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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

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

Full Question List & Answer Key

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

1. 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.

2. Calculate the value of ⁶P₃.

  1. 120
  2. 20
  3. 720
  4. 18
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Answer: A

⁶P₃ = 6! / (6 - 3)! = 6! / 3! = 6 × 5 × 4 = 120.

3. 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.

4. Find the number of signals that can be made using 3 flags of different colors placed one above the other from a set of 7 available flags.

  1. 210
  2. 35
  3. 5040
  4. 840
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Answer: A

Order matters, so use permutation: ⁷P₃ = 7 × 6 × 5 = 210.

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

  1. 180
  2. 360
  3. 120
  4. 60
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Answer: A

The last digit must be even (2, 4, or 6): 3 choices. The remaining 3 digits are chosen from the 5 remaining numbers: ⁵P₃ = 60. Total = 3 × 60 = 180.

6. Find the number of ways to arrange all letters in the word 'SUCCESS'.

  1. 420
  2. 5040
  3. 840
  4. 210
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Answer: A

Total letters = 7, with S repeating 3 times and C repeating 2 times. Permutations = 7! / (3! × 2!) = 50406 × 2 = 420.

7. 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.

8. If ⁿC₃ = ⁿC₅, find the value of n.

  1. 8
  2. 15
  3. 2
  4. 10
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Answer: A

Using property ⁿC_r = ⁿC_(n-r): n - 3 = 5 => n = 8.

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. In how many ways can the letters of the word 'VECTOR' be arranged such that the vowels are always together?

  1. 240
  2. 120
  3. 720
  4. 480
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Answer: A

Vowels in 'VECTOR' are E and O (2 vowels). Group (E, O) as 1 unit. Units to arrange: (EO), V, C, T, R = 5 units. Arrangements of 5 units = 5! = 120. Internal arrangements of E and O = 2! = 2. Total = 120 × 2 = 240.

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