In a hotel, four rooms are available. Six persons are to be accommodated in these four rooms in such

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

Professor PC Thomas Classes · Answer

(1080) : We can have 4 groups of person such that 2 groups contain 1 person each and other 2 groups contain 2 persons each.\ i.e., $6=2+2+1+1$ Number of ways to form the groups\ $=\frac{6!}{2!2!1!1!} 	imes \frac{1}{2!} \cdot \frac{1}{2!}=\frac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{2 \cdot 2 \cdot 2 \cdot 2}=45$\ They can be assigned 4 rooms in $4!$ ways. So the number of ways\ $=45 	imes 4!=45 	imes 24=1080$\