10 questions · Form 5 Additional Mathematics Bab 4: Permutation and Combination
A test paper consists of 10 questions divided into Section A (5 questions) and Section B (5 questions). A candidate needs to answer 6 questions, selecting at least 2 from each section. How many different selections can be made?
Prefer reading to quizzing? All 10 questions are listed below with the answer and explanation under each one.
1. A test paper consists of 10 questions divided into Section A (5 questions) and Section B (5 questions). A candidate needs to answer 6 questions, selecting at least 2 from each section. How many different selections can be made?
Answer: A
Possible distributions (Sec A, Sec B): (2, 4) = ⁵C₂ × ⁵C₄ = 10 × 5 = 50. (3, 3) = ⁵C₃ × ⁵C₃ = 10 × 10 = 100. (4, 2) = ⁵C₄ × ⁵C₂ = 5 × 10 = 50. Total = 50 + 100 + 50 = 200.
2. Calculate the value of ⁶P₃.
Answer: A
⁶P₃ = 6! / (6 - 3)! = 6! / 3! = 6 × 5 × 4 = 120.
3. 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.
4. Calculate the value of ⁸C₅.
Answer: A
⁸C₅ = 8! / (5! × 3!) = 8 × 7 × 63 × 2 × 1 = 56.
5. How many 4-digit even numbers can be formed using the digits 1, 2, 3, 4, 5, and 6 without repetition?
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. 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. How many triangles can be formed using the vertices of an octagon?
Answer: A
An octagon has 8 vertices. Choosing 3 vertices forms a triangle: ⁸C₃ = 8 × 7 × 63 × 2 × 1 = 56.
8. If ⁿC₃ = ⁿC₅, find the value of n.
Answer: A
Using property ⁿC_r = ⁿC_(n-r): n - 3 = 5 => n = 8.
9. 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.
10. A team of 3 students is to be chosen from a group of 8 students. How many different teams can be formed?
Answer: A
Order does not matter, so use combination: ⁸C₃ = 8 × 7 × 63 × 2 × 1 = 56.