The first room can be filled in ways (counting eachOriginally Posted by jamesinsc
permutation as a distinct way of filling the room. Thus for each combination
we have counted 5! permutations, so the number of ways of filling the first
room is .
Similaly for the next room except that we have only the left overs to use,
so there are now .
There are no options for who is allocated to room three once the first
two rooms are filled.
So our answer is: