Chapter 7 Full Solutions - WordPress.com · Web view(iii) Since two dogs are already selected, we...

9 Permutation and Combination

9 Permutation and CombinationReview Exercise 9 (p. 9.3)1. Colour Pattern Possible choices

From the tree diagram, there are 12 possible choices of T-shirts.

2. Boy Girl Possible choices

From the tree diagram, there are 8 possible choices.

3. 1st digit 2nd digit 3rd digit Possible numbers

From the tree diagram, there are 8 possible numbers.

ActivityActivity 9.1 (p. 9.18)1. For the first letter, we can select either A, B, C or D.

∴ The number of choices for the first letter =For the second letter, since one letter has been selected, the number of remaining letters is .∴ The number of choices for the second letter =

1st 2ndNumber of choices 4 3 By the multiplication rule of counting, the number of permutations of 4 different letters taken 2 at a time =

2. The number of choices for the first letter =The number of choices for the second letter =The number of choices for the third letter =By the multiplication rule of counting, the number of permutations of 4 different letters taken 3 at a time =

3. (i)(ii)(iii)

ClassworkClasswork (p. 9.15)(a) 3! = 3 2 1 = 6(b) 6! = 6 5 4 3 2 1 = 720(c) 0! 4! = 1 4 3 2 1 = 24(d) 2! 3! 1! = 2 1 3 2 1 1 = 12

Classwork (p. 9.29)(a) combination (b) permutation(c) combination (d) permutation(e) permutation

Quick PracticeQuick Practice 9.1 (p. 9.5)By the addition rule of counting,the number of possible choices = 4 + 6 = 10

Quick Practice 9.2 (p. 9.6)(a) By the addition rule of counting,

the number of possible ways = 12 + 9 = 21(b) By the addition rule of counting,

the number of possible ways = 10 + 11 + 9 = 30(c) By the addition rule of counting,

the number of possible ways = 12 + 10 + 11 + 9 = 42

Quick Practice 9.3 (p. 9.7)(a) By the addition rule of counting,

the required number of ways = 4 + 12 = 16(b) By the addition rule of counting,

the required number of ways = 12 + 26 – 6 = 32

Quick Practice 9.4 (p. 9.8)By the multiplication rule of counting,the number of possible choices = 4 × 2 = 8

Quick Practice 9.5 (p. 9.9)(a) By the multiplication rule of counting,

the number of possible choices = 3 × 4 = 12(b) By the multiplication rule of counting,

the number of possible choices = 2 × 3 × 2 × 4 = 48

97

NSS Mathematics in Action 5B Full Solutions

Quick Practice 9.6 (p. 9.10)The number of ways to choose one book = 20 + 8 = 28The number of ways to choose two books of different subjects= 20 × 8 = 160By the addition rule of counting,the total number of ways to choose one or two books = 28 + 160 = 188

Quick Practice 9.7 (p. 9.11)(a) The number of favourable choices for each character

= 26 + 26 = 52By the multiplication rule of counting,the number of PINs that contain letters only = 52 × 52 × 52 = 140 608

(b) The number of favourable choices for each character= 52 + 10 = 62The number of possible PINs = 62 × 62 × 62 = 238 328The number of PINs that do not contain any digits= the number of PINs that contain letters only= 140 608The number of PINs that contain at least one digit= 238 328 – 140 608= 97 720

Quick Practice 9.8 (p. 9.16)(a) 7! – 2! = (7 × 6 × 5 × 4 × 3 × 2 × 1) – (2 × 1) = 5038(b) From (a), we have 7! – 2! = 5038, but

(7 – 2)! = 5! = 120∴ a! – b! = (a – b)! is not always true.

Quick Practice 9.9 (p. 9.16)

(a)

(b)

(c)

Quick Practice 9.10 (p. 9.20)(a)

(b)

Quick Practice 9.11 (p. 9.21)(a) The number of permutations is

7! = 5040(b) The number of permutations is

Quick Practice 9.12 (p. 9.22)(a) The number of 3-digit numbers formed = (b) (i) The last digit must be 5.

There are ways to choose the remaining two digits.∴ The number of 3-digit numbers formed that are

multiples of 5 = 12

(ii) The last digit must be 1, 3 or 5. For each of these cases, there will be ways to choose the remaining two digits.By the multiplication rule of counting,the number of 3-digit odd numbers formed= 3 × 12 = 36

Quick Practice 9.13 (p. 9.23)Since the first digit cannot be ‘0’, there are 9 ways to choose this digit.In addition, there are ways to choose the remaining

digits, and ways to choose the letters.Consequently, by the multiplication rule of counting,the number of license plates formed = 9 × 504 × 650 = 2 948 400

Quick Practice 9.14 (p. 9.24)(a) There are 4! = 24 ways of arranging the Chinese books on

the left, and 2! = 2 ways of arranging the English books on the right.By the multiplication rule of counting,the number of possible arrangements = 24 × 2 = 48

(b) We can treat all the English books as one unit. In this case, there are 5! = 120 arrangements. However, the English books can also be arranged in 2! = 2 ways.By the multiplication rule of counting,the number of possible arrangements = 120 × 2 = 240

Quick Practice 9.15 (p. 9.24)Dogs: D D DCats: The number of permutations for the dogs = 3!The number of permutations for the cats =∴ The required number of ways

Alternative SolutionThe total number of permutations without restrictions = (3 + 2)! = 120The number of permutations with the cats standing together = 4! × 2! = 48∴ The number of permutations with no cats standing

together = 120 – 48 = 72


(a)

(b)

(c)


98


The number of combinations is

Quick Practice 9.18 (p. 9.31)The number of combinations is

Quick Practice 9.19 (p. 9.32)(a) The number of ways of picking 3 girls from 12 girls

= The number of ways of picking 2 boys from 10 boys = By the multiplication rule of counting,the number of possible combinations = = 9900

(b) Since 2 particular girls and 1 particular boy cannot be included, there are (12 – 2) = 10 girls and (10 – 1) = 9 boys remained for selection.The number of ways of picking 3 girls from 10 girls= The number of ways of picking 2 boys from 9 boys = By the multiplication rule of counting,the number of possible combinations = = 4320

Quick Practice 9.20 (p. 9.33)(a) The total number of mathematicians and physicists

= 8 + 6 = 14The number of committees formed without restrictions =

(b) Exactly 2 mathematicians and 3 physicists must be included.The number of possible committees =

(c) Number of committees without mathematicians = Number of committees with at least one mathematician = 2002 – 6 = 1996

(d) Number of committees with four physicists = Number of committees with at most three physicists = 2002 – 120 – 6 = 1876

Quick Practice 9.21 (p. 9.34)(a) The girls can be permuted in 4! = 24 ways and the boys in

6! = 720 ways.∴ The number of possible arrangements

= 24 × 720 = 17 280(b) (i) This is the same as a permutation over 10 distinct

positions.∴ The number of possible arrangements

= 10! = 3 628 800(ii) There are ways of selecting a boy to stand in

the same row as the girls. If the girls stand in the front row,the number of permutations of the 5 boys in the back row = 5! = 120the number of permutations of the 4 girls and 1 boy in the front row = 2! × 4! = 48The situation is similar if the girls stand in the back row.∴ The total number of arrangements

= 6 × 2 × 120 × 48 = 69 120

99


Further PracticeFurther Practice (p. 9.7)1. By the addition rule of counting,

the number of choices = 6 + 3 = 9

2. Number of ways of selecting an even number = 50Number of ways of selecting a number ends with 5(i.e. 5, 15, 25, 35 etc.) = 10By the addition rule of counting,the required number of ways = 50 + 10 = 60

3. (a) Number of ways of choosing a child = 3 + 5 = 8Number of ways of choosing a male = 1 + 5 = 6Number of ways in common = 5∴ The required number of ways = 8 + 6 – 5 =

(b) Number of ways of choosing a parent = 1 + 1 = 2Number of ways of choosing a female = 1 + 3 = 4Number of ways in common = 1∴ The required number of ways = 2 + 4 – 1 =

Further Practice (p. 9.11)1. By the multiplication rule of counting,

the number of ways = 3 × 4 = 12

2. By the multiplication rule of counting,the number of choices = 3 × 2 × 5 = 30

3. (a) By the addition rule of counting,the number of choices = 2 + 3 + 5 = 10

(b) By the multiplication rule of counting,the number of choices = 2 × 3 × 5 = 30

(c) The number of ways for taking tennis and volleyball courses = 2 × 3 = 6The number of ways for taking volleyball and squash courses = 3 × 5 = 15The number of ways for taking tennis and squash courses = 2 × 5 = 10By the addition rule of counting,the number of choices = 6 + 15 + 10 = 31

Further Practice (p. 9.25)1. (a) The number of 4-letter strings formed =

(b) The string must begin with A and end with O, or vice versa. ∴ The number of arrangements for the first and the

last letters is 2. Then, 3 letters remains.Number of ways of arranging the 2nd and the 3rd letters = = 6Thus, by the multiplication rule of counting,the number of 4-letter strings formed = 2 × 6 = 12

2. (a) The number of ways = 7! = 5040(b) Number of ways of placing dolls in dresses on the

left = 4! = 24Number of ways of placing dolls in trousers on the right = 3! = 6Hence, by the multiplication rule of counting,the number of arrangements = 24 × 6 = 144

100


3. (a) (i) The number of arrangements = (4 + 5)! = 9! = 362 880

(ii) Number of ways of arranging the first and last singers = Number of ways of arranging the remaining singers = (9 – 2)! = 7! = 5040Hence, by the multiplication rule of counting,the number of arrangements = 12 × 5040 = 60 480

(b) Number of ways of arranging the first singer = 3Number of ways of arranging the remaining singers = (6 – 1)! = 5! = 120Hence, by the multiplication rule of counting,the number of arrangements = 3 × 120 = 360

4. (a) We can treat the red balls as one unit. Then, we have 6! = 720 ways of permutation. In each of these arrangements, there are 3! = 6 ways of arranging the red balls.By the multiplication rule of counting,the number of arrangements = 720 × 6 = 4320

(b) There are 5! = 120 ways of arranging the blue balls.

Each of the red balls may be placed into one of the 6 places as shown above. ∴ Number of permutation =By the multiplication rule of counting,the number of arrangements = 120 × 120 = 14 400

Further Practice (p. 9.34)1. (a) The number of selections =

(b) The numbers 1 to 14 are less than 15.The number of selections

2. There are (18 – 10) = 8 students not wearing glasses.(a) The number of ways =

(b) The number of ways =(c) We first select 3 from students that wear glasses, and

then select 2 from students that do not wear glasses.The required number of ways =

3. Total number of animals in the shop = 5 + 6 = 11(a) (i) The required number of ways =

(ii) The required number of ways = (iii) Since two dogs are already selected, we only

need to select (6 – 2) = 4 more animals from a total of (11 – 2) = 9 animals.The required number of ways =

(iv) The number of ways of choosing 3 dogs =

The number of ways of choosing 4 dogs =

The number of ways of choosing 5 dogs =

∴ The total number of ways of choosing at least 3 dogs = 200 + 75 + 6 = 281

(b) The number of ways of choosing the animals = The number of ways of arranging the chosen animals = (4 + 2)! = 720By the multiplication rule of counting,the required number of ways = 75 × 720 = 54 000

ExerciseExercise 9A (p. 9.12)Level 11. By the addition rule of counting,

the number of possible selections = 12 + 7 = 19

2. By the multiplication rule of counting,the number of possible selections = 8 × 3 = 24

3. By the multiplication rule of counting,the number of possible selections = 5 × 6 = 30

4. By the addition rule of counting,the number of possible selections = 6 + 4 = 10

5. By the addition rule of counting,the number of possible selections = 23 + 28 + 35 = 86

6. By the addition rule of counting,the number of possible selections =12 + 23 + 11 + 25 = 71

7. Tossing a coin can end up either with a head or with a tail.By the multiplication rule of counting,the number of possible outcomes = 2 × 2 × 2 × 2 = 16

8. Throwing a dice can end up with one, two, three, four, five or six.By the multiplication rule of counting,the number of possible outcomes = 6 × 6 × 6 = 216

9. By the multiplication rule of counting,the number of ways = 2 × 2 × 2 × 2 × 2 × 2 = 64

10. By the multiplication rule of counting,the number of choices = 6 × 3 = 18

11. By the multiplication rule of counting,the number of choices = 2 × 3 = 6

12. By the addition rule of counting,(a) the number of choices = 12 + 22 = 34(b) the number of choices = 12 + 22 – 2 = 32

13. No. If none of the students takes two or more classes, then the number of registered students is 1510, which exceeds 1500. However, if 10 or more students attend more than one class, the claim will be incorrect.

Level 214. By the addition rule of counting,

(a) the number of ways = 4 + 4 = 8(b) the number of ways = 4 + 26 – 2 = 28(c) the number of ways = 20 + 13 – 5 = 28(d) the number of ways = 13 + 13 + 12 – 3 – 3 = 32

101


15. (a) By the multiplication rule of counting,the number of ways = 13 × 13 = 169

(b) After a ‘Queen’ is drawn, there are only 47 cards, which are not ‘King’s, left in the deck. By the multiplication rule of counting,the number of ways = 4 × 47 = 188

16. (a) By the multiplication rule of counting,the number of ways = 210 = 1024

(b) By the multiplication rule of counting,the number of ways = 46 = 4096

(c) By the multiplication rule of counting,the number of ways = 210 × 46 = 4 194 304

17. By the multiplication rule of counting,(a) the number of ways = 5 × 3 = 15(b) the number of ways = 5 × 3 × 5 × 4 = 300

18. (a) By the multiplication rule of counting,the number of possible zip codes = 105 = 100 000

(b) By the multiplication rule of counting,the number of possible zip codes = 9 × 104 = 90 000

(c) There are 10 choices for the first number. The second number must not be the same as the first, so, there are 9 choices. Similarly, there are 8 choices for the third number, 7 choices for the fourth and 6 choices for the fifth.By the multiplication rule of counting,the number of possible zip codes = 10 × 9 × 8 × 7 × 6 = 30 240

19. (a) There are m choices for the first ball and (m – 1) choices for the second.By the multiplication rule of counting,the number of possible ways = m × (m – 1) = m ( m – 1)

(b)

20. (a) By the addition rule of counting,the number of choices = 16 + 23 = 39

(b) The number of Chinese books left = 16 – 1 = 15By the multiplication rule of counting,the number of choices = 15 × 23 = 345

(c) The number of English books left = 23 – 1 = 22By the multiplication rule of counting,the number of choices = 16 × 22 = 352

21. (a) By the multiplication rule of counting,the number of possible strings = 2 × 2 × 2 × 2 = 16

(b) By the multiplication rule of counting,the number of possible strings = 2 × 2 = 4

(c) By the multiplication rule of counting,the number of possible strings = 2 × 2 × 2 = 8

(d) The number of strings that start with ‘10’ and end with ‘1’ = 2By the addition rule of counting,the required number of possible strings = 4 + 8 – 2

= 10

22. (a) By the multiplication rule of counting,the number of ways = 4 × 3 = 12

(b) The number of ways for taking one finance course and one language course = 3 × 6 = 18The number of ways for taking one accounting course and one language course = 4 × 6 = 24By the addition rule of counting,the number of ways = 12 + 18 + 24 = 54

(c) By the multiplication rule of counting,the number of ways = 4 × 3 × 6 = 72

23. (a) By the addition rule of counting,(i) the number of possible choices = 23 + 20 = 43(ii) the number of possible choices = 23 + 16 = 39

(b) The number of ways to pick two candidates, one from Group 1 and one from Group 2 = 23 × 20 = 460The number of ways to pick two candidates, one from Group 2 and one from Group 3 = 20 × 16 = 320The number of ways to pick two candidates, one from Group 1 and one from Group 3 = 23 × 16 = 368By the addition rule of counting,the number of possible choices = 460 + 320 + 368 = 1148

24. (a) The first digit cannot be 0. Thus, there are 6 choices for it. After the first digit has been picked, there are 6 choices for the second digit, including 0. Finally, there are 5 choices for the last digit.By the multiplication rule of counting,the number of 3-digit numbers formed = 6 × 6 × 5 = 180

(b) To ensure that the number formed is even, the last digit must be 0, 2 or 4. Case 1: last digit = 0

Number of 3-digit numbers formed = 6 × 5 × 1 = 30

Case 2: last digit = 2 or 4Number of 3-digit numbers formed = 5 × 5 × 2 = 50

By the addition rule of counting,the number of 3-digit numbers formed = 30 + 50 = 80

(c) The first digit can only be 2, 3 or 4.By the multiplication rule of counting,the number of 3-digit numbers formed = 3 × 6 × 5 = 90

(d) To ensure that the number formed is odd, the last digit must be 3, 5, 7 or 9. Case 1: last digit = 3

Number of 3-digit numbers formed = 4 × 5 × 1 = 20

Case 2: last digit = 5, 7 or 9Number of 3-digit numbers formed = 3 × 5 × 3 = 45

By the addition rule of counting,the number of 3-digit numbers formed = 20 + 45 = 65

Exercise 9B (p. 9.26)Level 1

1.

2.

102


3.

4.

5.

6.

7.

8.

9.

10.

11. There are 4! = 24 strings. They are listed as follows:CART, CATR, CRAT, CRTA, CTAR, CTRA, ACRT, ACTR, ARCT, ARTC, ATCR, ATRC, RCAT, RCTA, RACT, RATC, RTCA, RTAC, TCAR, TCRA, TACR, TARC, TRCA, TRAC

12. The number of arrangements = 6! = 720

13. The number of ways = 8! = 40 320

14. There are 14 letters from A to N.∴ The number of 5-letter strings formed

=

15. We can consider the election as permuting the candidates into the four posts.∴ The number of possible ways =

16. We can consider listing the names of the representatives as a permutation.∴ The number of ways =

17. The number of possible rankings =

18. (a) The number of arrangements for 10 balls = 10! = 3 628 800

(b) The number of arrangements for 6 balls =

19. Although both are permutation problems, the letter ‘L’ in

(a) is repeated while all the objects in (b) are distinct. Therefore only the numbers of arrangements in (b) is 8!.

103


Level 220. (a) The number of 3-digit numbers formed =

(b) The units digit is 5.The number of ways of arranging the tens and hundreds digits = 5 4 = 20∴ The numbers of 3-digit numbers = 20 1 = 20

21. (a) The number of ways for stacking 3 blocks =

(b) The number of ways for stacking 4 blocks = ∴ The number of ways is increased by

120 – 60 = 60

22. (a) The number of ways for different teams to get the prizes =

(b) By the multiplication rule of counting,the number of ways = 62 × 62 × 62 = 238 328

23. (a) The number of ways = 8! = 40 320(b) The layout of the chairs does not alter the fact that 8

distinct persons are seated in 8 distinct chairs. ∴ The required number of ways is also 40 320.

24. (a) We can treat the string AB as a unit.The number of possible strings = 4! = 24

(b) The possible strings are listed below:ABCDE, ABCED, ABDCE, ABDEC, ABECD, ABEDC, CABDE, CABED, DABCE, DABEC, EABCD, EABDC, CDABE, CEABD, DCABE, DEABC, ECABD, EDABC, CDEAB, CEDAB, DCEAB, DECAB, ECDAB, EDCABThe number of strings is indeed 24.

25. Suppose the 3 red balls are arranged on the left of the 3 green balls.

3 red balls 3 green ballsNumber of ways of arranging 3 red balls =

Number of ways of arranging 3 green balls = ∴ The number of ways of arranging the red balls on the

left = = 7200∴ The required number of ways = 7200 2 = 14 400

26. (a) The total number of arrangements without restrictions = 6! = 720

(b) We can treat Susan and Carmen as a unit. The queue can be arranged in 5! = 120 ways, while Susan and Carmen themselves can be permuted in 2! = 2 ways.By the multiplication rule of counting,the number of arrangements for Susan and Carmen to be standing together = 120 × 2 = 240

(c) The number of arrangements for Susan and Carmen not standing together = 720 – 240 = 480

27. The situation can be restructured as follows:Arrange 5 women and 5 men, M1, M2, M3, M4 and M5, in a row as shown below:

(F) (F) (F) (F) (F)

There are 5 places for the 5 men to stand.∴ The number of ways of pairing up = = 120

28. (a) The total number of arrangements without restrictions = 7! = 5040

(b) We may pick either parent to sit at either end of the row. There are (2 × 2) = 4 possibilities for this choice. The rest of the family can be seated in 6! = 720 ways.By the multiplication rule of counting,the number of arrangements = 4 × 720 = 2880

(c) We can treat all the sons as a unit. There are 5! = 120 ways of accommodating the family, while the boys themselves can be permuted in 3! = 6 ways.By the multiplication rule of counting, the number of arrangements = 120 × 6 = 720

(d) There are 2! = 2 ways of seating the parents at the two ends.(i) The children can be seated in 5! = 120 ways.

By the multiplication rule of counting,the number of arrangements = 2 × 120 = 240

(ii) If no sons can sit next to each other, then a boy must be seated next to a girl. The boys can be permuted in 3! = 6 ways, and the girls in 2! = 2 ways.By the multiplication rule of counting,the number of arrangements = 2 × 6 × 2 = 24

29. Because the two zeros are not distinct, double counting occurs. For example, 0011 will be counted twice as 0011 and 0011, etc. In fact, the six possible strings are 0011, 1001, 1100, 0101, 1010 and 0110.

Exercise 9C (p. 9.35)Level 1

1.

2.

3.

4.

5.

∴

6.

104


7.

8. (a) Listed below are all the possible combinations:67, 68, 69, 78, 79, 89

(b) The number of combinations =

9. (a) Listed below are all the possible combinations:RBG, RBY, RBP, RBV, RGY, RGP, RGV, RYP, RYV, RPV, BGY, BGP, BGV, BYP, BYV, BPV, GYP, GYV, GPV, YPV

(b) The number of combinations =

10. The number of committees formed =

11. The required number of ways =

12. The number of different teams =

13. The number of different groups =

14. The number of ways =

15. (a) This is a combination, because the order of invitation does not matter.

(b) This is not a combination, because there are two distinct prizes and the order is important.

16. The required number of ways =

Level 217. There are 17 players and 3 goalkeepers in the team.

∴ The required number of ways =

18. (a) The number of selections = (b) Only 7 letters can be selected.

∴ The number of selections = (c) Only 2 more letters can be selected from the

remaining 6 letters.∴ The number of selections =

19. (a) We only need to pick 2 more typists from the remaining 19 typists.∴ The number of choices =

(b) There are 19 eligible typists from which we have to select 3.∴ The number of choices =

20. (a) The number of different committees =(b) We still have 7 eligible economists.

∴ The number of different committees =

(c) We only need to choose 2 more accountants out of 6, and 1 more economist out of 7.∴ The number of different committees =

21. (a) We need to choose 2 from 8 S4 players and 2 from 5 S5 players.∴ The number of teams that can be formed =

(b) There are (8 + 5) = 13 players.The number of teams that can be formed without restrictions =The number of teams that can be formed without any S4 players =As a result, the number of teams with at least one S4 player = 715 – 5 = 710

22. (a) The required number of ways =

(b) The required number of ways =

(c) The required number of ways =

23. (a) We need to nominate 2 more participants from the rest of 30 scouts.∴ The number of ways that the team can be

formed =(b) There are 30 eligible scouts from which we must

nominate 4.∴ The number of ways that the team can be

formed =(c) We need to nominate 3 scouts from the rest of 30

scouts.∴ The number of ways that the team can be

formed =

24. (a) First, we assign 6 out of the 12 buses to the first group. Then, we assign 6 out of the 6 buses left to the second group.∴ The number of ways =

(b) First, we assign 4 out of the 12 buses to the first group. Then, we assign 4 out of the 8 buses left to the second group. Finally, the 4 buses left are assigned to the third group.∴ The number of ways =

25. (a) 2, 4 and 5 musicians are assigned to bedrooms A, B and C respectively.∴ The required number of ways

=(b) The beds in bedroom A can be permuted in

2! = 2 ways. Similarly, bedroom B: 4! = 24 ways

and bedroom C: 5! = 120 ways∴ The required number of ways

= 6930 × 2 × 24 × 120 = 39 916 800

105


(c) The number of ways of arranging the 11 musicians in a row = 11! = 39 916 800These two answers are equal. Since the beds are distinct, assigning musicians to these beds is similar to arranging the musicians in a row. Thus, the coincidence should not be surprising.

26. (a) There are (9 + 7) = 16 articles available.∴ The required number of ways =

(b) (i) The required number of ways =

(ii) The Chinese articles can be arranged in 4! = 24 ways, and the English articles can be arranged in 2! = 2 ways.∴ The required number of arrangements

= 1260 × 24 × 2 = 60 480

27. (a) The required number of ways = (b) If those two friends are invited together, then

the number of ways to select the guests =Since those 2 friends cannot be invited together, the number of ways to select the guests = 84 – 35 = 49

(c) If Mr Chan do not want to invite those two friends at all, then the number of ways to select the guests =

∴ The required number of ways = 35 + 7 = 42

28. (a) (i) To obtain exactly one head, this head can be appeared in any one out of the n tosses.∴ The required number of ways =

(ii) To obtain exactly two heads, these two heads can be appeared in any two out of the n tosses.∴ The required number of ways =

(iii) Each toss can end up in two ways: either as head or tail.The total number of outcomes for any n tosses = 2n

There is only one way in which we can get no heads, that is, when every toss ends up in a tail.Consequently, the number of ways to obtain at least 2 heads = 2 n – n – 1

(b) (i) From (a)(ii), we have

n2 – n – 72 = 0(n – 9)(n + 8) = 0 n = 9 or n = –8 (rejected)

(ii) The number of ways of obtaining at least 2 heads = 29 – 9 – 1 = 502

Revision Exercise 9 (p. 9.40)Level 11. By the multiplication rule of counting,

the number of different outfits = 12 × 8 × 4 = 384

2. By the addition rule of counting,(a) the required number of ways = 4 + 4 = 8(b) the required number of ways = 13 + 16 – 4 = 25(c) the required number of ways = 13 + 26 – 13 = 26

Alternative Solution We note that all hearts are red. Therefore, we only need to consider the number of ways a red card can be selected, which is 26.

(d) the required number of ways = 12 + 26 – 6 = 32(e) the required number of ways = 4 + 4 + 12 = 20(f) the required number of ways = 26 + 16 – 8 = 34

3. By the multiplication rule of counting,(a) the number of possible choices = 3 × 2 × 4 = 24(b) the number of possible choices = 3 × 2 × 2 × 4 = 48

4. By the multiplication rule of counting,the number of different codes = 1 × 10 × 10 × 26 × 26 = 67 600

5. (a) By the multiplication rule of counting,the required number of 4-digit numbers = 4 × 10 × 10 × 10 = 4000

(b) By the multiplication rule of counting,the required number of 4-digit numbers = 9 × 10 × 10 × 5 = 4500

(c) By the multiplication rule of counting,the required number of 4-digit numbers = 4 × 10 × 10 × 5 = 2000

(d) By the addition rule of counting,the required number of 4-digit numbers = 4000 + 4500 – 2000 = 6500

6. (a) By the multiplication rule of counting,the number of different choices = 4 × 3 × 3 = 36

(b) The number of possible ways for taking 1 Japanese and 1 French courses = 4 × 3 = 12The number of possible ways for taking 1 Japanese and 1 German courses = 4 × 3 = 12The number of possible ways for taking 1 French and 1 German courses = 3 × 3 = 9By the addition rule of counting,the number of possible choices = 12 + 12 + 9 = 33

7. (a) By the addition rule of counting,the number of possible choices = 8 + 6 + 5 = 19

(b) By the multiplication rule of counting,the number of possible choices = 8 × 6 = 48

(c) The number of ways to read 1 novel and 1 history books = 8 × 5 = 40The number of ways to read 1 science and 1 history books = 6 × 5 = 30By the addition rule of counting,the number of possible choices = 48 + 40 + 30 = 118

(d) By the multiplication rule of counting,the number of possible choices = 8 × 6 × 5 = 240

8. (a) By the multiplication rule of counting,the required number of ways = 6 × 8 = 48

(b) By the addition rule of counting,the required number of ways = 6 + 8 = 14

(c) By the multiplication rule of counting,the required number of ways = 6 × 8 × 15 = 720

(d) The number of ways of joining an academic or a cultural club = 8 + 15 = 23∴ The required number of ways = 23 × 6 = 138Alternative Solution The number of ways of joining an academic club and a service group = 8 × 6 = 48The number of ways of joining a culture club and a service group = 15 × 6 = 90By the addition rule of counting,the required number of ways = 48 + 90 = 138

106


9. (a) The number of combinations with different flavours =

(b) The number of combinations with the same flavour = 25By the addition rule of counting,The number of combinations that can be made = 300 + 25 = 325

10. We can treat the triplets as a unit. The children can be arranged in 6! = 720 ways, while the triplets can themselves be permuted in 3!, i.e. 6 ways.By the multiplication rule of counting,the number of possible arrangements = 720 × 6 = 4 320

11. There are 3! ways of arranging D, E and F in a row.

A, B and C can stand in 3 of the 4 places as shown above.∴ The number of ways of arrangements =By the multiplication rule of counting,the number of ways the people can be arranged = 3! × 24 = 144

12. We need to choose 4 more violinists from the rest of 8 violinists.∴ The number of possible ways =

13. The process can be broken down into 3 steps:Step 1: Choose 3 out of 12 to form a group.Step 2: Choose 4 out of (12 – 3) = 9 to form another

group.Step 3: 9 – 4 = 5 left behind form the last group.∴ The required number of ways =

14. First, we assign 3 out of the 12 books to the first child. Then, we assign 3 out of the 9 books left to the second child. Then, we also assign 3 out of the 6 books to the third child. Finally, the 3 books left are assigned to the fourth child. ∴ The number of ways =

Level 215. (a) By the multiplication rule of counting,

the number of license plates = 26 × 26 × 9 × 10 × 10 × 10 = 6 084 000

(b) The number of license plates without ‘I’ or ‘O’ = 24 × 24 × 9 × 10 × 10 × 10 = 5 184 000The number of license plates without ‘I’ or ‘O’, and with all digits the same = 24 × 24 × 9 = 5184By the addition rule of counting,the number of possible license plates = 5 184 000 – 5184 = 5 178 816

16. There are (26 + 10) = 36 legal characters for the password.The number of passwords without restrictions = 366 = 2 176 782 336The number of passwords without a digit = 266 = 308 915 776∴ The number of possible passwords

= 366 – 266 = 1 867 866 560

17. (a) The number of possible seating arrangements = 7! = 5040

(b) The remaining 6 guests can be seated in the rest of the row.The number of possible seating arrangements = 6! = 720

18. (a) The number of ways of selecting 4 people = The number of ways these people can be seated = 4! = 24∴ The number of ways of assigning 7 people to 4

seats = 35 × 24 = 840(b) The number of ways of selecting 4 seats =

The number of ways these seats are assigned to the 4 people = 4! = 24∴ The number of ways of assigning 4 people to 7

seats = 35 × 24 = 840(c) The results in (a) and (b) are the same. We can think

that the people in (a) were seats and the seats were people. The problem then becomes exactly the same as that in (b).

19. (a) We can treat the scouts as a unit. The row can be arranged in 6! = 720 ways, while the scouts themselves can be permuted in 4! = 24 ways.By the multiplication rule of counting,the required number of arrangements = 720 × 24 = 17 280

(b) There are 5! = 120 ways of arranging the girl guides.

The scouts can stand in 4 of the 6 places as shown above.∴ The scouts can be arranged in ways.By the multiplication rule of counting,the required number of arrangements = 120 × 360 = 43 200

20. For every cat Tom has 2 options: to bring or not to bring.If there were no restrictions, the number of ways for Tom = 24 = 16Tom only has one way to bring less than a cat: he brings no cats to the park.∴ The number of ways for him to bring at least one cat

= 16 – 1 = 15

21. (a) The number of ways to assign presents to the first group = The number of ways to assign presents to the second group = The number of ways to assign presents to the third group = The number of ways to assign presents to the fourth group = The number of ways to assign presents to the fifth group = The groups can be ordered in 5! = 120 ways.However, since the groups are indistinguishable, the order of the groups does not affect the result.

107


∴ The required number of ways

=

(b) Let the two friends be A and B. Nancy has two choices for each present: whether to give it to A. The rest of the presents are assigned automatically to B.The number of ways Nancy can distribute her presents = 210 = 1024The number of ways to give no presents to A= The number of ways to give one present to A = The number of ways to give two presents to A = The number of ways to give A less than three presents = 1 + 10 + 45 = 56Similarly, the number of ways to give B less than three presents = 56∴ The number of ways to give A and B at least

three presents each= 1024 – 56 – 56 = 912

22. The number of juries with 7 men =

The number of juries with 8 men = ∴ The number of juries with more men than women

= 2016 + 210 = 2226

23. For ‘6’ to be the second largest number drawn, we must draw three balls from ‘1’ to ‘5’, and one ball from ‘7’ to ‘12’.The number of ways of drawing the first three balls =

The number of ways of drawing the last ball = ∴ The number of ways of drawing such that ‘6’ is the

second largest = 6 × 10 = 60

24. These are three possible cases of obtaining at least 6 points:Case I: 2 black marbles and 2 white marbles are drawn;

6 points obtainedCase II: 3 black marbles and 1 white marbles are drawn;

7 points obtainedCase III: 4 black marbles and 0 white marbles are drawn;

8 points obtainedThe number of ways for Case I =

The number of ways for Case II =

The number of ways for Case III = ∴ The number of ways of obtaining at least 6 points

= 6 + 4 + 1 = 11

25. The number of ways of arranging the athlete who can only row on the left = The number of ways of arranging the athletes who can only row on the right = The number of ways of arranging the rest of the athletes = 5! = 120

108


∴ The number of possible arrangements = 4 × 6 × 120 = 2880

26.

27.

28. (a) The number of ways of choosing one King = 4The number of ways of choosing the other cards = ∴ The number of required poker hands

= 4 × 194 580 = 778 320

(b) The number of possible poker hands = The number of hands without a King = ∴ The number of the required poker hand

= 2 598 960 – 1 712 304 = 886 656

(c) The number of possible ranks for the pair = 13The number of possible suit combination for the pair = Since there should be only one pair, there are now only 48 cards left for selection. Afterwards, every card picked removes 3 other cards of the same rank

from the pool. This ensures that there will be no more than one pair.The number of ways of choosing the other three cards

=

∴ The number of the required hand = 13 × 6 × 14 080 = 1 098 240

29. (a) (i) The number of different lottery tickets =

109


(ii) The number of different lottery tickets =

(b) (i) The number of different lottery tickets =

(ii) The number of different lottery tickets =

30. (a) (i) Total number of courses = (4 + 5) = 9The number of possible choices =

(ii) If the student had selected M1 and F2 already, he only needs to select 2 more courses out of the 7 courses left.The number of courses with M1 and F2 together = The number of possible choices = 126 – 21 = 105

(b) Number of choices with no Mathematics courses = Number of choices with no Finance courses = Number of possible choices = 126 – 5 – 1 = 120

31. (a) The required number of ways = (b) The number of ways for boys to leave the lift

= The number of ways for girls to leave the lift = 20∴ The required number of ways

= 380 × 20 = 7600

32. (a) 4 people take the first taxi. Then, the other 4 people will take the second taxi.∴ The number of arrangements =

(b) 2 adults and 2 children take the first taxi.∴ The number of arrangements =

(c) The number of ways of assigning 5 people to the first taxi and 3 to the second = Similarly, the number of ways of assigning 3 people to the first taxi and 5 to the second = 56∴ The number of arrangements without restrictions

= 70 + 56 + 56 = 182

(d) Case 1: 4 people take the first taxi, and the remaining 4 take the second taxi.The number of ways with no adults in the first taxi = 1

Case 2: 3 people take the first taxi, and the remaining 5 take the second taxi.The number of ways with no adults in the first taxi = = 4

∴ The number of ways with no adults in one of the taxis = 2 × (1 + 4) = 10

∴ The number of arrangements with at least one adult in each taxi = 182 – 10 = 172

33. (a) John has to make 7 moves in total: 3 moves northwards and 4 moves eastwards. In any one of the paths, he can make any 3 out of the 7 moves to be northwards moves, and the rest to eastwards moves.∴ The number of possible paths =

(b) John has to make n out of (m + n) moves to be northwards moves.

∴ The number of possible paths =

Alternative Solution John has to make m out of (m + n) moves to be eastwards moves.

∴ The number of possible paths =

34. Let m be the number of games in one of the sequences by one of the team. Then, m is at least n and at most (2n – 1). If the team has won n games out of the total m, there will be ways of these games.∴ The total number of possible outcomes

=

Multiple Choice Questions (p. 9.44)1. Answer: B

There are 2 choices for the rice and 3 choices for the noodle.∴ There are (2 + 3) choices for the main dish.There are 5 choices for the drink.∴ The number of choices for a customer

2. Answer: CThe black and the purple marbles must be placed at the ends of the row.The other four marbles are placed in the middle of the row.


3. Answer: BThe number of ways to arrange C and D =

Since A and B cannot stand together, there are 3 places, as shown above, for A and B.So, the number of ways to arrange A and B =


4. Answer: BThe number of different licence numbers can be formed

5. Answer: BIf a particular person must be included, then the 4 members of the committee can be chosen from the remaining 11 people.


110


6. Answer: AIf a particular player must be excluded, then 6 team

players can be selected from the remaining 9 players.


7. Answer: BThe number of outcomes without ‘6’ = 54

The number of outcomes which have at least one ‘6’

8. Answer: AEach letter can be placed in 4 post boxes.∴ The required number of ways =

9. Answer: DThe greatest number of seats can be labelled

10. Answer: BCase I: Car A carries 5 people and car B carries 3

people.The number of ways =

Case II: Car A carries 4 people and car B carries 4 people.The number of ways =


HKMO (p. 9.46)1. There are 25 multiples of 2 from 1 to 50.

There are 12 multiples of 4 from 1 to 50.There are 6 multiples of 8 from 1 to 50.There are 3 multiples of 16 from 1 to 50.There are only one multiple of 32 from 1 to 50.∴ The largest possible value of a

2. Since the two books selected are of the same language, they may be 2 English books or 2 Chinese books or 2 Japanese books.By the addition rule of counting,

3. Refer to the figure.

The ant moves from A to B in 4 moves.1 move: along the x-direction1 move: along the z-direction2 moves: along the y-direction∴ The number of permutations of these 4 moves

4. From the question, we form the following table for the different cases.

Case no. 1 2 3 4 5 6 7 8correct 4 3 3 2 2 2 1 1blank 0 1 0 2 1 0 3 2wrong 0 0 1 0 1 2 0 1Total 8 6 5 4 3 2 2 1

Case no. 9 10 11 12 13 14 15correct 1 1 0 0 0 0 0blank 1 0 4 3 2 1 0wrong 2 3 0 1 2 3 4Total 0 –1 0 –1 –2 –3 –4

The possible total marks for one candidate to solve 4 problems are:8, 6, 5, 4, 3, 2, 1, 0, –1, –2, –3, –4.There are 12 possible total marks. To ensure that 2 candidates will have the same scores, there should be at least (12 + 1) = 13 candidates in the competition. Similarly, to ensure that 3 candidates will have the same scores, there should be at least (12 2 + 1) = 25 candidates.

Investigation Corner (p. 9.47)(a) Suppose 5 colours are available.

Actions to be completed Number of waysTo colour region 1 5To colour region 2 5 – 1 = 4To colour region 3 5 – 2 = 3To colour region 4 (Note: There are two cases.)Case I : with same colour as region 2 1 To colour region 5 5 – 2 = 3Case II: with a colour different from

that of region 2 5 – 3 = 2

To colour region 5 5 – 3 = 2∴ The number of ways to fill the figure with colours

(b) Suppose 10 colours are available. Using the similar argument as in (a), the number of ways to fill the figure with colours

111


(c) Suppose n colours are available, where .The number of ways to fill the figure with colours

(d) The structures of the figures in (i) and (ii) are basically the same in light of the number of neighbouring regions of each region, i.e. the centre region adjoins 4 outer regions, and each outer region adjoins 3 regions.Moreover, different colouring sequences yield the same result.(i) The number of ways to fill the figure with colours

(ii) The number of ways to fill the figure with colours

112

Chapter 7 Full Solutions - WordPress.com · Web view(iii) Since two dogs are already selected, we...

Documents

Transcript of Chapter 7 Full Solutions - WordPress.com · Web view(iii) Since two dogs are already selected, we...