The permutation is defined as:
[tex]_nP_r=\frac{n!}{(n-r)!}[/tex]Then in this case we have:
[tex]\begin{gathered} _{16}P_4=\frac{16!}{(16-4)!} \\ =\frac{16!}{12!} \\ =\frac{16\cdot15\cdot14\cdot13\cdot12!}{12!} \\ =16\cdot15\cdot14\cdot13 \\ =43680 \end{gathered}[/tex]Therefore the answer is D.
