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 ______

$4 \text{ rooms},\qquad 6 \text{ persons}$

$\text{each room: at least } 1,\text{ at most } 2 \text{ persons}$

$\text{To find: the number of ways to accommodate all } 6 \text{ persons}$

$x \text{ rooms have } 2,\qquad 4-x \text{ have } 1$

$2x+(4-x)=6\ \Rightarrow\ x=2$

$\text{so 2 rooms of } 2 \text{ and 2 rooms of } 1$

$\binom{4}{2}=6\ \text{ways to pick the two double rooms}$

$\text{line up the } 6 \text{ persons}: 6! \text{ orders}$

$\text{first 2} \to \text{room 1},\ \text{next 2} \to \text{room 2},\ \text{last 2} \to \text{rooms 3,4}$

$\dfrac{6!}{2!\,2!}\ \text{ways to fill the rooms (rooms fixed)}$

$\dfrac{6!}{2!\,2!}=\dfrac{720}{4}=180$

$6\times 180=1080$

$\text{Answer: } 1080$