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0606/22 May/Jun 2025, Question 5: Combinations: Counting Selections

Cambridge IGCSE Additional Mathematics 0606 · Paper 22 (Calculator) · Calculator · 8 marks

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The question, in short

Count 4-digit numbers from digits 0, 2, 4, 5, 6, 8 with no repeats and no leading zero; then solve an equation involving combinations.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (ai) · 1 mark

Find the total number of valid 4-digit numbers using the given digits without repetition, not starting with 0.

  1. 1First digit: 5 choices (not 0)
  2. 2Remaining digits: 5 × 4 × 3
  3. 35 × 5 × 4 × 3 = 300

Answer300

Part (aii) · 2 marks

Find how many of these 4-digit numbers are even.

  1. 1Odd numbers end in 5: first digit 4 choices (not 0 or 5), then 4 × 3
  2. 2Ending in 5: 4 × 4 × 3 = 48
  3. 3Even = 300 − 48 = 252

Answer252

Part (aiii) · 2 marks

Find how many of these 4-digit numbers are divisible by 5.

  1. 1Divisible by 5 means the last digit is 0 or 5
  2. 2Ends with 0: 5 × 4 × 3 = 60
  3. 3Ends with 5: 4 × 4 × 3 = 48
  4. 4Total = 60 + 48 = 108

Answer108

Part (b) · 3 marks

Solve (n + 1) × ⁿ⁺¹C₁₂ = 33(n − 10) × ⁿC₁₀ for n.

  1. 1ⁿ⁺¹C₁₂ = (n + 1)!/(12!(n − 11)!) and ⁿC₁₀ = n!/(10!(n − 10)!)
  2. 2(n + 1)(n + 1)!/(12!(n − 11)!) = 33(n − 10)n!/(10!(n − 10)!)
  3. 3Divide by n!/(10!(n − 11)!): (n + 1)²/(12 × 11) = 33
  4. 4(n + 1)² = 33 × 12 × 11 = 4356
  5. 5n + 1 = 66, so n = 65

Answern = 65

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Also searched as: 0606/22/M/J/25 · 0606_s25_qp_22 · 0606 s25 22 · question 5.
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