+1 vote
202 views

Consider the following relation R(A1, A2,...A15) with (A1,A2, ... A6) of relation R are simple candidate key. The number of possible superkey in relation R is_

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0
we have to apply " principle of Mutual exclusion "
0

That's will be very difficult to apply here.

+5

we know n simple candidate key forms $2^{n}-1$(for this -https://gateoverflow.in/290395/gatebook-2019-grand-test-dbms-4) super keys and here n=6,

$2^{6}-1$= 63

and if we'll add other 9(non prime attributes) elements in these keys then it will also form super key.

so $63*2^{9}=32256$

Made easy gave this solution and it's quite logical.

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0

(2^6-1)total cand key possible *2^9.
by (333 points)