The range of the function $f\left(x\right)={ }^{7-x}{P}_{x-3}$ is

We have $f\left(x\right)= { }^{7-x}{P}_{x-3}=\frac{\left(7-x\right)!}{\left(10-2x\right)!}$

We must have $7-x>0, x\ge 3, \mathrm{and} 7-x\ge x-3.$ Therefore,

$\begin{array}{l} x<7,x\ge 3, \mathrm{and} x\le 5\\ \mathrm{or} 3\le x\le 5\\ \mathrm{or} x=3,4,5\end{array}$

$\begin{array}{l}\mathrm{Now}, f\left(3\right)=\frac{4!}{4!}=1, f\left(4\right)=\frac{3!}{2!}=3, f\left(5\right)=\frac{2!}{0!}=2.\\ \mathrm{Hence}, R_f=\left\{1,2,3\right\}.\end{array}$