10 questions · Form 5 Additional Mathematics Bab 4: Permutation and Combination
10 questions · plays in fullscreen
Prefer reading to quizzing? All 10 questions are listed below with the answer and explanation under each one.
1. Find the value of n if ⁿP₂ = 42.
Answer: A
ⁿP₂ = n(n - 1) = 42 => n² - n - 42 = 0 => (n - 7)(n + 6) = 0. Since n must be positive, n = 7.
2. 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?
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.
3. Find the number of ways to arrange 6 people around a circular table.
Answer: A
Circular permutation = (n - 1)! = (6 - 1)! = 5! = 120.
4. If ⁿC₃ = ⁿC₅, find the value of n.
Answer: A
Using property ⁿC_r = ⁿC_(n-r): n - 3 = 5 => n = 8.
5. Calculate the value of ⁸C₅.
Answer: A
⁸C₅ = 8! / (5! × 3!) = 8 × 7 × 63 × 2 × 1 = 56.
6. Given that ⁿC₂ = 28, find the value of n.
Answer: A
ⁿC₂ = nn - 12 = 28 => n(n - 1) = 56 => n² - n - 56 = 0 => (n - 8)(n + 7) = 0. Thus, n = 8.
7. 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.
Answer: A
Order matters, so use permutation: ⁷P₃ = 7 × 6 × 5 = 210.
8. Find the number of ways to arrange 5 distinct beads on a necklace.
Answer: A
For a necklace where clockwise and counter-clockwise arrangements are identical: (n - 1)! / 2 = (5 - 1)! / 2 = 242 = 12.
9. In how many ways can the letters of the word 'VECTOR' be arranged such that the vowels are always together?
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.
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?
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.