Thread: combinations

25 wrestlers attend a tournament.

In each 5 wrestlers who are chosen randomly, there is at least one wrestler who played with the other four wrestlers.

How many wrestlers who play with the all wrestlers who attend the tournament can be there? (Minimum!)

There is a way for the "5 wrestlers chosen randomly -> 1 of them played each of the other four" condition to be true and have each wrestler except one play exactly 23 others, and that one wrestler play 24. So the answer is either one or zero. To expand on this, arrange all of the wrestlers in a circle and make each wrestler play any wrestler that is not adjacent in the circle. Then, going around the circle, have two adjacent wrestlers play each other, and then the next pair of adjacent wrestlers not play each other, and continue alternating in this fashion. At the end of this process, each wrestler except for one will play 23 others, except one, who will play all the other 24. Every group of five wrestlers will contain at least one wrestler that plays the other four, and there will be several groups of five in which this is true for all five wrestlers. If the number of people in the tournament was even, the answer would be zero minimum as long as at least 6 participate, by the above logic.

Dear, icemanfan!

Namely,

Is the answer 1?