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Quiz Chapter 4: Permutation and Combination

10 questions · Form 5 Additional Mathematics Bab 4: Permutation and Combination

Question 1 of 10Score: 0

Find the number of ways to arrange 6 people around a circular table.

Full Question List & Answer Key

Prefer reading to quizzing? All 10 questions are listed below with the answer and explanation under each one.

1. Find the number of ways to arrange 6 people around a circular table.

  1. 120
  2. 720
  3. 360
  4. 24
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Answer: A

Circular permutation = (n - 1)! = (6 - 1)! = 5! = 120.

2. A team of 3 students is to be chosen from a group of 8 students. How many different teams can be formed?

  1. 56
  2. 336
  3. 24
  4. 112
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Answer: A

Order does not matter, so use combination: ⁸C₃ = 8 × 7 × 63 × 2 × 1 = 56.

3. How many triangles can be formed using the vertices of an octagon?

  1. 56
  2. 336
  3. 24
  4. 112
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Answer: A

An octagon has 8 vertices. Choosing 3 vertices forms a triangle: ⁸C₃ = 8 × 7 × 63 × 2 × 1 = 56.

4. Evaluate ⁴P₄ + ⁴C₄.

  1. 25
  2. 24
  3. 16
  4. 28
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Answer: A

⁴P₄ = 4! = 24. ⁴C₄ = 1. Total = 24 + 1 = 25.

5. How many straight lines can be formed by joining any two of 10 points on a plane where no 3 points are collinear?

  1. 45
  2. 90
  3. 120
  4. 20
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Answer: A

To form a straight line, choose 2 points from 10: ¹⁰C₂ = 10 × 92 = 45.

6. In how many ways can 5 boys and 3 girls be seated in a row such that no two girls sit next to each other?

  1. 14400
  2. 7200
  3. 40320
  4. 2880
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Answer: A

First arrange the 5 boys: 5! = 120 ways. The 5 boys create 6 gap spaces (_ B₁ _ B₂ _ B₃ _ B₄ _ B₅ _). Choose 3 spaces for the girls and arrange them: ⁶P₃ = 6 × 5 × 4 = 120. Total ways = 120 × 120 = 14400.

7. Calculate the value of ⁸C₅.

  1. 56
  2. 6720
  3. 336
  4. 28
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Answer: A

⁸C₅ = 8! / (5! × 3!) = 8 × 7 × 63 × 2 × 1 = 56.

8. Find the number of ways to arrange 5 distinct beads on a necklace.

  1. 12
  2. 24
  3. 120
  4. 60
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Answer: A

For a necklace where clockwise and counter-clockwise arrangements are identical: (n - 1)! / 2 = (5 - 1)! / 2 = 242 = 12.

9. Given that ⁿC₂ = 28, find the value of n.

  1. 8
  2. 7
  3. 9
  4. 14
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Answer: A

ⁿC₂ = nn - 12 = 28 => n(n - 1) = 56 => n² - n - 56 = 0 => (n - 8)(n + 7) = 0. Thus, n = 8.

10. A box contains 5 red balls and 4 blue balls. How many ways can 3 balls be selected such that at least 2 red balls are chosen?

  1. 50
  2. 40
  3. 10
  4. 80
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Answer: A

Case 1: 2 red, 1 blue = ⁵C₂ × ⁴C₁ = 10 × 4 = 40. Case 2: 3 red, 0 blue = ⁵C₃ × ⁴C₀ = 10 × 1 = 10. Total ways = 40 + 10 = 50.

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