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Q3). Consider the Relation $R(A,B,C,D,E)$ and $F.D's$ are $AB\rightarrow CD,CD\rightarrow E,E\rightarrow AB$ then the total number of super key are:-

(A). $10$

(B). $12$

(C). $18$

(D). None

The answer given is :

$18$ Superkeys.

${\{AB\}}^+$is key$\rightarrow$ superkey $\{ABC,ABD,ABE,ABCD,ABCE,ABDE,ABCDE\}$

$\{E \}^*$ is key$\rightarrow$ superkey $\{EA,EB,EC,,ED,EAC,EAD,EBC,EBD,ECD,EACD,EBCD\}$

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AB, CD and E are the candidate keys.

Take AB- pair it with any subset of {C, D, E} - 8 super keys. (4 including E)

Take CD- same as above- 8 super keys. But ABCD and ABCDE are counted twice. So, 6 more. (3 of them including E)

Take E- with {A, B, C, D} we get 16 super keys. But 4+3 = 7 are already counted. So, 9 more.

So, totally 8 + 6 + 9 = 23.
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