To determine how many different pairs of captains the coach can choose, we first need to identify which players are eligible based on the given criteria: being either a junior (Jr) or a senior (Sr) and having played on the team for more than 1 year.
Let's analyze the data provided in the table.
Eligible Players:
- Player 1: Sr, 1 year (not eligible)
- Player 2: Fr, 1 year (not eligible)
- Player 3: Jr, 2 years (eligible)
- Player 4: So, 2 years (not eligible)
- Player 5: So, 2 years (not eligible)
- Player 6: Sr, 3 years (eligible)
- Player 7: Sr, 4 years (eligible)
- Player 8: Jr, 3 years (eligible)
- Player 9: Fr, 1 year (not eligible)
- Player 10: Sr, 1 year (not eligible)
- Player 11: Sr, 3 years (eligible)
- Player 12: Sr, 4 years (eligible)
- Player 13: So, 2 years (not eligible)
- Player 14: Jr, 3 years (eligible)
- Player 15: Jr, 2 years (eligible)
- Player 16: Sr, 1 year (not eligible)
- Player 17: So, 2 years (not eligible)
- Player 18: Sr, 2 years (eligible)
List of Eligible Players:
- Player 3 (Jr, 2 years)
- Player 6 (Sr, 3 years)
- Player 7 (Sr, 4 years)
- Player 8 (Jr, 3 years)
- Player 11 (Sr, 3 years)
- Player 12 (Sr, 4 years)
- Player 14 (Jr, 3 years)
- Player 15 (Jr, 2 years)
- Player 18 (Sr, 2 years)
Total Eligible Players:
We find that there are 9 eligible players.
Choosing 2 Captains:
The formula for choosing 2 players from a group of 9 is given by the combination formula \( C(n, k) \), where \( n \) is the total number of items (players, in this case) and \( k \) is the number of items to choose (2 players).
\[ C(9, 2) = \frac{9!}{2!(9-2)!} = \frac{9 \times 8}{2 \times 1} = 36 \]
Final Answer:
The coach can choose 36 different pairs as captains.