Answer
$450$
Work Step by Step
If 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.
The order doesn't matter here in the questions, thus we have to use combinations. Thus $_{10}C_{8}\cdot_{5}C_{3}=\frac{10!}{(10-8)!8!}\frac{5!}{(5-3)!3!}=45\cdot10=450$
