Factorials — A/L Combined Mathematics

How do I practise Factorials?

2 questions with worked solutions

Check before you answer: Rigorous 0! = 1 Justification: Applying 0! = 1 consistently in combination definitions ⁿC₀ = n! / (0! n!) = 1 and ⁿC_n = 1.

1. What is the exact numerical value of the factorial fraction 6! / (4! · 2!)?
   1. A) 20
   2. B) 15
   3. C) 30
   4. D) 12
   Check my answer

   Correct answer: B) 15

   Expanding factorials: 6! / (4! · 2!) = (6 × 5 × 4!) / (4! × 2 × 1) = (6 × 5) / 2 = 30 / 2 = 15.

   Show worked solution

   Step 1: 6! = 6 × 5 × 4!. Step 2: 6! / (4! × 2!) = (6 × 5) / 2 = 15. Answer: 15.
   Modelled on: the A/L Combined Mathematics examination pattern
2. Find the positive integer n that satisfies (n + 2)! / n! = 56.
   1. A) 6
   2. B) 8
   3. C) 5
   4. D) 7
   Check my answer

   Correct answer: A) 6

   Since (n + 2)! = (n + 2)(n + 1) · n!, the fraction (n + 2)! / n! equals (n + 2)(n + 1). Setting (n + 2)(n + 1) = 56 gives n² + 3n - 54 = 0, that is (n + 9)(n - 6) = 0. As n is a positive integer, n = 6.

   Show worked solution

   Step 1: Write (n + 2)! = (n + 2)(n + 1) · n!. Step 2: Cancel n!: (n + 2)! / n! = (n + 2)(n + 1). Step 3: Solve (n + 2)(n + 1) = 56: n² + 3n - 54 = 0, so (n + 9)(n - 6) = 0. Step 4: n must be a positive integer, so reject n = -9. Answer: n = 6.
   Modelled on: the A/L Combined Mathematics examination pattern

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Where does Factorials fit in the syllabus?

Year 2 · Term 1