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Question

A database relation has10 attributes {A,BCD,E,F,J} “CDEF” is known to be a candidate key of relation . There may be other candidate keys, which are unknown. What is the maximum number of candidate keys that could simultaneously have?

Answer 1

In total it has 21 $C_k$

Comments

can I take 1  attribute of c,d,e,f, and 1 attribute of a,b,g,h,i,j for forming ck

then 1 attribute with any 2 attributes of the other set

I think more ck will be possible for this

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This answer is correct.

As consider R(A,B,C,D,E) and we have to find out the maximum no. of candidate keys that could simultaneously have.

For this there is a formula : nCn/2, but here a CK is given already, i.e. CDEF. So the remaining attributes are {A,B,G,H,I,J}.

Now applying the formula for n=6, #simultaneous CKs = 20.

So maximum no. of candidate keys that could simultaneously have = 20+1 = 21.

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No bro correct answer is 247

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Max no of candidate keys possible with 10 keys are 10c5=252. Since cdef is a ck so some of these 5 cks will be superkeys,so we have to remove them.

So combinations will cdef with remaining 6 attributes. So 6 need to be removed.

So ans is 252-6+1(for cdef since it is a c]candidate key).

ans is 247

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20 and 21 are wrong. I have posted the correct answer. Pls check if anyone has doubt.

Answer 2

As we know the maximum number of simultaneous candidate keys can be given by the set having exactly n/2 attributes, where n is the no. of attributes present in the relation.

So according to the above statement, max no. of simultaneous CKs possible = nCn/2.

Here n = 10, nCn/2 = 252.

But here the given relation has CK : {CDEF}. The cardinality of the CK is 4. If we add any one of the remaining attributes (A,B,G,H,I,J), it’ll give us a Super Key, of cardinality 5. The possible Super Keys,of cardinality 5, are : { ACDEF, BCDEF,GCDEF,HCDEF,ICDEF,JCDEF}.

So all the above mentioned keys are SKs not CKs. Therefore we gotta remove these.

Hence, (252 – 6) + 1 = 246 + 1 = 247 (1 is added for the given CK i.e. CDEF).