Question

Five schools are going to send their baseball teams to a tournament in which each team must play each other team exactly once. How many games are required?

Short answer

10 games are required for each team to play every other team once.

Step-by-step solution

Understanding the Problem

We need to find the total number of games when each team plays every other team exactly once. This is a typical combination problem where we choose 2 teams out of a total of 5 to play a game.

Identify the Formula

The number of ways to choose 2 items from a set of 5 items can be calculated using the combination formula.\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]where \(n\) is the total number of teams and \(k\) is the number of teams to choose for each match.

Apply the Formula

To find the number of games, plug in \(n = 5\) and \(k = 2\) into the combination formula:\[\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10\]This means there are 10 games in total.

Verify the Solution

Verify that the computation is correct by thinking about pairing each team with every other team. Each team plays 4 other teams exactly once (since there are 5 teams total), ensuring that all possible games are counted, which confirms our result of 10 games.

Key concepts

Combination formula

In combinatorics, the combination formula is a tool used to determine the number of different ways you can choose items from a larger set without regard to the order of selection. It's particularly useful in scenarios like the one described in the exercise, where each team plays another team exactly once.

The combination formula is given as:

• \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]

Where:

• \(n\) is the total number of items.
• \(k\) is the number of items to choose.
• \(!\) (factorial) denotes the product of an integer and all the integers below it down to 1. For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1\).

In our exercise, we have 5 teams and we're choosing 2 teams for each game. Applying the formula yields 10 possible games, showing the power of combinations to simplify complex calculations involving pairs.

Discrete mathematics

Discrete mathematics is a field of study focused on mathematical structures that are countable or otherwise distinct and separable. It encompasses various topics like logic, set theory, graph theory, and combinatorics, which is what our original exercise falls under.

In discrete mathematics, problems often involve:

• Exact counting, such as the combination problem in our exercise.
• Logical analysis of finite systems, which helps us understand complex systems through simple, discrete steps.
• Using algorithms for efficient problem solving.

With its distinct, stepwise problem-solving approach, discrete mathematics is particularly applicable to fields like computer science, cryptography, and network modeling. The example of determining the number of games unique for each pair of teams in a tournament perfectly highlights the discrete nature since it involves a finite, countable number of games.

Permutation and combination

Permutation and combination are two essential concepts in discrete mathematics used to solve problems involving arrangements and selections of objects.

**Permutations** consider the order of arrangement of objects. If order matters, as in ranking participants or arranging books on a shelf, permutations are applicable. The formula for permutations is:

• \[P(n, k) = \frac{n!}{(n-k)!}\]

**Combinations**, however, are used when the order of selection doesn't matter, like choosing teams for a game where being the first or second team in the pair doesn't change the match.

• The combination formula is \(\binom{n}{k}\), as detailed in our previous section.

Using combinations simplifies solving problems where order doesn't matter, such as picking teams for games in a tournament. By focusing on sets rather than sequences, it allows for efficient counting and exploration of possibilities.