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Question

In a three-storey building, there are four rooms on the ground floor, two on the first and two on the second floor. If the rooms are to be allotted to six persons, one person occupying one room only, the number of ways in which this can be done so that no floor remains empty is


A
8P62(6!)
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B
8P6
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C
8P5(6!)
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D
None of these
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Solution

The correct option is A $$^{8}P_6 - 2(6!)$$
Total number of allocation methods $$= ^{ 8 }{ P }_6$$
Among these, there would be cases where a floor may be empty. We can only keep the 1st or 2nd floors empty. Thus, there are 2 cases.
In that case, the remaining 6 rooms are all occupied in $$^{ 6 }{ P }_6$$ ways. 
Thus, the answer is $$^{ 8 }{ P }_6 - 2\times ^{ 6 }{ P }_6$$$$=^{8}P_{6} - 2(6!)$$.
Hence, (A) is correct.

Mathematics

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