Question Number 228102 by mr W last updated on 21/Mar/26
$${there}\:{are}\:\mathrm{32}\:{students}\:{in}\:{a}\:{class}.\:{for} \\ $$$${each}\:{competition}\:{in}\:{a}\:{sport}\:{event}\: \\ $$$${in}\:{the}\:{school}\:{each}\:{class}\:{can}\:{send} \\ $$$${a}\:{team}\:{with}\:{three}\:{students}.\:{if}\:{no} \\ $$$${two}\:{students}\:{may}\:{be}\:{in}\:{the}\:{same} \\ $$$${team}\:{for}\:{more}\:{than}\:{one}\:{time},\:{in} \\ $$$${how}\:{many}\:{different}\:{competitions}\: \\ $$$${can}\:{this}\:{class}\:{participate}? \\ $$
Answered by vnm last updated on 28/Mar/26
![There are a total of 32×31/2=496 pairs of students and no more than [496/3] =165 teams can be formed from these pairs. Since the number of the students is even there will be at least one other student for each student with whom they will not be in the same team and the minimum number of pairs of students who will not be in the same team is 32/2=16. Therefore the maximum number of pairs from which teams can be formed is 496−16=480 and the maximum possible number of teams is 480/3=160 (may be less than 160, but 160 is an upper limit). All that remains is to give an example of how this can be done. This problem has already been posted here on 22/6/25 where I demonstrated the solution.](https://www.tinkutara.com/question/Q228124.png)
$$ \\ $$$$\mathrm{There}\:\mathrm{are}\:\mathrm{a}\:\mathrm{total}\:\mathrm{of}\:\mathrm{32}×\mathrm{31}/\mathrm{2}=\mathrm{496}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{students}\:\mathrm{and}\:\mathrm{no} \\ $$$$\mathrm{more}\:\mathrm{than}\:\left[\mathrm{496}/\mathrm{3}\right]\:=\mathrm{165}\:\mathrm{teams}\: \\ $$$$\mathrm{can}\:\mathrm{be}\:\mathrm{formed}\:\mathrm{from}\:\mathrm{these}\:\mathrm{pairs}.\:\mathrm{Since}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{the}\:\mathrm{students}\:\mathrm{is}\:\mathrm{even}\:\mathrm{there}\:\mathrm{will}\:\mathrm{be}\: \\ $$$$\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{other}\:\mathrm{student}\:\mathrm{for}\:\mathrm{each}\:\mathrm{student}\:\mathrm{with}\:\mathrm{whom}\:\mathrm{they}\:\mathrm{will}\:\mathrm{not}\:\mathrm{be}\:\mathrm{in}\:\mathrm{the}\:\mathrm{same}\:\mathrm{team} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{number}\:\mathrm{of} \\ $$$$\mathrm{pairs}\:\mathrm{of}\:\mathrm{students}\:\mathrm{who}\:\mathrm{will}\:\mathrm{not} \\ $$$$\mathrm{be}\:\mathrm{in}\:\mathrm{the}\:\mathrm{same}\:\mathrm{team}\:\mathrm{is}\:\mathrm{32}/\mathrm{2}=\mathrm{16}.\: \\ $$$$\mathrm{Therefore}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{number}\:\mathrm{of}\:\mathrm{pairs}\:\mathrm{from}\:\mathrm{which}\:\mathrm{teams}\:\mathrm{can}\:\mathrm{be}\:\mathrm{formed}\:\mathrm{is}\: \\ $$$$\mathrm{496}−\mathrm{16}=\mathrm{480}\:\mathrm{and}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{possible}\:\mathrm{number}\:\mathrm{of}\:\mathrm{teams}\:\mathrm{is}\:\mathrm{480}/\mathrm{3}=\mathrm{160} \\ $$$$\left({may}\:{be}\:{less}\:{than}\:\mathrm{160},\:{but}\:\mathrm{160}\:{is}\:{an}\:{upper}\:{limit}\right). \\ $$$$\mathrm{All}\:\mathrm{that}\:\mathrm{remains}\:\mathrm{is}\:\mathrm{to}\:\mathrm{give}\:\mathrm{an}\:\mathrm{example}\:\mathrm{of}\:\mathrm{how}\:\mathrm{this}\:\mathrm{can}\:\mathrm{be}\:\mathrm{done}. \\ $$$$\mathrm{This}\:\mathrm{problem}\:\mathrm{has}\:\mathrm{already}\:\mathrm{been}\:\mathrm{posted}\:\mathrm{here}\:\mathrm{on} \\ $$$$\mathrm{22}/\mathrm{6}/\mathrm{25}\:\mathrm{where}\:\mathrm{I}\:\mathrm{demonstrated} \\ $$$$\mathrm{the}\:\mathrm{solution}. \\ $$