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MS-Sol-EE-C01 - CHAPTER 1 PERMUTATIONS AND COMBINATIONS...

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CHAPTER 1 PERMUTATIONS AND COMBINATIONS EXERCISE 1.1 Section 1.1 The multiplication principle of counting (page 5) 1. The number of different routes = 5 × 4 = 20 2. The number of different outfits = 6 × 7 = 42 3. The number of choices = 7 × 8 = 56 4. The number of pairs = 10 × 12 = 120 5. The number of ways = 3 × 5 = 15 6. The number of ways = 7 × 6 × 2 = 84 7. The number of different computer systems = 5 × 4 × 7 = 140 8. The number of different arrangements = 20 × 7 × 11 = 1 540 9. The number of dinner choices = 4 × 5 × 3 × 6 = 360 10. The number of ways = 3 × 4 × 5 × 2 × 7 = 840 11. (a) The number of possible PINs = 10 × 10 × 10 × 10 × 10 × 10 = 10 6 (b) The number of possible PINs = 10 × 9 × 8 × 7 × 6 × 5 = 151 200 12. The number of ways to travel via T = 4 × 5 = 20 The number of ways to travel from X to Y = 3 + 20 = 23 13. (a) The required number of ways = 2 × 2 × 2 × 2 × 2 = 32 (b) The required number of ways = 4 10 (c) The required number of ways = 32 × 4 10 = 2 25 14. (a) The number of choices = 7 + 5 + 3 = 15 (b) If she reads a book and watches a video, the number of choices = 7 × 5 = 35. If she reads a book and goes for a drink, the number of choices = 7 × 3 = 21. If she watches a video and goes for a drink, the number of choices = 5 × 3 = 15. The total number of choices = 35 + 21 + 15 = 71 (c) The number of choices = 7 × 5 × 3 = 105 1
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C HAPTER 1 P ERMUTATIONS AND C OMBINATIONS 15. The number of symbols = 2 + 2 2 + 2 3 + 2 4 + 2 5 = 62 16. Number of facial appearances with one feature only = 8 + 3 + 4 = 15 Number of facial appearances with two features only = 8 × 3 + 3 × 4 + 4 × 8 = 68 Number of facial appearances with three features = 8 × 3 × 4 = 96 The total number of facial appearances possible = 15 + 68 + 96 = 179 17. (a) As the first digit cannot be zero, there are 6 ways of choosing the first digit. The number of possible numbers = 6 × 6 × 5 = 180 (b) When the last digit is 0, the number of ways = 6 × 5 × 1 = 30. When the last digit is 6 or 8, the number of ways = 5 × 5 × 2 = 50. The number of even numbers = 30 + 50 = 80 (c) When the number formed is less than 500, the first digit is either 1 or 3. The required number of ways = 2 × 6 × 5 = 60 18. (a) The number of different plates = 26 × 26 × 10 × 10 × 10 × 10 = 6 760 000 (b) The number of different plates = 24 × 24 × 10 × 10 × 10 × 10 = 5 760 000 (c) If the digit 8 is excluded, the number of different plates = 24 × 24 × 9 × 9 × 9 × 9 = 3 779 136. The required number = 5 760 000 - 3 779 136 = 1 980 864 EXERCISE 1.2 Section 1.2 Permutations (page 13) 1. ! 2 ! 6 = 6 × 5 × 4 × 3 = 360 2. ! 13 ! 15 = 15 × 14 = 210 3. 1 2 3 4 6 7 8 9 = ! 5 ! 4 ! 9 × × × × × × = 126 4. 1 2 3 9 10 11 = ! 8 ! 3 ! 11 × × × × = 165 5. 5 5 P = 5! = 120 6. P 4 7 = 4)! 7 ( ! 7 - = 7 × 6 × 5 × 4 = 840 7. 4)! + ( 6)! + ( n n = 4)! + ( 4)! + ( 5) + ( 6) + ( n n n n = ( n + 6) ( n + 5) 2
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S ECTION 1.2 P ERMUTATIONS 8. P n 3 = 3)! ( ! - n n = n ( n - 1) ( n - 2) 9. (a) The possible permutations are: AJQK, JAQK, QAJK, KAJQ, AJKQ, JAKQ, QAKJ, KAQJ, AQJK, JQAK, QJAK, KJAQ, AQKJ, JQKA, QJKA, KJQA, AKJQ, JKAQ, QKAJ, KQAJ, AKQJ, JKQA, QKJA, KQJA. (b) The number of different permutations = 24 10. The number of ways = 5! = 120 11. The number of ways = 6! = 720 12. The number of ways = 7! = 5 040 13. (a) (i) The possible outcomes are: GD, DG, AG, EG, GA, DA, AD, ED, GE, DE, AE, EA.
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