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?
Best Answer
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.