Accommodation of Six Persons into Four Rooms

To find the total number of distinct ways to accommodate $6$ distinct persons into $4$ distinct hotel rooms such that each room contains at least $1$ person and at most $2$ persons, we can model the situation mathematically.

Let $x_i$ represent the number of persons accommodated in room $i$ for $i \in \{1, 2, 3, 4\}$.

According to the given conditions:

1. Every room must contain at least $1$ person and at most $2$ persons: $1 \le x_i \le 2 \quad \text{for each } i \in \{1, 2, 3, 4\}$
2. The total number of persons is $6$: $x_1 + x_2 + x_3 + x_4 = 6$

Since $x_i \in \{1, 2\}$, the only partition of $6$ into $4$ parts satisfying $1 \le x_i \le 2$ is: $\{x_1, x_2, x_3, x_4\} = \{2, 2, 1, 1\}$

This means exactly two rooms will accommodate $2$ persons each, and the other two rooms will accommodate $1$ person each.

Step 1: Select the rooms according to their capacities

Choose $2$ out of the $4$ available rooms to accommodate $2$ persons each. The remaining $2$ rooms will automatically accommodate $1$ person each. $\text{Number of ways to choose the rooms} = \binom{4}{2} = 6$

Step 2: Assign the 6 persons to the rooms

Having fixed which two rooms have capacity $2$ and which two rooms have capacity $1$:

1. Choose $2$ persons out of $6$ for the first room with capacity $2$: $\binom{6}{2} = 15$
2. Choose $2$ persons out of the remaining $4$ for the second room with capacity $2$: $\binom{4}{2} = 6$
3. Choose $1$ person out of the remaining $2$ for the first room with capacity $1$: $\binom{2}{1} = 2$
4. Assign the remaining $1$ person to the last room with capacity $1$: $\binom{1}{1} = 1$

Step 3: Compute the total number of ways

Multiplying the number of choices from each step, we get: $\text{Total Ways} = \binom{4}{2} \times \binom{6}{2} \times \binom{4}{2} \times \binom{2}{1} \times \binom{1}{1}$

$\text{Total Ways} = 6 \times 15 \times 6 \times 2 \times 1 = 1080$

Alternatively, we can divide the $6$ distinct persons into $4$ groups of sizes $2, 2, 1, 1$: $\text{Number of ways to form groups} = \frac{6!}{(2!)^2 (1!)^2 \cdot 2! \cdot 2!} = \frac{720}{4 \times 1 \times 4} = 45$

Then, arrange these $4$ distinct groups among the $4$ distinct rooms: $\text{Total Ways} = 45 \times 4! = 45 \times 24 = 1080$

Thus, the total number of all possible ways is 1080.