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 ______. [2020]

Smart Achievers · Accepted Answer

(1080)

Groups can be possible in only 2, 2, 1, 1 way.

Number of ways of dividing persons in group

$= \frac{6!}{{(2!)}^{2} {(1!)}^{2} {(2!)}^{2}}$

Number of ways after arranging rooms $= \frac{6!}{{(2!)}^{4}} \cdot 4! = 1080$

Solution

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 ______. [2020]

Ans.
(1080)

Groups can be possible in only 2, 2, 1, 1 way.

Number of ways of dividing persons in group

$= \frac{6!}{{(2!)}^{2} {(1!)}^{2} {(2!)}^{2}}$

Number of ways after arranging rooms $= \frac{6!}{{(2!)}^{4}} \cdot 4! = 1080$

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Solution

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 ______. [2020]

Ans. (1080) Groups can be possible in only 2, 2, 1, 1 way. Number of ways of dividing persons in group =6!(2!)2(1!)2(2!)2 Number of ways after arranging rooms =6!(2!)4·4!=1080

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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 ______. [2020]

Ans.
(1080)

Groups can be possible in only 2, 2, 1, 1 way.

Number of ways of dividing persons in group

$= \frac{6!}{{(2!)}^{2} {(1!)}^{2} {(2!)}^{2}}$

Number of ways after arranging rooms $= \frac{6!}{{(2!)}^{4}} \cdot 4! = 1080$