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).