Does the permutation of the objects within the three positions matter? If so, then you multiply the number of positions that the first object could be in (12), by the number of positions the second object could be in (11), and so on, until you've run out of objects. If differing permutations within the three used positions aren't counted as unique, then you must divide it by n! in order to erase the non-unique permutations, where n is the number of objects being placed.