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 _____

Total rooms available = 4 (say $$R_1 , R_2 , R_3 , R_4$$) and total persons = 6, all distinct.

Each room must have at least 1 and at most 2 persons. Hence the only possible occupancy pattern that sums to 6 is

$$2,\;2,\;1,\;1$$    (two rooms house 2 persons each, the remaining two rooms house 1 person each).

Step 1: Choose which rooms get 2 persons
Number of ways = $$\binom{4}{2}=6$$.

Suppose the chosen rooms are $$R_a$$ and $$R_b$$; the other two (with one person) are $$R_c$$ and $$R_d$$.

Step 2: Distribute the 6 distinct persons

• Choose 2 out of 6 persons for room $$R_a$$: $$\binom{6}{2}=15$$ ways.
• Choose 2 of the remaining 4 persons for room $$R_b$$: $$\binom{4}{2}=6$$ ways.
• Choose 1 of the remaining 2 persons for room $$R_c$$: $$\binom{2}{1}=2$$ ways.
• The last person automatically goes to room $$R_d$$: $$1$$ way.

Total ways for this fixed choice of rooms
$$15 \times 6 \times 2 \times 1 = 180$$.

Step 3: Combine both steps
Total admissible arrangements
$$6 \times 180 = 1080$$.

Hence, the number of ways to accommodate the six persons is 1080.