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A Boolean function $F$ is called self dual if and only if $F(x_{1}, x_{2},.....x_{n}) = F(\bar{x}_{1}, \bar{x}_{2},....\bar{x}_{n})$. How many Boolean functions of degree $n$ are self-dual ? 

  1. $2^{n}$
  2. $(2)^{2^{n}}$
  3. $(2)^{n^{2}}$
  4. $(2)^{2^{n-1}}$
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option D.

A function is self-dual if F = FD. A boolean function with n variables is self-dual if it is a neutral function AND the terms in it are mutually exclusive. 

number of mutually exclusive pairs: 2n / 2    ->   2n-1

out of a pair I have to select 1 term, so I have 2 choices. Hence option D.

(2)2n−1

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ans is D

A function F is called self dual if it has equal number of minterms and maxterms, also mutually exclusive terms should not be included. 
the number of mutually exclusive terms is (pair wise) is 2n/2 = 2n-1
Number of functions possible by taking any of the one term from above mentioned mutually exclusive pair is =
(2) 2n-1

Answer:

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