - 2 1
6! _ 6 - 5 4
3 ! 3 ! ~
6!
3!3!
6!
3!3!
Second section:
1 ! 1 ! ^ 1-1
_2!^
1!1!
3 2 1
5 4
20
= 2

2 3
21
2!
2!2!
2!2!
Using the Fundamental Counting Principle:
20 • 2 • 6 = 2 4 0
She can walk 2 4 0 different v/ays.
This problem can also be solved using a pathway
diagram;
5. a) Order does not matter, so this is a combination
problem. There are (5 2) ways to choose t w o boys
from five, a n d (6 2) ways to choose two giris from six.
2 i! / I
f
f, ,
f ,
• 4 - 3 ! 6-5--
'••2 'th'
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 ic f: t. 1 •;•>.!• committee, there
c i c nine fl. f.pl- ioi ttjifidiimig t w o positions:
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
There 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 + 5^1
-1
60 + 5
There are 65 committees that can be made with more
boys than giris.

B. a) 001, 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. if you pluf.o t h r f and K as required. Ihrs uan
li.;jp[)on only O O P w a y for oaf:h position. Fhu
romaininp jsovon letters can then t)o arranged in
between, keeping in mind that there are repealed
IHtom- two 72s, two S's. and two O's-
1.1 " -iO''hlV"
. 2I2'2! I 212-2
l^2'2'2'; ^ 2 2
U!2!2!j i 4
1,1^1= 630
l 2 ! 2 ! 2 ! j
There are 6 3 0 different w a y s to arrange the letters
with the given conditions
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

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