In how many different ways can six players be arranged in a line such that two of them, Ajit and Mukherjee, are never together?

To find the number of ways Ajit and Mukherjee are never together, we subtract the number of ways they ARE together from the total possible arrangements.

Step 1: Total arrangements of 6 players The total number of ways to arrange 6 players is 6 factorial, which is 6 x 5 x 4 x 3 x 2 x 1 = 720.

Step 2: Arrangements where Ajit and Mukherjee are together Treat Ajit and Mukherjee as a single unit or block. Now we have 5 units to arrange (the block + the 4 other players). The number of ways to arrange these 5 units is 5 factorial, which is 5 x 4 x 3 x 2 x 1 = 120. Within that block, Ajit and Mukherjee can swap places in 2 ways (Ajit-Mukherjee or Mukherjee-Ajit). So, total arrangements where they are together = 120 x 2 = 240.

Step 3: Arrangements where they are never together Subtract the together arrangements from the total arrangements: 720 - 240 = 480.

Therefore, there are 480 different ways the players can be arranged such that Ajit and Mukherjee are never together. Correct Option is D.