Given $3$ children and $3$ different rooms, what's the probability that two children are placed in one room and the third is placed in another? Given 3 different children and 3 different rooms, what's the probability of randomly putting 2 children in one room, 1 in the second one, and 0 in the third one, provided that each room can contain 0 to 3 children? According to my calculation, the sample space is 324; the required probability would, therefore, be 18/324. Thanks
 A: The first child can be put in any room. The second child can be put in the same room, so that the third child must be put in one of the remaining rooms, or can be put in one of the two other rooms, so that the third child must be put in one of the previously used rooms. As such, the probability of putting two children in the same room and a third child in a different room, equals:
$$\frac{3 \cdot 1 \cdot 2 + 3 \cdot 2 \cdot 2}{3^3} = \frac{18}{27}$$
For completeness, one can also calculate the probability for the other cases. The probability of putting the three children in three different rooms equals:
$$\frac{3!}{3^3} = \frac{6}{27}$$
The probability of putting all children in the same room equals:
$$\frac{3 \cdot 1 \cdot 1}{3^3} = \frac{3}{27}$$
Indeed, $\frac{18 + 6 + 3}{27} = 1$.
A: The randomly means the probabilities are equally likely.
Without loss of generality we may assume the first child $(C1)$ is in the first room $(R1)$.
The success cases:
$$P(\color{blue}{C2R1}\cap C3R2|\color{blue}{C1R1})+P(\color{blue}{C2R1}\cap C3R3|\color{blue}{C1R1})+$$
$$P(C2R2\cap \color{red}{C3R1}|\color{red}{C1R1})+P(\color{red}{C2R2}\cap \color{red}{C3R2}|C1R1)+$$
$$P(C2R3\cap \color{green}{C3R1}|\color{green}{C1R1})+P(\color{green}{C2R3}\cap \color{green}{C3R3}|C1R1)=6\cdot \frac19=\frac23.$$
