How many boolean functions $F(x, y, z)$

Question

Question:
How many boolean functions $F(x, y, z)$ are there so that $F(\bar{x}, y, z) = F(x, \bar{y}, z) = F(x, y, \bar{z})$ for all values of the Boolean variables $x, y,$ and $z$?

I'm at loss on where to start.

Answer 1

Given : $F(\bar{x}, y, z) = F(x, \bar{y}, z) = F(x, y, \bar{z})$

complementing $y/\bar{y}$ & $z/\bar{z}$ we get,

$F(\bar{x}, \bar{y}, \bar{z}) = F(x, y, \bar{z}) = F(x, \bar{y}, z)$

merging both equation we get,

$F(\bar{x}, y, z) = F(x, \bar{y}, z) = F(x, y, \bar{z})=F(\bar{x}, \bar{y}, \bar{z})$

similarly,

$F(x, \bar{y}, \bar{z}) = F(\bar{x}, y, \bar{z}) = F(\bar{x}, \bar{y}, z)=F(x, y, z)$

That means there are only 2 independent entries in the truth table.

Hence, total no. of different boolean functions $ = 2*2 = 4 $

Answer 2

HINT

Once you know

1. $F(1,0,0)$, then you know the value of $F(0,1,0)$ and $F(0,0,1)$.
2. $F(1,0,1)$, then you know the value of $F(0,1,1)$ and $F(0,0,0)$.

Likewise think of all the other values that will be obtained in this way. Now bear in mind that $F$ can take only two possible values at each input.