How many distinct arrangements are possible using the letters of (PERMUTATION)?

Correct answer: (19958400)

The word contains eleven letters: P, E, R, M, U, T, A, T, I, O, and N. Every letter occurs once except T, which occurs twice. When repeated letters are present, simply using \\(11!\\) would count the same visible arrangement twice, because interchanging the two identical T letters does not create a new arrangement. We must therefore divide by \\(2!\\).

The number of distinct arrangements is \\(\\frac{11!}{2!}\\). Since \\(11!=39,916,800\\) and \\(2!=2\\), the result is \\(19,958,400\\). Thus option A is correct. The division is needed only for the repeated T letters; all other letters are distinct. This reasoning counts each different arrangement exactly once rather than counting identical copies separately.