how many who complete w/ all others can be at tournament?

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kingdomofthenight

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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!)
 
Re: RELATIVELY-URGENT HELP ...

Hello Night Kingdom:

If I correctly understand this, then there cannot be more than four players in the group of 25 who have not each played with all of the others.

If there were, then it would be possible that a random selection would choose five players who have not each played with all of the others. This event would violate the given condition that at least one of the five has played with all of the others.

Does it make sense that the minimum asked for would be the difference between 25 and 4?

:!: Consider carefully because I'm not sure that I correctly understand the exercise. :idea:

~ Mark :)
 
Re: URGENT HELP !

Hello mmm4444bot,

No, I don't think so.
You hadn't understand the question.

I think that the answer must be between 1 and 10.
 
Re: URGENT HELP !

I think the answer is 1.

Pair all the wrestlers up (except #25). Every wrestler plays every other wrestler *except* their partner. Since we select at random 5 wrestlers (an odd number), there must be at least one wrestler without their partner in the 5 - in which case, he will have played everyone else.

Agree?
 
Re: URGENT HELP !

kingdomofthenight said:
Every wrestler plays every other wrestler *except* their partner.
Click to expand...

Oh, I did not realize that the exercise regards a tournament schedule.

I was thinking that the exercise described a relation between wrestlers that exists before they come together. (Good grief! I am so dense, sometimes. :lol: )

I am now confident that I cannot assist you.

(I should stay off the Statistics board. :oops: )

Somebody else around here will most likely comment on your reasoning.

Cheers,

~ Mark :)
 
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