The coach can choose the batting lineup in approximately 6.49 x 10^3 ways.
To solve this, we need to consider the different cases based on the number of juniors in the lineup:
Case 1: No juniors (0 juniors):
The coach can choose any 9 players from the remaining 13 (4 freshmen + 4 sophomores + 4 seniors + 1 junior).
Combinations of 9 players from 13: 13C9 = 715
Case 2: One junior (1 junior):
The coach can choose 1 junior in 5 ways.
For the remaining 8 spots, the coach can choose any 8 players from the remaining 12 (4 freshmen + 4 sophomores + 4 seniors).
Combinations of 8 players from 12: 12C8 = 495
Total ways with 1 junior: 5 * 495 = 2475
Case 3: Two juniors (2 juniors):
The coach can choose 2 juniors in 10 ways (5C2).
For the remaining 7 spots, the coach can choose any 7 players from the remaining 11 (4 freshmen + 4 sophomores + 3 seniors).
Combinations of 7 players from 11: 11C7 = 330
Total ways with 2 juniors: 10 * 330 = 3300
Total ways with no more than 2 juniors:
715 + 2475 + 3300 = 6490
Therefore, the coach can choose the batting lineup in approximately 6.49 x 10^3 ways.