League Matches Calculation

The problem asks for the total number of matches when 5 distinct teams play each other exactly once in a league format. This is a problem of combinations, as the order of teams playing does not matter (Team A vs Team B is the same match as Team B vs Team A).

Determining Matches Needed

We need to find the number of ways to choose 2 teams out of 5 to play a match. The formula for combinations is:

$ C(n, k) = \binom{n}{k} = \frac{n!}{k!(n-k)!} $

Where:

• $n$ is the total number of teams ($n=5$).
• $k$ is the number of teams participating in one match ($k=2$).

Applying the Combination Formula

Substitute the values into the formula:

$ C(5, 2) = \binom{5}{2} = \frac{5!}{2!(5-2)!} $

$ C(5, 2) = \frac{5!}{2!3!} $

Calculate the factorials:

$ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 $

$ 2! = 2 \times 1 = 2 $

$ 3! = 3 \times 2 \times 1 = 6 $

Now, substitute the factorial values back into the formula:

$ C(5, 2) = \frac{120}{(2)(6)} = \frac{120}{12} = 10 $

Alternative Method: Summation

Alternatively, we can list the matches systematically:

• Team 1 plays against Teams 2, 3, 4, 5 (4 matches).
• Team 2 has already played Team 1, so it plays against Teams 3, 4, 5 (3 matches).
• Team 3 has already played Teams 1 and 2, so it plays against Teams 4, 5 (2 matches).
• Team 4 has already played Teams 1, 2, and 3, so it plays against Team 5 (1 match).
• Team 5 has already played all other teams.

Total matches = $4 + 3 + 2 + 1 = 10$.

Conclusion

Both methods confirm that 10 matches are required to complete the league round where 5 teams play each other exactly once.