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Q.86 The number of possible boolean functions that can be defined for $n$ boolean variables over $n$-valued boolean algebra is
(a) $2^{2^n}$
(b) $2^{n^2}$
(c) $n^{2^n}$
(d) $n^{n^n}$

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7 Answers

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Best answer
$2^n$ possible combinations in truth table

All these combinations can either take value $0$ or $1$ (if a variable takes any other value, its no longer boolean)

So no of boolean func=$2^{2^n}$
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4 Comments

what does n-valued mean here ??
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Pooja I think your answer is wrong.2^2^n is the no. of boolean functions when n variables can take only two values-0 and 1, but here in the question it is specified that n variables can take n values not only two values So answer will be 2^n^n which is 2^n^2
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I am getting a bit confusion here. Lets take two propositional variables say p, q . For these the possible combinations are 4

p    q
T    T   -------------(0,1)
T    F----------------(0,1)
F   T ---------------(0,1)
F   F  --------------(0,1)

with the above mapping we are getting 8 mappings only right. But as per formula we need to get 16

please clarfify me..
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Siva Tarun   You have $4(2^{2}=4)$combinations in the truth table.

All these combinations can either take value 0 or 1.

p q f0 f1 f2 f3 f4 f5 f6 f7 f8 f9 f10 f11 f12 f13 f14 f15
0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1
0 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1
1 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1
1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

f0=0,    f1= x.y,     f2=x.y',    f3= x,    f4=x'.y,   f5=y,    f6=x xor y,    f7= x +y,   f8=(x+y)' ,   f9= x xnor y,   f10=y',   f11=x+y', f12 = x',   f13= x'+y,    f14=(x.y)',   f15=1

Number of functions = 16

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2 votes
2 votes

i think the answer would be .. n^{2^n}.

as there are 2^n possible entries in table  and  function is n valued so it will go power to n.

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option (C ) is correct.

Because the nos of combinations with n nos boolean variable is 2^n.

Again for each such combination there will n value (as compared to 0,1 as binary values).

hENCE the possible boolean function is  n*n*n*n*n-------upto 2^n times=n^2^n which is the option (C).

1 comment

If a variable can take n values who'll call it a boolean variable?
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Answer will be (a) 

Consider 2^2^n as a^b^c then we have a = 2, b = 2, c = n.

Now, remember this shortcut:

  • a represents no of variables
  • b represents type of variables
  • c represents type of functions.

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