Hi, can someone show me how to approach this problem-

Incomplete table:
Team A - won 4 games
Team B - won 7 games
Team C - won 9 games
Team D - won 2 games
Team E - won 2 games
Team X - ?

if each of the 6 teams in the league played each of the other teams exactly twice and there were no ties, how many games did team C win? (only 2 teams play in a game.)

Count down the list:

Team A played 10 games in total (2 games with each of the other 5 teams)
Team B played 8 games. (Don't count the games it played with A again)
Team C played 6.

In this way, you arrive at 30 games played total.
Count the number of wins (the data supplied adds up to 24 games won).
Since no games were a draw, team X must have won the remaining games. Which comes to (30 - 24) = 6.

Hope this helps!

Guest, another approach to calculate the number of games played is to approach this as a combinatorics problem. There are six teams, and we are solving for the number of combinations that include two of the teams. Thus, for each team to play each other once, there are 6!/2!4! or 15 combinations (i.e. games). Simply double this number to calculate the number of games for each team to play each of the other teams twice.

Then, simply subtract 24 (the total number of games played by teams other than X) from 30 (the total number of games played) and 6 is the correct result.

-dan