Question

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 1

Two rooms accommodate two persons each other two room have one person each.
Total ways = $\frac{6!}{(2!)^2(1!)^{2}2!2!}\times 4!=1080$