Your answer is correct, but the simplest tool to use here is formation of unlabeled teams, which, written in factorial form, would be
$\frac{5!}{5!} + \frac{5!}{4!1!} +\frac{5!}{3!2!} + \frac{5!}{3!1!1!}/2! + \frac{5!}{2!2!1!}/2! + \frac{5!}{2!1!1!1!}/3! = 51$
Notice particularly the divisions where teams have identical numbers of people.
If you prefer, you can use the choose function instead, e.g.
$\binom55 + \binom54 + \binom53 + \binom53\binom21/2! +...$
but I feel that less errors will be committed using the factorial form, although it is longer, because it spells out the full details.
Re your doubts
You can use inclusion-exclusion, firstly assuming that the rooms are distinct, and then erasing their identity by dividing by $k!$ where $k$ is the nmber of occupied rooms, and you will have to do it separately for $1,2,3,4$ rooms occupied and add up.
Similarly, Stirling numbers of the second kind, of course, can be used, but again in summation form i.e. $$\sum_{k=1}^4{5\brace {k}}$$
So while both these tools are available, I wonder whether they are worthwhile, unless you have a handy table of Stirling numbers with you !
How to use PIE here
We shall compute for $1,2,3,4,5$ boxes full and add them up. Remember that we need to erase the identity of the boxes, so
$\frac{1^5}{1!} + \frac{2^5-\binom21\cdot1^5}{2!} +\frac{3^5-\binom31\cdot2^5+\binom32\cdot1^5}{3!} + \frac{4^5-\binom41\cdot3^5 +\binom42\cdot2^5 -\binom43\cdot1^5}{4!} $
which yields the answer of $\boxed{51}$
You might agree that the first method given was the simplest amd most transparent !
