Let  $f\left(x\right)=x^2-3x+3$ and $x_1,\text{\hspace{0.33em}}{x}_2,x_3,x_4$  be the solutions of equation $f\left(f\left(x\right)\right)=x$ , then the number of arrangements of  $x_1,\text{\hspace{0.33em}}{x}_2,x_3$ and $x_4$  taken all at a time is 

$f\left(f\left(x\right)\right)=x$$\therefore 3\&1arerootsoff\left(x\right)=xalso$$f\left(x\right)=x\Rightarrow x^2-4x+3=0\Rightarrow x=3or1$

$\Rightarrow {\left(x^2-3x+3\right)}^2-3\left(x^2-3x+3\right)+3=x$

$\Rightarrow x^4-6x^3+12x^2-10x+3=0$

$\therefore 3\&1\text{\hspace{0.33em}}areroots\text{}off\left(x\right)=x,theyarerootsoff\left(f\left(x\right)\right)=xalsoand$

$\therefore f\left(f\left(x\right)\right)-x=\left(x-3\right)\left(x-1\right){\left(x-1\right)}^2$$=\left(x-3\right){\left(x-1\right)}^3$

$\therefore 3,1,1,1\text{\hspace{0.33em}}aresolutionsoff\left(f\left(x\right)\right)=x$