In a hotel , four rooms are available , Six persons are to be accommodated in these four rooms in such a way that each of these rooms contains at least one person and at most two persons . Then the number of all possible ways in which this can be done is ______

Answer

To solve the problem of accommodating 6 persons in 4 rooms such that each room has at least 1 person and at most 2 persons, we can follow these steps:

### Step 1: Determine the distribution of persons in rooms
Since we have 4 rooms and need to accommodate 6 persons, the only valid distribution that satisfies the conditions (at least 1 person and at most 2 persons in each room) is:
- 2 rooms with 2 persons each
- 2 rooms with 1 person each

### Step 2: Choose the rooms for 2 persons
We need to select 2 out of the 4 rooms to accommodate 2 persons. The number of ways to choose 2 rooms from 4 is given by the combination formula:
\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \]

### Step 3: Distribute persons in the selected rooms
Now, we need to distribute the 6 persons into the rooms. We first choose 2 persons to go into the first selected room (which can hold 2 persons). The number of ways to choose 2 persons from 6 is:
\[ \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \]

After placing 2 persons in the first room, we have 4 persons left. We then choose 2 persons from these 4 to go into the second selected room:
\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \]

### Step 4: Assign the remaining persons to the other rooms
Now, we have 2 persons left, and they will each go into the remaining 2 rooms (1 person per room). The number of ways to assign these 2 persons to the 2 remaining rooms is:
\[ 2! = 2 \]

### Step 5: Calculate the total arrangements
Now we can multiply all the ways together:
\[ \text{Total arrangements} = \text{Ways to choose rooms} \times \text{Ways to choose persons for first room} \times \text{Ways to choose persons for second room} \times \text{Ways to assign remaining persons} \]
\[ = 6 \times 15 \times 6 \times 2 \]
\[ = 6 \times 15 = 90 \]
\[ 90 \times 6 = 540 \]
\[ 540 \times 2 = 1080 \]

Thus, the total number of ways to accommodate the 6 persons in the 4 rooms is **1080**.