There are $3$ different rooms and $6$ people. How many different ways are there to put the $6$ people into the $3$ rooms if each room has to have at least $1$ person?
I am not sure I am right. I figure there are $2^6-2 = 62$ different ways to put $6$ people into $2$ rooms without having either of the $2$ rooms being empty, and there are 3 different ways to put all 3 people into a single room, so for the answer I got $3^6-3(62)-3=540$. Is this right? Is there another way to do this more directly?