
First Room P(9,4)= 126 [9 people with 4 beds]
Second Room P(5,3)= 10 [since the first room has 4 beds, there are only 5 people left and this room contains 3 beds]
Third Room P (2,2)= 1 [7 out of the 9 people have already their own beds in room 1 and 2. Therefore, there are only 2 people left in the room that contains 2 beds.
126*10*1= 1260

consider the rooms first.
Room 1: 9*8*7*6 ways for guests to be there
Room 2: 5*4*3 ways to be there
room 1: 2 ways to be there.
Now consider the beds: 4!3!2!=ways to arrange the beds in rooms.
(multipy the first three, divide by the ways beds can be arraged.
9!/(4!3!2!)=1260

Anonymous has the best response with regards to the Nelson Math of Data Management textbook which has this exact question relating to combinations. However, instead of P(9,4) x P(5,3) x P(2,2), it is actually C(9,4) x C(5,3) x C (2,2)