Answer
$4845$
Work Step by Step
f we want to choose $k$ elements out of $n$ disregarding the order, not allowing repetition, we can do this in $_{n}C_k=\frac{n!}{(n-k)!k!}$ ways. T
The order doesn't matter here when choosing actors, and we choose $4$ out of $20$, thus we have to use combinations.
Thus the number of possibilities is: $_{20}C_{4}=\frac{20!}{(20-4)!4!}=\frac{20\cdot19\cdot18\cdot17}{4\cdot3\cdot2\cdot1}=4845$
