Answer
$302400$.
Work Step by Step
If we have two groups with $n$ and $m$ objects and we choose $r(r\leq n)$ and $s(s\leq m)$ objects from the two groups where the order doesn't matter then the number of combinations is: $C(n,r)C(m,s).$ (Similarly for more than $2$ groups.)
We know that $C(n,r)=\frac{n(n-1)(n-2)...(n-k+1)}{r!}$. Also $C(n,0)=1$ by convention. Also $C(n,r)=C(n,n-r).$
Hence here $C(10,5)C(10,3)C(5,3)=\frac{10\cdot9\cdot8\cdot7\cdot6}{5\cdot4\cdot3\cdot2\cdot1}\frac{10\cdot9\cdot8}{3\cdot2\cdot1}\frac{5\cdot4\cdot3}{3\cdot2\cdot1}=252\cdot120\cdot10=302400$.
