Answer
$126$
Work Step by Step
The order doesn't matter here, thus we use combinations.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 $C(2,1)C(3,2)C(7,2)=\frac{2}{1}\frac{3\cdot2}{2\cdot1}\frac{7\cdot6}{2\cdot1}=2\cdot3\cdot21=126$.
