Super Keys

Question

How many Super keys Possible for R(A,B,C,D,E) with

1. {A,BC,DE} as the keys ?

2. {A,BC,CDE} as the keys ?

Comments

For first part if i take A as a single bit BC and DE as two different bits Total possiblity = 2^3 = 8

i need to subtract 1 case where everybody will be absent = 8 - 1 = 7   is it correct ?

For second i am confused

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1)23

2) 21

yes

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1. I don't know answer of first one

2. Answer given is 21

Answer 1

1) R (A,B,C,D,E)

 Candidate key :  {A,BC,DE}

  = super-key(A) + super-key(BC) + super-key(DE) - super-key(ABC) - super-key(ADE) - super-key(BCDE) + super-key(ABCDE)

  = 2⁴ + 2³ + 2³ -2²-2² -2¹ + 2⁰

  = 23

 

2)

R (A,B,C,D,E)

   Candidate key :  {A,BC,CDE}

  = super-key(A) + super-key(BC) + super-key(CDE) - super-key(ABC) - super-key(ACDE) - super-key(BCDE) + super-key(ABCDE)

  = 2⁴ + 2³ + 2² -2² - 2¹ -2¹ + 2⁰

  = 21

 

Comments

See the basic concept is this ..if A is a candidate key then how many Super keys are possible ??

you can choose {B C D E} in 2 x 2 x 2 x 2 = 16 ways right ??

if BC is a candidate key then {A D E }  in 2 x 2 x 2 = 8 ways

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Magma in first one cant i consider A,BC and CDE as three bits so total cases are 8 i subtract 1 here so 7 (because where every bit is 0 so phie which is not super key )

Why this is wrong ?