Calculating Tournament Matches Played

The question asks for the total number of matches played in a tournament where fourteen teams participate, and every team plays exactly one match against every other team.

This type of problem involves combinations because the order of the teams in a match does not matter (Team A playing Team B is the same match as Team B playing Team A). We need to find the number of ways to choose 2 teams out of the total 14 teams.

The formula for combinations of choosing $r$ items from a set of $n$ items is given by:

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

In this problem:

• Total number of teams, \(n = 14\)
• Number of teams per match, \(r = 2\)

We need to calculate the number of combinations of choosing 2 teams out of 14, which is \(\binom{14}{2}\).

\( \binom{14}{2} = \frac{14!}{2!(14-2)!} \)

\( \binom{14}{2} = \frac{14!}{2!12!} \)

Now, let's expand the factorials:

\( 14! = 14 \times 13 \times 12 \times 11 \times \dots \times 1 \)

\( 2! = 2 \times 1 = 2 \)

\( 12! = 12 \times 11 \times \dots \times 1 \)

So, the calculation becomes:

\( \binom{14}{2} = \frac{14 \times 13 \times 12!}{2 \times 1 \times 12!} \)

We can cancel out the \(12!\) from the numerator and the denominator:

\( \binom{14}{2} = \frac{14 \times 13}{2 \times 1} \)

\( \binom{14}{2} = \frac{182}{2} \)

\( \binom{14}{2} = 91 \)

Thus, a total of 91 matches will be played in the tournament where every team plays every other team exactly once.

Let's look at the options provided:

1. 66
2. 101
3. 91
4. 78

Our calculated number of matches, 91, matches option 3.

Understanding the Logic for Tournament Matches

Consider a smaller example. If there were only 4 teams (A, B, C, D) and each played every other team once:

• Team A plays B, C, D (3 matches)
• Team B has already played A, so B plays C, D (2 matches)
• Team C has already played A and B, so C plays D (1 match)
• Team D has already played A, B, and C (0 new matches)

Total matches = 3 + 2 + 1 = 6.

Using the formula with \(n=4\): \(\binom{4}{2} = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = \frac{12}{2} = 6\). The formula works.

For 14 teams, Team 1 plays 13 other teams. Team 2 has already played Team 1, so it plays 12 new teams. Team 3 plays 11 new teams, and so on, until the last team which plays 0 new teams. The total number of matches is the sum of an arithmetic series: \(13 + 12 + 11 + \dots + 1\). The sum of the first \(n-1\) positive integers is \(\frac{(n-1)n}{2}\). For \(n=14\), this is \(\frac{(14-1) \times 14}{2} = \frac{13 \times 14}{2} = 13 \times 7 = 91\). This confirms the combination formula result.

Revision Table: Tournament Match Calculation

Additional Information: Tournament Formats

Understanding different tournament formats can help clarify why combinations are used here.

• Round-Robin Tournament: In a round-robin tournament, every participant plays against every other participant once (or twice in a double round-robin). The problem describes a single round-robin format. The total number of games in a single round-robin with \(n\) teams is indeed \(\binom{n}{2}\).
• Knockout Tournament: In a knockout (or single-elimination) tournament, a loss eliminates a team. If there are \(n\) teams, there will be \(n-1\) matches played to determine a single winner. For example, with 14 teams, there would be 13 matches (7 in round 1, 4 in round 2, 2 in round 3, 1 final). This is different from the 'every team plays every other team' scenario.

The question specifically states "every team plays one match with every other team", which perfectly matches the definition of a single round-robin tournament, requiring the use of the combinations formula \(\binom{n}{2}\).