Propositional function doubt

Question

The Number of n-variable propositional function  ?

a. 2^2^n.

b.2^n^2

c.2^n

d. n^2

Can any one please explain the what should be the answer and how to get it?

Answer 1

With n variable, we can have $2^n$ outputs as each output can be true or false.

Now we need to see how many functions can be made out of $2^n$ outputs.

 Let us say m=$2^n$ // number of outputs

Now with m ,I can have 2^m functions as each value of m can be 0 or 1.

So $2^m" = 2^2^n, a is the answer

Comments

actually this answer is wrong for the question asked. if the question would have asked " Number of non equivalent propositional function" than  the answer is true. but there are way too many propositional functions possible for n variables (check for just 2 variables with connectives(and,or,imnplies,double implies and negation it would give around 20). thus if Fn is the number of propositional function for n propositional variables then Fn=big omega(2^2^n).