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

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  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Two rooms accommodate two persons each other two room have one person each. Total ways (6 !/(2 !)2(1 !)2 2 ! 2 !) × 4 !=1080

Q. 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 ______

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

Q. 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 ______

Two rooms accommodate two persons each other two room have one person each. Total ways (2!)2(1!)22!2!6!​×4!=1080

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