n∙(n-1)∙(n-2)∙...∙(n-k+1) = $\frac{\text{n!}}{\text{(n-k!)}}$
$\left(\genfrac{}{}{0ex}{}{\mathrm{n}}{\mathrm{k}}\right)$ = $\frac{\text{n!}}{\text{k! (n-k!)}}$

6 MATH PRB 2:nPr 4


6 MATH PRB 3:nCr 4