Maths olympiad question · P5–P6★★★★ Expert
Combinations: olympiad question 11
An expert question on combinations for P5–P6, with a step-by-step solution and the answer. Counting arrangements and choices appears in the Hua Xia Cup from the P1 and P2 first rounds to the P5 finals.
The question
8 pupils, Ming and May among them, reach the final of an inter-school maths contest.
There are 1 gold, 1 silver and 4 bronze awards; each pupil gets one award at most, and awards of the same kind are alike.
Their teacher expects Ming to win one (gold, silver or bronze).
How many different results are possible?
Show the hints (3)
- Hint 1: Ignore Ming first: choose the gold winners, then silver from those left, then bronze. Awards of a kind are alike, so the chosen ones aren't put in order.
- Hint 2: "Ming wins" is hard to count directly. What is its opposite?
- Hint 3: All 840, less the results where Ming gets nothing: every award goes to the other 7.
Worked solution
Have a go first, then read the solution step by step.
- 1.Without the condition: gold, 1 of 8: 8; silver, 1 of the 7 left: 7; bronze, 4 of the 6 left: 6 × 5 × 4 × 3 ÷ (4 × 3 × 2 × 1) = 15. In all 8 × 7 × 15 = 840.
- 2.The opposite: Ming gets nothing, so the awards go among the other 7: 7 × 6 × 5 = 210.
- 3.Wanted = 840 − 210 = 630.
- ⚠️Common mistake: treating awards of a kind as different (a first and second silver), multiplying by orders; or missing cases when adding "Ming gold / silver / bronze" directly.
Answer: 630 results
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