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There are four different attributes ,each one can not identify a tuple uniquely but a combination of two attribute(except room number and roll no) can identify a tuple uniquely so to select 2 items from four items we have 6 ways,So there are 6-1=5 candidates key.
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Consider R(Last_name, Rank, Room_number, shift) as R(ABCD)

We can derive following FD’s by the given relation

B→ C , AB- >CD, AC → BD,AD→ BC, BD ->AC,CD→ AB

if we find closure for the above , we get CK ={ AB, AC, AD, BD, CD}  

Therefore, number of Candidate keys are 5.

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