- #1

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## Homework Statement

In a round-robin tournament each team plays every other team once, find the number of different outcomes possible for ##n## teams.

e.g. for 4 teams the possible outcomes are:

|3-0 | 3-0 | 2-1 | 2-1

| 2-1 | 1-2 | 2-1 | 2-1

| 1-2 | 1-2 | 1-2 | 2-1

| 0-3 | 1-2 | 1-2 | 0-3

so there are 4 possible outcomes.

## Homework Equations

Combinations: ##\frac{n!}{r!(n-r)!}##

## The Attempt at a Solution

I don't know much about how to solve this question so I will just list what I do know. ##n(n-1)/2## games will take place, this gives ##2^{n(n-1)/2}## possible outcomes for the games. The total number of wins must always be equal to ##(n-1)(n-2)/2## so if it's useful the question can be turned into how many unique ways can you create ##(n-1)(n-2)/2## from ##n## digits using the digit ##(n-1)## at most once since only one team could theoretically go flawless and digits ranging from ##0\leq d < n##. Unfortunately I'm terrible at combinatorics so I really have no clue how to approach solving this.

For 2 teams:

1-0 |

0-1 |

For 3 teams:

2-0 | 1-1

1-1 | 1-1

0-2 | 1-1

If someone could point me in the right direction that would help a lot!