How Many Games in a Tournament: Solving the 7-Player Challenge and 10-Team Scenario

The question of how many games will take place in a tournament where each player plays every other player once is a classic example used to illustrate the concept of combinations. In this article, we'll delve into the logic and mathematics behind this problem, first with a 7-player tournament, then with a 10-team scenario. We will use combination formulas to solve these problems accurately and efficiently.

7-Player Tournament

Consider a tournament with 7 players. Each player must play every other player exactly once. To find the total number of games, we use the combination formula, which is specifically designed for such scenarios. The formula for combinations is given by:

( binom{n}{2} frac{n(n-1)}{2} )

In this context, ( n ) represents the number of players, which is 7. Plugging in the value, we get:

( binom{7}{2} frac{7(7-1)}{2} frac{7 times 6}{2} 21 )

Therefore, the total number of games played in this 7-player tournament is 21 games.

10-Team Tournament

For a tournament with 10 teams, the calculation is similar. Each team plays against every other team exactly once. Let's break it down step by step:

Team 1 plays 9 matches (against teams 2 through 10). Team 2 plays 8 matches (excluding the match against team 1, and against teams 3 through 10). Team 3 plays 7 matches (excluding matches against teams 1 and 2, and against teams 4 through 10). Team 4 plays 6 matches (excluding matches against teams 1 through 3, and against teams 5 through 10). Team 5 plays 5 matches (excluding matches against teams 1 through 4, and against teams 6 through 10). Team 6 plays 4 matches (excluding matches against teams 1 through 5, and against teams 7 through 10). Team 7 plays 3 matches (excluding matches against teams 1 through 6, and against teams 8 and 9, and team 10). Team 8 plays 2 matches (excluding matches against teams 1 through 7, and against team 9, and team 10). Team 9 plays 1 match (against team 10). Team 10 does not play any further matches.

Adding up the total number of matches, we get: 9 8 7 6 5 4 3 2 1 45. However, since every match is counted twice (once for each team), the actual number of matches is:

( frac{10 times 9}{2} 45 )

This can also be derived using the combination formula:

( binom{10}{2} frac{10(10-1)}{2} frac{10 times 9}{2} 45 )

Thus, the total number of games played in a 10-team tournament, where each team plays every other team exactly once, is 45 games.

Combinatorial Approach

The combination formula provides a more straightforward method to solve these problems. The formula ( binom{n}{r} ) is defined as:

( binom{n}{r} frac{n!}{r!(n-r)!} )

In the context of a 10-team tournament, we are choosing 2 teams out of 10 to play a match:

( binom{10}{2} frac{10!}{2!(10-2)!} frac{10 times 9}{2 times 1} 45 )

This confirms that the total number of games is 45.

Conclusion

Both the step-by-step method and the combinatorial approach effectively solve the problem of determining the number of games in a tournament where each player or team plays every other exactly once. For 7 players, the answer is 21 games, and for 10 teams, the answer is 45 games.

Reference Materials

Combinations and Permutations (Math is Fun)

Stack Exchange: How Many Games in a Tournament where Every Team Plays Every Other Team Once?