250 Answers NEL

iii) If point (0, 6) is not on a graph, then the equation of the graph is not y 5 3x 2 1 6. True.

iv) If the equation is not y 5 3x 2 1 6, then point (0, 6) is not on the graph. False. Point (0, 6) is on graphs with different equations, such as y 5 2x 1 6.

7. D. 8. e.g., The inverse is false, because its truth is always the same as that of the

converse. The contrapositive is true because its truth is always the same as that of the conditional statement.

Chapter 3 Test Prep, page 84

Q1: • For a set and its complement, n(A) 1 n(A9) 5 n(U ). • For two disjoint sets, n(A < B) 5 n(A) 1 n(B) and n(A > B) 5 0 • For two non-disjoint sets, the Principle of Inclusion and Exclusion states

n(A < B) 5 n(A) 1 n(B) 2 n(A > B)or n(A < B) 5 n(A \ B) 1 n(B \ A) 1 n(A > B)

• For three non-disjoint sets, the Principle of Inclusion and Exclusion statesn(A < B < C ) 5 n(A) 1 n(B) 1 n(C ) 2 n(A > B) 2 n(A > C ) 2 n(B > C )

1 n(A > B > C )Q2: A conditional statement is false when the hypothesis is true and the

conclusion is false. Otherwise, the conditional statement is true.Q3: A1: • Write the converse by exchanging the hypothesis and conclusion

of the conditional statement. • For the inverse, negate the hypothesis and conclusion of the

conditional. • For the contrapositive, exchange and negate the hypothesis and

conclusion of the conditional. A2: Converse: q 1 p inverse: ¬p 1 ¬q contrapositive: ¬q 1 ¬p

Chapter 3 Test, page 85

1. A. 2. C. 3. B. 4. A. 5. 22 passengers; 13 passengers; 17 passengers 6. 277 7. 125 students; 136 students; 58 students 8. e.g., “Egypt,” “Jordan,” “Pyramids,” “Petra”; by combining two or more

of these terms, Lyn can search for the intersection of web pages related to these terms. For example, “Egypt” and “Petra” are more likely to give useful information than either of these terms on its own, or without the “and.”

9. 7 members 10. 14 people 11. 12 customers 12. e.g., If y 1 6 $ 3, then y $ 23. This is in the form p 1 q, where the

hypothesis, p, is y 1 6 $ 3 and the conclusion, q, is y , 23. The inverse is q 1 p. Inverse: If y $ 23, then y 1 6 $ 3. The converse is ¬p 1 ¬q. Converse: If y 1 6 , 3, then y , 23. The contrapositive is ¬q 1 ¬p. Contrapositive: If y , 23, then y 1 6 , 3. All of these statements are true. The conditional statement is biconditional, so y 1 6 $ 3 if and only if y $ 23.

Chapter 4Getting Started, page 88

1. a) ii) b) iv) c) iii) d) i) 2. a) e.g.,

2 3, 5,...4, 6,...

UE P

0, 1, 9,...

Lesson 3.5, page 76

1. a) p 5 “I am reading”; q 5 “I am studying” b) false; e.g., I might be reading for pleasure, not studying. c) If I am studying, then I am reading; false; e.g., I might be studying

music by practising. 2. a) p 5 “a number ends in 0”; q 5 “a number is divisible by 5” b) true c) If it is fall, then the deciduous trees are changing colour; true 3. a) true b) “If an integer is divisible by 3, then it is divisible by 9”; false;

e.g., 15 is divisible by 3 but not by 9. 4. a) e.g., If a polygon is a square, then it has four equal sides. b) e.g., If a polygon has four equal sides, then it is a square. c) true; false d) no; e.g., The conditional statement is true but the converse is not. 5. A. 6. B. 7. false; e.g., I could pass with less than 75%.

Lesson 3.6, page 80

1. a) 5x 1 37 Z 59 b) The flowers are not red. c) Spring does not follow winter. 2. e.g., no; “I will not do that chore tomorrow.” 3. a) Converse: If C is the same distance from every point on a circle then

C is at the centre of a circle. Inverse: If C is not at the centre of a circle, then C is not the same distance from every point on the circle. Contrapositive: If C is not the same distance from every point on a circle, then C is not at the centre of a circle.

b) Converse: If 3y is divisible by 3, then y is a whole number. Inverse: If y is not a whole number, then 3y is not divisible by 3. Contrapositive: If 3y is not divisible by 3, then y is not a whole number.

4. a) Converse: e.g., If a polygon is a hexagon, then it has six sides. Inverse: e.g., If a polygon does not have six sides, then it is not a hexagon.

b) Yes 5. a) no; 27 is also a possible value for x. b) yes; 72 5 49 c) yes; If x 2 Z 49, then x Z 7. d) no; e.g., If x Z 7, then it is still possible that x 2 5 49. 6. a) i) True. ii) If you are in the Northwest Territories, then you are in

Yellowknife. False. You might be in another community. iii) If you are not in Yellowknife, then you are not in the Northwest

Territories. False. You might be in another community. iv) If you are not in the Northwest Territories, then you are not in

Yellowknife. True. b) i) True. ii) If a cat is female, then it has had kittens. False. A female cat might

not have had kittens. iii) If a cat has not had kittens, then it is not female. False. A female

cat might not have had kittens. iv) If a cat is not female, it has not had kittens. True. c) i) True. ii) If Oleg has a racquet in his hand, then he is playing badminton.

False. He might be playing squash or tennis. iii) If Oleg is not playing badminton, then he does not have a racquet

in his hand. False. He might be playing squash with a racquet. iv) If Oleg does not have a racquet in his hand, then he is not playing

badminton. True. d) i) False. Point (0, 6) is on many graphs, such as y 5 2x 1 6. ii) If the equation of a graph is y 5 3x 2 1 6, then point (0, 6) is on

the graph. True.

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b) e.g., There are 3 ways to choose the colour, AND 2 ways to choose the finish, AND 2 ways to choose the trim. So there are 3 ? 2 ? 2 5 12 choices altogether.

3. a) yes; e.g., choosing the options are separate tasks related by AND. b) no; e.g., the options are related by OR. c) yes; e.g., the choices are separate tasks related by AND. d) no; e.g., what you roll on one die affects what you must roll on the

other. 4. a) 16 choices b) 32 ways c) 18 choices d) 5 ways 5. a) 800 combinations b) 720 combinations 6. a) 30 ways b) 20 ways 7. 23 choices; e.g., The choices are mutually exclusive. 8. B. 9. a) 96 salads b) 24 salads

Lesson 4.2, page 94

1. a) 40 320 b) 5040 2. a) 120 b) 5040 c) 120 d) 420 3. a) n 1 1 b) n 1 12 c) n! d)

1n2 1 n

4. n!

(n 2 2)(n 2 3)!5

n(n 2 1)(n 2 2) . . . (3)(2)(1)(n 2 2)(n 2 3)(n 2 4)(n 2 5) . . . (3)(2)(1)

, or

n( n 2 1 )

So n(n 2 1) 5 20 n2 2 n 5 20 n2 2 n 2 20 5 0 (n 2 5)(n 1 4) 5 0 n 2 5 5 0 or n 1 4 5 0 n 5 5 or n 5 24

Lesson 4.3, page 96

1. a) 60 b) 210 2. a) 120 b) 360 c) 840 3. a) 8P8 5 40 320 b) 8P5 5 6720 c) 8P3 5 336 d) 8P1 5 8 4. a) 20 ways b) 60 ways 5. a) 2 b) 1 c) 22 6. a) 720 exchanges b) 999 exchanges 7. a) 2730 ways b) 3375 ways 8. a) e.g., No; the value of the expression equals

1n 1 1

or at most 12

, which is not a whole number.

b) e.g., No; the expression would have to represent the number of permutations of a set of objects chosen from a smaller set of objects, which is impossible.

9. a) e.g., Case 1: 5 letters and 3 digits; Case 2: 6 letters and 2 digits; Case 3: 7 letters and 1 digit

b) 7200 passwords 10. a) n 5 5; n $ 1 b) r 5 3; 0 # r # 5 11. 9P6; e.g., Both permutations are products of 3 consecutive natural

numbers, but each number is greater for 9P6.

Lesson 4.4, page 100

1. a) 35 b) 210 c) 1120 d) 5040 2. a) 1260 arrangements b) 280 arrangements c) 980 arrangements 3. 15 120 rearrangements 4. a)

11123456789

11111111

b) x 1 y routes c) 165 routes

5. a) 140 routes b) 95 routes 6. a) 17 153 136 ways b) 6 ways 7. a) e.g.,

Number of Objects, n 3 4 5 6 7 8

Number of Arrangements 1 4 20 120 840 6720

e.g., The number of arrangements is multiplied by 4, then by 5, then by 6, etc.

b) e.g.,

1S,1C

2S, 2C,...1H, 1D

WA B

2H, 2D,...

3. a) 16 b) 39 4. a) 467 visitors b) 215 visitors 5. a) J♥, Q♥, K♥, J♦, Q♦, K♦

b) 12, 13, 14, 21, 23, 24, 31, 32, 34, 41, 42, 43 c) 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 30, 36 6. a) {}, {h}, {e}, {s},{h, e}, {h, s}, {e, s},{h, e, s} b) {}, {25} c) {}, {A}, {B}, {C}, {D}, {A, B}, {A, C}, {A, D}, {B, C}, {B, D}, {C, D},

{A, B, C}, {A, B, D},{A, C, D}, {B, C, D}, {A, B, C, D} 7. a) e.g.,

AC

B

b) A and C are not disjoint. 8. a) A9 5 {2, 3, 4, 6, 7, 9, 10, 11, 12, 14, 15}; e.g., B9 is the set of odd

numbers in S; A < B 5 {1, 2, 4, 5, 6, 8, 10, 12, 13, 14, 16}; A > B 5 {8, 16}

b) A9 < B9 5 {1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15}; A9 > B9 5 {3, 7, 9, 11, 15}

c)

AS

B

10 12 14

6428

161351

1511973

Lesson 4.1, page 90

1. a) Black Milk Cream

Sugar

No sugar 6 ways

b) e.g., There are 3 ways to decide on black, milk, or cream, AND 2 ways to decide on sugar or no sugar. So there are 3 ? 2 5 6 ways Bryce can take his coffee.

2. a) e.g.,

mg

m

ic

ic

lgg

m

ic

ic

pgg

m

ic

ic

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252 Answers NEL

7. a) 5C3 b) 4C2, 4C3 c) 5C3 5 4C2 1 4C3 10 5 6 1 4

d) 1 1 1 1 2 1 1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

8. a) n 5 7; n $ 2 b) n 5 11; n $ 2 c) r 5 2 or 6; 0 # r # 8 9. B. 10. a) 34 650 ways b) 369 600 ways

Lesson 4.7, page 110

1. a) permutations; e.g., It is important who gets which position, so the order matters.

b) combinations; e.g., The order in which the flavours are chosen does not matter.

c) permutations; e.g., Even though the 1 symbols are identical and the 3 symbols are identical, the order in which they are selected changes the sequence.

2. a) 100 000 possibilities b) 30 240 possibilities 3. a) permutations; e.g., In this case, the order of the cards matters for the

sequence. b) 14 515 200 sequences 4. a) combinations; e.g., Once the passengers for each vehicle are

determined, the order in which they are seated does not matter. b) Fundamental Counting Principle; e.g., Seating the squad in the

vehicles is a collection of tasks that can be linked by the word AND. c) 1 441 440 ways 5. 1800 collages 6. 546 000 collections 7. 17 rolls 8. 120 ways 9. a) 840 arrangements b) 480 arrangements 10. a) e.g., direct reasoning, as there are fewer cases than with indirect

reasoning b) 1 299 480 hands

Chapter 4 Test Prep, page 114

Q1: • AND often indicates two or more separate tasks. If one task can be performed in a ways and another can be performed in b ways, then by the Fundamental Counting Principle, both tasks can be performed in a ? b ways.

• OR often indicates tasks or sets that may or may not be mutually exclusive.

Q2: • The number of permutations of n different objects is n!. • The number of permutations of r objects chosen from n different

objects is

n Pr 5

n!

(n 2 r)!

Q3: • For a set of n objects of which a are identical, another b are identical, another c are identical, and so on, the number of permutations is given by

n!

a! ? b! ? c! …

Q4: • Use combinations in situations where order does not matter. • The number of combinations of r objects chosen from n different

objects is

n Cr 5 a n

rb 5

n!

r!(n 2 r)!

b) e.g.,Number of Objects, n 6 7 8 9

Number of Arrangements 20 140 1120 10 080

e.g., The number of arrangements is multiplied by 7, then by 8, then by 9, etc.

8. C. 9. D. 10. a) 120 b) 120 c) 912 11. e.g., How many different arrangements are there of the letters A, A, A, B,

B, C, and D?

Lesson 4.5, page 104

1. a) 60 permutations b) 6 ways c) 10 combinations 2. a) 120 permutations b) 1 combination c) e.g., The order does not matter, so the set of all 5 symbols is the only

combination possible. 3. a) abc acb bac bca cab cba

abd adb bad bda dab dbaacd adc cad cda dac dcabcd bdc cbd cdb dbc dcb24 permutations

b) abc abd acd bcd; 4 combinations

c) 6; e.g., For each combination there are 6 ways to arrange the letters, contributing 6 to the total number of permutations.

4. 495 ways

5. a) e.g., 7!

(7 2 4)! b) 4!

c) e.g., 7!

4! 17 2 4 2 !Lesson 4.6, page 106

1. a) s, h, e s, e, h h, s, e h, e, s e, s, h e, h, ss, h, n s, n, h h, s, n h, n, s n, s, h n, h, ss, e, n s, n, e e, s, n e, n, s n, s, e n, e, sh, e, n h, n, e e, h, n e, n, h n, h, e n, e, h

b) s, h, e s, h, n s, e, n h, e, n

c) e.g., The number of permutations is 6 times the number of combinations, because each combination can be arranged in 6 different ways.

2. a) 10 c) 10!

5! ? 5!, or 252

b) 7!

3! ? 4!, or 35 d)

13!

1! ? 12!, or 13

3. a) 1716 ways b) 210 ways 4. a) 210 ways b) 840 ways c) 1050 ways

5. a) n!

(n 2 r)!(n 2 (n 2 r))!

b) e.g., n!

(n 2 r)! ? (n 2 (n 2 r))!5

n!(n 2 r)! ? (n 2 n 1 r)!

n!

(n 2 r)! ? (n 2 (n 2 r))!5

n!(n 2 r)! ? r!

n!

(n 2 r)! ? (n 2 (n 2 r))!5

n!r! ? (n 2 r)!

a nn 2 r

b 5 anrb

6. 10 192 combinations

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2.

3. a) 720 b) 21 c) 336 d) 604 800 e) 10 f ) 113 4. a) 1 2 3 4 5 6

1 0 1 2 3 4 52 1 0 1 2 3 43 2 1 0 1 2 34 3 2 1 0 1 25 4 3 2 1 0 16 5 4 3 2 1 0

b) 29

c) 1

36

d) 1

18 5. a)

b)

c)

6. 90 090 ways 7. 9 979 200 ways

Lesson 5.1, page 120

1. e.g., adding one bracelet of some colour other than blue

2. a) fair, since both have a 3

16 chance of winning

b) not fair, since Sabrina has only a 14

chance of winning

3. A.

Chapter 4 Test, page 115

1. C. 2. B. 3. B. 4. C. 5. B. 6. D. 7. a) 10 000 codes b) 5040 codes 8. a) 10 possibilities b) 16 possibilities 9. a) 720 ways b) 35 ways 10. 30 240 arrangements 11. 5 269 017 601 sequences 12. 30 240 groups 13. a) 21 triangles b) 2 097 152 patterns 14. a)

24 outcomes b) e.g., total number of outcomes

5 outcomes for die ? outcomes for loonie ? outcomes for toonie 5 6 ? 2 ? 2, or 24

c) 20 outcomes d) e.g., number of outcomes

5 outcomes with multiple of 3 1 outcomes with at least one tail 2 outcomes with multiple of 3 and at least one tail

5 8 1 18 2 6, or 20 15. a) e.g., grid 2, because there are two ways to go across the join between

the two parts of the grid b) grid 1: 400 routes; grid 2: 600 routes 16. 1 237 792 hands

Chapter 5Getting Started, page 118

1. a) intersection b) combination c) sample space d) experimental e) Fundamental Counting Principle f ) permutation g) theoretical

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