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

We have four rooms and six people.
By grouping, $6\to (1,1,2,2)$
$\therefore$ Required number of ways
$=\frac{6!}{1!1!2!2!2!2!}\times 4!$
$=\frac{720}{2\times 2\times 2\times 2}\times 24$
$=1080$