Let $\mathrm{f}\left(\mathrm{n}\right)=\left|\begin{array}{ccc}\mathrm{n}& \mathrm{n}+1& \mathrm{n}+2\\ { }^{\mathrm{n}}{\mathrm{P}}_{\mathrm{n}}& { }^{\mathrm{n}+1}{\mathrm{P}}_{\mathrm{n}+1}& { }^{\mathrm{n}+2}{\mathrm{P}}_{\mathrm{n}+2}\\ { }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}& { }^{\mathrm{n}+1}{\mathrm{C}}_{\mathrm{n}+1}& { }^{\mathrm{n}+2}{\mathrm{C}}_{\mathrm{n}+2}\end{array}\right|\text{, }$where the symbols have their usual meanings. then f(n) is divisible by

$\mathrm{f}\left(\mathrm{n}\right)=\left|\begin{array}{ccc}\mathrm{n}& \mathrm{n}+1& \mathrm{n}+2\\ \mathrm{n}!& (\mathrm{n}+1)!& (\mathrm{n}+2)!\\ 1& 1& 1\end{array}\right|$
[Applying ${\mathrm{C}}_3\to {\mathrm{C}}_3-{\mathrm{C}}_2\text{ and }{\mathrm{C}}_2\to {\mathrm{C}}_2-{\mathrm{C}}_1$]

=$=\left(n+1\right)\left(n+2\right)\left|\begin{array}{ccc}\mathrm{n}& 1& 1\\ \mathrm{n}!& n.n!& \left(n+1\right).(\mathrm{n}+1)!\\ 1& 0& 0\end{array}\right|$

Expanding the determinant along with R₃ then