This kind of problem is solved using permutation without repetition. This means the order matters and an item may only be used once.

In the question, we have to arrange six persons in such a way that two of them, Ajit and Mukherjee, are never together. We can consider Ajit and Mukherjee as one unit, that leaves us with 5 units (A&M, and the four other players) to arrange, which can be arranged in 5! (5 factorial) ways, or 120 ways. Because Ajit and Mukherjee can also arrange among themselves in 2! (2 factorial) ways, we multiply the two results to get our final answer.

Option 1: 120 ways, suggests only the arrangement of five units (still considering Ajit and Mukherjee as one unit) without considering the arrangement between Ajit and Mukherjee.

Option 2: 240 ways, is the result of slightly miscalculated multiplication.

Option 3: 360 ways, is another incorrect multiplication of the two factorials.

Option 4: 480 ways, is indeed correct because it correctly multiplies the 5! (120 ways) and 2! (2 ways), getting 240.

Therefore