0606/13 May/Jun 2026, Question 7: Combinations: Counting Selections
Cambridge IGCSE Additional Mathematics 0606 · Paper 13 (Non-calculator) · Non-calculator · 4 marks
اقرأ هذا الحل مشروحاً بالعربي ←
The question, in short
Given that 20Cr equals 4/3 times 20C(r+1), find the value of r.
Summarised in our own words. Use the official paper for the exact wording and any diagrams.
Solution · 4 marks
- 1²⁰C_r = (4/3) × ²⁰C_(r+1)
- 220!/(r!(20 − r)!) = (4/3) × 20!/((r + 1)!(19 − r)!)
- 31/(20 − r) = (4/3) × 1/(r + 1)
- 43(r + 1) = 4(20 − r), 7r = 77
- 5r = 11
Answerr = 11
Stuck on a step? Read the free lesson: Combinations: Counting Selections
Practise more combinations: counting selections questions
Real 0606 past-paper questions, marked on your working like the exam, with a worked solution for every part.
Practise free on LemmaraMore past-paper questions on combinations: counting selections
Also searched as: 0606/13/M/J/26 · 0606_s26_qp_13 · 0606 s26 13 · question 7.
Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education. The worked solutions are Lemmara’s own.