Made easy mock test 1 DBMS super keys

Question

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_
 

Comments

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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Answer may be 32256

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it is uesd here for three variable https://www.geeksforgeeks.org/dbms-number-possible-superkeys/

Answer 1

There are m simple candidate keys 

So every candidate have 2 chance : they can Present in super key  OR Absent , But at least one of in 'm'  should  present .

                      No. of chance to appear m candidate key in Superset is = (2^m)-1

                   and  all remaining  attribute 'n-m' have to chance they can present Or absent in Super key.

                   No .of chance all remaining attributes to appear in Super Key =2^(n-m)

                                         So ,     Total no of Super Key Possible=((2^m)-1)(2^(n-m))

       ( 2^6 -1)( 2^(15-6)) = 32256

Answer 2

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