1. **State the problem:** We want to find the number of ways to arrange two games of doubles tennis from a group of eight players. 2. **Understand the problem:** Each doubles game requires 4 players, split into 2 teams of 2 players each. We need to arrange two such games, so all 8 players are used. 3. **Step 1: Choose players for the first game.** Since all 8 players are used, the first game will have 4 players and the second game will have the remaining 4 players. The number of ways to choose 4 players out of 8 is given by the combination formula: $$\binom{8}{4} = \frac{8!}{4!\times(8-4)!} = \frac{8!}{4!4!}$$ 4. **Calculate the combination:** $$\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$$ 5. **Step 2: Arrange teams within each game.** For each group of 4 players, we split them into 2 teams of 2 players each. The number of ways to split 4 players into two teams of 2 is: - Choose 2 players for the first team: $$\binom{4}{2} = 6$$ - The remaining 2 players form the second team automatically. 6. **Account for indistinguishable teams:** Since the two teams are indistinguishable (Team A vs Team B is the same as Team B vs Team A), we divide by 2 to avoid double counting: $$\frac{6}{2} = 3$$ 7. **Calculate total arrangements:** - Number of ways to choose players for the first game: 70 - Number of ways to split first game players into teams: 3 - Number of ways to split second game players into teams: 3 Multiply all together: $$70 \times 3 \times 3 = 630$$ 8. **Final answer:** There are **630** ways to arrange two games of doubles tennis from a group of eight players.