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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 ___________.

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answer is 1080.

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Detailed Solution

Number of persons in rooms will be 2, 2, 1, 1.
So, required number of ways $= \frac{6!}{2! 2! 1! 2! 2!} \times 4! = 1080$

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