I want to check my answer about: distributing $4$ people into $3$-distinct rooms, and each room can contain any number of people.
First consider the rooms indistinguishable, and list the size in decreasing order for each possible rooms summary(e.g. 4,0,0 means there is one room assigned 4 people, the others empty), and count the combination for each:
$$\begin{align} 4,0,0&: C(4,4) &=1\\ 3,1,0&: C(4,3)C(1,1) &=4\\ 2,2,0&: C(4,2)C(2,2)/2! &=3\\ 2,1,1&: C(4,2)C(2,1)C(1,1)/2!. &=6\\ \end{align}$$
Then compute the permutation each:
$$\begin{align} 4,0,0&: C(4,4)\cdot 3!/2! &=1\cdot 3\\ 3,1,0&: C(4,3)C(1,1)\cdot 3! &=4\cdot 6\\ 2,2,0&: C(4,2)C(2,2)/2!\cdot 3!/2! &=3\cdot 3\\ 2,1,1&: C(4,2)C(2,1)C(1,1)/2!\cdot 3!/2!. &=6\cdot 3\\ \end{align}$$
Is the answer 3+24+9+18=54 correct?
If both the rooms and the people are distinguishable, then each person has 3 rooms to choose from, so the answer is $3^4=81$.
Your mistake is the $/2!$ in the last two lines of the first calculation. Or, specifically, that when you suddenly make a distinction between the rooms, you don't take them away. For instance, you don't make a distinction between $$ ab, c,d $$ and $$ ab,d,c $$ in the final calculation (where $a,b,c,d$ are the people and $a$ and $b$ share room).
Similarly, you don't make a distinction between $$ ab,cd, $$ and $$ cd,ab, $$ This change makes the sum into $3+24+18+36=81$.