Maths olympiad question · P5–P6★★★★ Expert
Inclusion–exclusion: olympiad question 8
An expert question on inclusion–exclusion for P5–P6, with a step-by-step solution and the answer. Common in the Hua Xia Cup's P3 to P6 qualifying rounds and finals; harder ones have three overlapping groups, or ask who is in exactly two.
The question
A school has 130 pupils and 4 after-school activities (a pupil may join several, or none): basketball 62, choir 61, chess 115, science 86.
At least how many pupils must be in three or more activities?
Show the hints (3)
- Hint 1: Add the activities' numbers: 324 places in all. If nobody took more than two, how many places could there be at most?
- Hint 2: A pupil in three or more can be in up to 4. Say x pupils are; everyone else takes at most two.
- Hint 3: 130 pupils at two each give only 260 places, 64 short. Each pupil in all 4 gives 2 more than two.
Worked solution
Have a go first, then read the solution step by step.
- 1.The activities add to 62 + 61 + 115 + 86 = 324 places.
- 2.If nobody took three or more, 130 pupils would give at most 130 × 2 = 260 places — 64 short of 324.
- 3.Say x pupils take three or more, at most 4 each — at most 2 more than two. So 2 × x ≥ 64, x ≥ 64 ÷ 2, so x is at least 32.
- 4.And 32 is possible: let these 32 take (nearly) every activity, covering the extra 64 places, and spread the rest two or fewer each to fill every activity — no activity has more than 130, so it can be arranged.
- 5.So at least 32 pupils are in three or more.
- ⚠️Common mistake: assuming such pupils take exactly three, getting 64; they can take up to 4, each making up 2 places, so fewer are needed.
Answer: 32 pupils
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