hotel occupancy counting problem The multiple-choice question:
A hotel has four vacant rooms. Each room can accommodate a maximum of four people. In how many different ways can six people be accommodated in the four rooms?
(A) 4020 (B) 4068 (C) 4080 (D) 4096.
My attempt:
If there were no 'four person' constraint, the six people could be accommodated in $4^6=4096$ ways. however, this number includes combinations with 5 or 6 people in one room, which should be excluded.
If one room has 5 occupants, one other room has 1 occupant. The number of '1,5' combinations is ${{4} \choose {2}}$=12. If one room has 6 occupants, the other three rooms have no occupants, and there are four such combinations.
I believe the answer should be $4096-12-4=4080$, but am not confident. I'd be grateful for any comments on my attempt.

Seems perfectly fine. A slightly-different way: besides the 4 ways of having $6$ people in a room, for the case of $5$-to-a-room, there are 4 ways of choosing the room for the 5 people, and 3 ways to choose where the $6$th person will stay, for a total of $4(3)$=12. So you do get $4096-4-12=4080$ , like you said.
