Solution
The different orders is given as
[tex]_nP_r=_{14}P_3[/tex]
This is simplified as
[tex]\begin{gathered} _{14}P_3=\frac{14!}{(14-3)!} \\ =\frac{14!}{11!} \end{gathered}[/tex]
This further give
[tex]\frac{14!}{11!}=\frac{14\times13\times12\times11!}{11!}=2184[/tex]
Therefore the answer is 2184
