Answer
$=n\cdot(n-1)\cdot...\cdot4\cdot3=24$
Work Step by Step
If we want to choose $k$ elements out of $n$ regarding the order, not allowing repetition, we can do this in $_{n}P_k=\frac{n!}{(n-k)!}$ ways.
Hence here $_{n}P_{n-2}=\frac{n!}{(n-(n-2)!}=\frac{n!}{2!}=n\cdot(n-1)\cdot...\cdot4\cdot3=24$
