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

Find the number of all possible ways in which this can be done is:

To accommodate six people in four rooms we can have $1,1,2,2$ people in each room respectively.

So,

The total possible combinations are

$\Rightarrow \frac{6!}{2!2!}\times \frac{4!}{2!2!}=1080\left[\because n!=n\left(n-1\right)\left(n-2\right)....\right]$

Hence, the correct answer is $1080$.