Answer
$P(4, 3)=24$
Refer to the list below.
Work Step by Step
If we want to choose $r$ elements out of $n$ regarding the order, not allowing repetition, we can do this in $P(n, r)=\frac{n!}{(n-r)!}$ ways.
Hence,
$P(4, 3)=\dfrac{4!}{(4-3)!}=\dfrac{4!}{1!}=4\cdot3\cdot2\cdot1=24$
The list of all arrangements:
$123,124,132,134,142,143\\
213,214,231,234,241,243\\
312,314,321,324,341,342\\
412,413,421,423,431,432$
