1. ## counting problem

How do I show that 8 people played in the tournament to the following problem:
At a chess tournament, each person played exactly 1 game with every other person. 28 games were played. How many people played in the tournament?

2. Hello sarahh
Originally Posted by sarahh
How do I show that 8 people played in the tournament to the following problem:
At a chess tournament, each person played exactly 1 game with every other person. 28 games were played. How many people played in the tournament?
If there are $\displaystyle n$ players in the tournament, and each players plays 1 game against everyone else, then the number of different games is the number of ways of choosing $\displaystyle 2$ from $\displaystyle n$; which is:
$\displaystyle \binom{n}{2} = \tfrac12n(n-1)$
Equate this to $\displaystyle 28$; solve the quadratic equation in $\displaystyle n$, and ignore the negative root.

3. Hi Grandad--that certainly works! I'm not sure how you deduced the formula though.. I though of doing the ordered pairings but if you didn't know it was 8 it would be hard to deduce--is there a more elementary way of solving it too?

4. Hello sarahh
Originally Posted by sarahh
...I'm not sure how you deduced the formula though..
If everyone plays everyone else, then all possible pairs of players must be included in the set of games. So we simply need the number of possible pairs that can be formed from the available players.

If there are $\displaystyle n$ players, the number of ways of choosing the first player to make a pair to play one another is $\displaystyle n$; and, since there are then $\displaystyle (n-1)$ players left to choose from, there are $\displaystyle (n-1)$ ways of choosing his/her opponent. So the number of possible ways of choosing two players in order is $\displaystyle n(n-1)$.

But the order doesn't matter: (A, B) is the same match as (B, A). Since there are $\displaystyle 2$ ways of arranging the two players, each pair in the $\displaystyle n(n-1)$ choices will occur twice. So we must divide by $\displaystyle 2$ to get the number of different matches.

This is simply what $\displaystyle \binom{n}{2}$, or $\displaystyle {^nC_2}$, means: the number of ways of choosing $\displaystyle 2$ from $\displaystyle n$. And, of course, the formula for this is:
$\displaystyle \binom{n}{2}=\frac{n!}{2!(n-2)}$
$\displaystyle =\tfrac12n(n-1)$
I though of doing the ordered pairings but if you didn't know it was 8 it would be hard to deduce--
I don't think I've assumed it's $\displaystyle 8$, have I? I let it be $\displaystyle n$, and then set up an equation, the solution of which was $\displaystyle n = 8$.
is there a more elementary way of solving it too?
Trial and error, perhaps? Start with 2 players: that's 1 match. Then 3 players: 3 matches. And so on... until you get 28 matches.