# (Enumeration) In how many ways can 5 people be placed in 8 rooms

In how many ways can 5 people be put into 8 rooms if only 2 of the people (very dubious characters) can’t share a room with anyone? Note that there is no maximum capacity for the rooms.

My attempt: 2 separated people = $P(8,2) = 56$,

6 rooms remaining for the other people to choose from,

Case 1: Remaining people: 2 in one room and one in the other = $P(6,2) = 30$,

Case 2: Remaining people all separated = $P(6,3) = 120$,

Total number of ways = $(6 + 30 + 120) \times 56 = 8736$

• In case 1, the cases are not symmetric, you've selected two rooms, but there are two ways to make this arrangement, 2 people in the "first" room and 1 in the other room, or the other way around. Also, which two people are put into which room? – Michael Burr Oct 15 '15 at 21:53

You're right up to the $56$, but then you made some mistakes, as Michael pointed out in a comment. At this point you just have to put each of the remaining $3$ people independently into one of the remaining $6$ rooms, so the total number is $56\cdot6^3=12096$.