Statistics Question: (combination/permutation)

A certain baseball team has 21 players. Only nine can be on the field at a time. Each of the nine players on the field has a distinct field position: pitcher, catcher, first baseman, second baseman, third baseman, shortstop, left field, right field, or center field. Assume for the moment that every player is qualified to play every position.

1.) How many ways are there to assign field positions to a given set of nine players?

9!/(3! x (9-3)!) = 84

3 field positions, 9 players.

I got that... but after rereading the question, I'm now utterly confused... I'm not sure if the formula I used is correct now... also, I didn't take into account that every player can play every position when I did this problem last night.

Thanks!

the q asks for assignments for a *given* set of 9 players,

so the simple answer is 9P9 = 9! = 362,880 <------

the data given about the total pool of 21 is only to confuse !