Form 5 : Chapter 6 Permutations and Combinations 6 permutations and combinations ...

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<ul><li><p>CHAPTER 6 PERMUTATIONS AND COMBINATIONS FORM 5 </p><p>116 </p><p>PAPER 1 </p><p>1. A committee of 3 boys and 3 girls are to be formed from 10 boys and 11 girls. </p><p>In how many ways can the committee be formed? </p><p> [2 marks] </p><p>2. How many 4-letter codes can be formed using the letters in the word 'GRACIOUS' without </p><p>repetition such that the first letter is a vowel? </p><p> [2 marks] </p><p>3. Find the number of ways of choosing 6 letters including the letter G from the word </p><p>'GRACIOUS'. </p><p> [2 marks] </p><p>4. How many 3-digit numbers that are greater than 400 can be formed using the digits 1, 2, 3, 4, </p><p>and 5 without repetition? </p><p> [2 marks] </p><p>5. How many 4-digit even numbers can be formed using the digits 1, 2, 3, 4, and 5 without </p><p>repetition? </p><p> [2 marks] </p><p>6. Diagram shows 4 letters and 4 digits. </p><p>A code is to be formed using those letters and digits. The code must consists of 3 letters </p><p>followed by 2 digits. How many codes can be formed if no letter or digit is repeated in each </p><p>code ? </p><p>[3 marks] </p><p>7. A badminton team consists of 8 students. The team will be chosen from a group of 8 boys and 5 girls. Find the number of teams that can be formed such that each team consists of </p><p>a) 5 boys, b) not more than 2 girls. </p><p>[4 marks] </p><p>8. Diagram shows five cards of different letters. </p><p>a) Find the number of possible arrangements, in a row, of all the cards. b) Find the number of these arrangements in which the letters A and N are side by side. </p><p>[4 marks] </p><p>A B C D 5 6 7 8 </p><p>R A J I N </p></li><li><p>CHAPTER 6 PERMUTATIONS AND COMBINATIONS FORM 5 </p><p>117 </p><p>9. A debating team consists of 6 students. These 6 students are chosen from 2 monitors, 3 assistant monitors and 5 prefects. </p><p>a) there is no restriction, </p><p>b) the team contains only 1 monitor and exactly 3 prefects. </p><p>[4 marks] </p><p>10. Diagram shows seven letter cards. </p><p>A five-letter code is to be formed using five of these cards. Find </p><p>a) the number of different five-letter codes that can be formed, </p><p>b) the number of different five-letter codes which end with a consonant. </p><p>[4 marks] </p><p>11. How many 5-digit numbers that are greater than 50000 can be formed using the digits 1, 2, 3, 4, </p><p>5, 6, 7, 8, and 9 without repetition? </p><p> [4 marks] </p><p>12. How many 4-digit even numbers can be formed using the digits 1, 2, 3, 4, 5 and 6 without any </p><p>digit being repeated? </p><p> [4 marks] </p><p>13. A coach wants to choose 9 players consisting of 6 boys and 3 girls to form a squash team. </p><p>These 9 players are chosen from a group of 8 boys and 6 girls. Find </p><p> (a) the number of ways the team can be formed, </p><p> (b) the number of ways the team members can be arranged in a row for a group photograph, </p><p> if the 6 boys sit next to each other. [4 marks] </p><p>14. 2 girls and 8 boys are to be seated in a row of 5 chairs. Find the number of ways they can be </p><p>seated if no two persons of the same sex are next to each other. </p><p> [3 marks] </p><p>15. Diagram shows six numbered cards. </p><p> A four-digit number is to be formed by using four of these cards. </p><p> How many </p><p>a) different numbers can be formed? b) different odd numbers can be formed? </p><p> [4 marks] </p><p>R O F I N U M </p><p>9 8 7 5 4 1 </p></li><li><p>CHAPTER 6 PERMUTATIONS AND COMBINATIONS FORM 5 </p><p>118 </p><p>ANSWERS ( PAPER 1 ) </p><p>1. 3</p><p>10C x 311C 1 </p><p> = 19800 </p><p>1 </p><p>2. 1</p><p>4 p x 37 p 1 </p><p> = 840 </p><p>1 </p><p>3. 1 x 57C 1 </p><p> = 21 </p><p>1 </p><p>4. 1</p><p>2 p x 24 p 1 </p><p> = 24 </p><p>1 </p><p>5. 3</p><p>4 p x 12 p 1 </p><p> = 48 </p><p>1 </p><p>6. 34 p x 2</p><p>4 p 2 </p><p> = 288 </p><p>1 </p><p>7. a) 58C x 3</p><p>5C = 560 1 </p><p>b) If the team consists of 8 boys and 0 girl 88C x 0</p><p>5C = 1 </p><p> If the team consists of 7 boys and 1 girl 78C x 1</p><p>5C = 40 </p><p> If the team consists of 6 boys and 2 girl 68C x 2</p><p>5C = 280 </p><p>1 </p><p> The number of teams that can be formed = 1 + 40 + 280 1 </p><p> = 321 </p><p>1 </p><p>8. a) 5! = 120 1 </p><p> b) 4! x 2! 2 </p><p> = 48 </p><p> 1 </p><p>9. a) 610C = 210 2 </p><p>b) 12C x 3</p><p>5C x 23C = 60 2 </p><p>10. a) 57 p = 2520 2 </p><p>b) 46 p x 1</p><p>4 p 1 </p><p> = 1440 </p><p> 1 </p><p>11. 1</p><p>5 p x 48 p 2 </p><p> = 8400 </p><p> 1 </p><p>12. 3</p><p>5 p x 13 p = 180 2 </p><p> = 180 </p><p> 1 </p></li><li><p>CHAPTER 6 PERMUTATIONS AND COMBINATIONS FORM 5 </p><p>119 </p><p>13. a) 68C 3</p><p>6C 1 </p><p> = 560 1 </p><p>b) 6! x 4! 1 </p><p> = 17280 1 </p><p>14. 38 P x 2</p><p>2 P 2 </p><p> = 672 1 </p><p>15. a) 6P4 = 360 </p><p>1 </p><p>b) 5P3 x </p><p>4P1 2 </p><p> = 240 1 </p></li></ul>