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
A rooming house has three rooms that contains four beds, three beds and two beds respectively. In how many ways can nine quests be assigned to these rooms?
The book says it is 1260. I have no idea how they got this. I know it requires combinations. Could someone please explain it to me. Thanks a lot.
3 answers
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
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
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)