group of $12$ people go into hotel room problem with $3$ of them in a room A group of $12$ people including Nicole, Jennifer and Caitlyn are going to stay in a camp. There are $3$ different rooms in the camp, where each room can accommodate up to $4$ people. If Nicole, Jennifer and Caitlyn have to stay in the same room, how many ways are there to allocate this group of people to the rooms?
Attempt 1:
The three of them have to be in the same room so there so no choice, then pick one from the remaining nine. Of the remaining eight, pick $4$ of them using combinations. The last $4$ have no choice.
Then the three rooms can interchange, so $6$ arrangements.
$C_3^3 \cdot C_1^9 \cdot C_4^8 \cdot C_4^4 \cdot P_3^3 = 3780$
Attempt 2:
From the nine people, pick $4$ to go into a room.  Then from the $5$ remaining, pick another $4$ to go into a room.  The last one go with the trio, no choice.  Then interchange the rooms.
$C_4^9 \cdot C_4^5 \cdot P_3^3 = 3780$
Both attempts give me same answer, but the suggested answer says $1890$. Could anyone tell me why please?
 A: Notice that $1890 \times 2 = 3780$, and this is exactly what you double-count in the question!
The reasoning in both attempts are very sound, but you misunderstood the way that the rooms are assigned (it is a little bit ambiguous in the question, actually)
Indeed, the rooms can "interchange", but a better formulation is that the rooms are indistinguishable.
So say Nicole, Jennifer and Caitlyn are in room 1, then the remaining two groups of 4 people, (call them group $A$ and $B$) are in room 2 and room 3 respectively. It doesn't make any difference if group $A$ is in room 3, group $B$ is in room 2.
But in both of your attempts, you treat the above case as two ways of assigning them, although, in reality, this could only count as one way.
This applies to every case, that's why your answer is exactly 2 times the answer provided.
A: There are three ways to assign Nicole, Jennifer, and Caitlyn to the same room and nine ways to select their roommate.  Of the remaining eight people, there are two ways to assign a room to the youngest person.  There are $\binom{7}{3}$ ways to select the roommate of that person.  The other four people must be assigned to the remaining room.  Hence, the number of distinguishable room assignment which can be made if Nicole, Jennifer, and Caitlyn are all assigned to the same room is
$$\binom{3}{1}\binom{9}{1}\binom{2}{1}\binom{7}{3} = 1890$$
When you assigned four of the remaining eight people to a room, you counted each assignment twice, once when you assigned a group to a room and once when you assigned its complement to the other room.  For instance, suppose Nicole, Jennifer, Caitlyn, and the oldest of the remaining people have been assigned to room 1.  You would get the same distribution of people if you assigned the four youngest remaining people to room 2 as you would if you assigned the four oldest of the remaining people to room 3.
A: Choose one room among $3$ rooms for the mentioned people by $C(3,1) $,then now select $1$ people among the $9$ to stay with them in same room by $C(9,1)$. After that ,  choose $4$ people among the remaining $8$ people for one of the remaining rooms and the rest is for the other remaining room by $C(4,4)$. I think you confused for placing the rest $8$ people , you do not need to select any room for them , just select $4$ of them and place a random room.
Hence ,answer is $C(3,1) \times C(9,1) \times C(8,4) \times C(4,4)=3 \times 9 \times 70 \times 1 = 1890$
