Pošto kod funkcije [inlmath]f\left(x_1,x_2,\ldots,x_n\right)[/inlmath] svaki od argumenata [inlmath]x_1,x_2,\ldots,x_n[/inlmath] može imati neku od dve vrednosti, [inlmath]0[/inlmath] ili [inlmath]1[/inlmath], tada je broj načina na koji se vrednosti argumenata mogu rasporediti jednak broju varijacija s ponavljanjem od [inlmath]2[/inlmath] elementa [inlmath]n[/inlmath]-te klase i iznosi [inlmath]2^n[/inlmath]. Za svaki od tih [inlmath]2^n[/inlmath] rasporeda vrednosti argumenata, funkcija može dati neku od dve moguće vrednosti, [inlmath]0[/inlmath] ili [inlmath]1[/inlmath]. Znači, broj načina na koji možemo definisati logičku funkciju jednak je broju varijacija s ponavljanjem od [inlmath]2[/inlmath] elementa [inlmath]2^n[/inlmath]-te klase, a to iznosi [inlmath]2^{2^n}[/inlmath].
Evo i ilustracije – prikazano je na koje sve načine možemo definisati logičke funkcije od [inlmath]n[/inlmath] argumenata:
[dispmath]\begin{matrix}
\left.\begin{array}{c}
f_1\left(0,0,\ldots0,0\right)=0\\
f_1\left(0,0,\ldots0,1\right)=0\\
f_1\left(0,0,\ldots1,0\right)=0\\
\vdots\\
f_1\left(1,1,\ldots1,0\right)=0\\
f_1\left(1,1,\ldots1,1\right)=0
\end{array}\;\right\}2^n & \left.\begin{array}{c}
f_2\left(0,0,\ldots0,0\right)=0\\
f_2\left(0,0,\ldots0,1\right)=0\\
f_2\left(0,0,\ldots1,0\right)=0\\
\vdots\\
f_2\left(1,1,\ldots1,0\right)=0\\
f_2\left(1,1,\ldots1,1\right)=1
\end{array}\;\right\}2^n & \left.\begin{array}{c}
f_3\left(0,0,\ldots0,0\right)=0\\
f_3\left(0,0,\ldots0,1\right)=0\\
f_3\left(0,0,\ldots1,0\right)=0\\
\vdots\\
f_3\left(1,1,\ldots1,0\right)=1\\
f_3\left(1,1,\ldots1,1\right)=0
\end{array}\;\right\}2^n\\
\\
& \vdots\\
\\
\left.\begin{array}{c}
f_{m-2}\left(0,0,\ldots0,0\right)=1\\
f_{m-2}\left(0,0,\ldots0,1\right)=1\\
f_{m-2}\left(0,0,\ldots1,0\right)=1\\
\vdots\\
f_{m-2}\left(1,1,\ldots1,0\right)=0\\
f_{m-2}\left(1,1,\ldots1,1\right)=1
\end{array}\;\right\}2^n & \left.\begin{array}{c}
f_{m-1}\left(0,0,\ldots0,0\right)=1\\
f_{m-1}\left(0,0,\ldots0,1\right)=1\\
f_{m-1}\left(0,0,\ldots1,0\right)=1\\
\vdots\\
f_{m-1}\left(1,1,\ldots1,0\right)=1\\
f_{m-1}\left(1,1,\ldots1,1\right)=0
\end{array}\;\right\}2^n & \left.\begin{array}{c}
f_m\left(0,0,\ldots0,0\right)=1\\
f_m\left(0,0,\ldots0,1\right)=1\\
f_m\left(0,0,\ldots1,0\right)=1\\
\vdots\\
f_m\left(1,1,\ldots1,0\right)=1\\
f_m\left(1,1,\ldots1,1\right)=1
\end{array}\;\right\}2^n
\end{matrix}[/dispmath]
[dispmath]\Rightarrow\quad m=2^{2^n}[/dispmath]
Evo ti i primer za logičke funkcije dve promenljive:
[dispmath]\begin{matrix}
\begin{array}{c}
f_1\left(0,0\right)=0\\
f_1\left(0,1\right)=0\\
f_1\left(1,0\right)=0\\
f_1\left(1,1\right)=0
\end{array} & \begin{array}{c}
f_2\left(0,0\right)=0\\
f_2\left(0,1\right)=0\\
f_2\left(1,0\right)=0\\
f_2\left(1,1\right)=1
\end{array} & \begin{array}{c}
f_3\left(0,0\right)=0\\
f_3\left(0,1\right)=0\\
f_3\left(1,0\right)=1\\
f_3\left(1,1\right)=0
\end{array} & \begin{array}{c}
f_4\left(0,0\right)=0\\
f_4\left(0,1\right)=0\\
f_4\left(1,0\right)=1\\
f_4\left(1,1\right)=1
\end{array}\\
\\
\begin{array}{c}
f_5\left(0,0\right)=0\\
f_5\left(0,1\right)=1\\
f_5\left(1,0\right)=0\\
f_5\left(1,1\right)=0
\end{array} & \begin{array}{c}
f_6\left(0,0\right)=0\\
f_6\left(0,1\right)=1\\
f_6\left(1,0\right)=0\\
f_6\left(1,1\right)=1
\end{array} & \begin{array}{c}
f_7\left(0,0\right)=0\\
f_7\left(0,1\right)=1\\
f_7\left(1,0\right)=1\\
f_7\left(1,1\right)=0
\end{array} & \begin{array}{c}
f_8\left(0,0\right)=0\\
f_8\left(0,1\right)=1\\
f_8\left(1,0\right)=1\\
f_8\left(1,1\right)=1
\end{array}\\
\\
\begin{array}{c}
f_9\left(0,0\right)=1\\
f_9\left(0,1\right)=0\\
f_9\left(1,0\right)=0\\
f_9\left(1,1\right)=0
\end{array} & \begin{array}{c}
f_{10}\left(0,0\right)=1\\
f_{10}\left(0,1\right)=0\\
f_{10}\left(1,0\right)=0\\
f_{10}\left(1,1\right)=1
\end{array} & \begin{array}{c}
f_{11}\left(0,0\right)=1\\
f_{11}\left(0,1\right)=0\\
f_{11}\left(1,0\right)=1\\
f_{11}\left(1,1\right)=0
\end{array} & \begin{array}{c}
f_{12}\left(0,0\right)=1\\
f_{12}\left(0,1\right)=0\\
f_{12}\left(1,0\right)=1\\
f_{12}\left(1,1\right)=1
\end{array}\\
\\
\begin{array}{c}
f_{13}\left(0,0\right)=1\\
f_{13}\left(0,1\right)=1\\
f_{13}\left(1,0\right)=0\\
f_{13}\left(1,1\right)=0
\end{array} & \begin{array}{c}
f_{14}\left(0,0\right)=1\\
f_{14}\left(0,1\right)=1\\
f_{14}\left(1,0\right)=0\\
f_{14}\left(1,1\right)=1
\end{array} & \begin{array}{c}
f_{15}\left(0,0\right)=1\\
f_{15}\left(0,1\right)=1\\
f_{15}\left(1,0\right)=1\\
f_{15}\left(1,1\right)=0
\end{array} & \begin{array}{c}
f_{16}\left(0,0\right)=1\\
f_{16}\left(0,1\right)=1\\
f_{16}\left(1,0\right)=1\\
f_{16}\left(1,1\right)=1
\end{array}
\end{matrix}[/dispmath]
Dakle, za [inlmath]n=2[/inlmath] imamo ukupno [inlmath]2^{2^2}[/inlmath], tj. [inlmath]16[/inlmath] mogučih funkcija.
Imali smo nedavno
vrlo sličan zadatak, samo ne s logičkim funkcijama, već s binarnim relacijama – tamo se u opštem slučaju dobije malko drugačiji rezultat, [inlmath]2^{n^2}[/inlmath]. Možeš i njega pogledati.