# Logical question (no variations, permutations or combinations)

## Homework Statement

How many contestants have on one chess tournament, if every person have played only one game with all of the other contestants separately, and there are 210 games played.

This problem should not be solved by variations, permutations or combinations. This problem should be solved by using the quadric equation. (we should put some condition, and out of there make some quadric equation).

## The Attempt at a Solution

I have no idea.

x- the number of players

I really don't know.

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The answer just popped up into my head. Don't know how to explain it

$$\sum^x_{N=1} N-1 = 210$$ x is the number of participants

Last edited:
HallsofIvy
Homework Helper
The answer just popped up into my head. Don't know how to explain it

$$\sum^x_{N=1} N-1 = 210$$ x is the number of participants
Then how do you know it is the answer?

If there are n contestants then, for each one, there are n- 1 "other contestants" and so each plays n-1 games. n people, each playing n-1 games, means there are n(n-1) games- except that each game involves two players: there are actually n(n-1)/2 games played. Since there were a total of 210 games, n(n-1)/2= 210.

(Which is, by the way, exactly what armis is saying!)

Oh, exactly. Thanks HallsofIvy

I understand. Thank you.