ABC->DE
D->AB
Candidate keys of these FD's are ABC and CD
Number of super keys with ABC are ABC _ _ =4
Number of superkeys with CD are _ _ CD_ =8
Total number of superkeys =n(ABC U CD)=n(ABC)+n(CD)-n(ABC ∩ CD)
since ABCD ,ABCDE are common in both keys so n(ABC ∩ CD) =2
= 4+8-2 =10
Therefore total number of superkeys are 10