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.

The count of $5+1$ needs to be updated.  You have $4$ ways to choose the room that $5$ will be in, $3$ ways to choose the room for $1$, and $6$ ways to choose the lone traveler.  So the number of ways to accommodate the guests is $4096-4-4\cdot 3 \cdot 6=4096-76=4020$
