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

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  • CHAPTER 6 PERMUTATIONS AND COMBINATIONS FORM 5

    116

    PAPER 1

    1. A committee of 3 boys and 3 girls are to be formed from 10 boys and 11 girls.

    In how many ways can the committee be formed?

    [2 marks]

    2. How many 4-letter codes can be formed using the letters in the word 'GRACIOUS' without

    repetition such that the first letter is a vowel?

    [2 marks]

    3. Find the number of ways of choosing 6 letters including the letter G from the word

    'GRACIOUS'.

    [2 marks]

    4. How many 3-digit numbers that are greater than 400 can be formed using the digits 1, 2, 3, 4,

    and 5 without repetition?

    [2 marks]

    5. How many 4-digit even numbers can be formed using the digits 1, 2, 3, 4, and 5 without

    repetition?

    [2 marks]

    6. Diagram shows 4 letters and 4 digits.

    A code is to be formed using those letters and digits. The code must consists of 3 letters

    followed by 2 digits. How many codes can be formed if no letter or digit is repeated in each

    code ?

    [3 marks]

    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

    a) 5 boys, b) not more than 2 girls.

    [4 marks]

    8. Diagram shows five cards of different letters.

    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.

    [4 marks]

    A B C D 5 6 7 8

    R A J I N

  • CHAPTER 6 PERMUTATIONS AND COMBINATIONS FORM 5

    117

    9. A debating team consists of 6 students. These 6 students are chosen from 2 monitors, 3 assistant monitors and 5 prefects.

    a) there is no restriction,

    b) the team contains only 1 monitor and exactly 3 prefects.

    [4 marks]

    10. Diagram shows seven letter cards.

    A five-letter code is to be formed using five of these cards. Find

    a) the number of different five-letter codes that can be formed,

    b) the number of different five-letter codes which end with a consonant.

    [4 marks]

    11. How many 5-digit numbers that are greater than 50000 can be formed using the digits 1, 2, 3, 4,

    5, 6, 7, 8, and 9 without repetition?

    [4 marks]

    12. How many 4-digit even numbers can be formed using the digits 1, 2, 3, 4, 5 and 6 without any

    digit being repeated?

    [4 marks]

    13. A coach wants to choose 9 players consisting of 6 boys and 3 girls to form a squash team.

    These 9 players are chosen from a group of 8 boys and 6 girls. Find

    (a) the number of ways the team can be formed,

    (b) the number of ways the team members can be arranged in a row for a group photograph,

    if the 6 boys sit next to each other. [4 marks]

    14. 2 girls and 8 boys are to be seated in a row of 5 chairs. Find the number of ways they can be

    seated if no two persons of the same sex are next to each other.

    [3 marks]

    15. Diagram shows six numbered cards.

    A four-digit number is to be formed by using four of these cards.

    How many

    a) different numbers can be formed? b) different odd numbers can be formed?

    [4 marks]

    R O F I N U M

    9 8 7 5 4 1

  • CHAPTER 6 PERMUTATIONS AND COMBINATIONS FORM 5

    118

    ANSWERS ( PAPER 1 )

    1. 3

    10C x 311C 1

    = 19800

    1

    2. 1

    4 p x 37 p 1

    = 840

    1

    3. 1 x 57C 1

    = 21

    1

    4. 1

    2 p x 24 p 1

    = 24

    1

    5. 3

    4 p x 12 p 1

    = 48

    1

    6. 34 p x 2

    4 p 2

    = 288

    1

    7. a) 58C x 3

    5C = 560 1

    b) If the team consists of 8 boys and 0 girl 88C x 0

    5C = 1

    If the team consists of 7 boys and 1 girl 78C x 1

    5C = 40

    If the team consists of 6 boys and 2 girl 68C x 2

    5C = 280

    1

    The number of teams that can be formed = 1 + 40 + 280 1

    = 321

    1

    8. a) 5! = 120 1

    b) 4! x 2! 2

    = 48

    1

    9. a) 610C = 210 2

    b) 12C x 3

    5C x 23C = 60 2

    10. a) 57 p = 2520 2

    b) 46 p x 1

    4 p 1

    = 1440

    1

    11. 1

    5 p x 48 p 2

    = 8400

    1

    12. 3

    5 p x 13 p = 180 2

    = 180

    1

  • CHAPTER 6 PERMUTATIONS AND COMBINATIONS FORM 5

    119

    13. a) 68C 3

    6C 1

    = 560 1

    b) 6! x 4! 1

    = 17280 1

    14. 38 P x 2

    2 P 2

    = 672 1

    15. a) 6P4 = 360

    1

    b) 5P3 x

    4P1 2

    = 240 1