Answer
a) $63$
b) $35$
c) $1$
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 $C(n,r)=C(n,n-r).$
a)Hence here $C(7,2)C(3,1)=\frac{7\cdot6}{2\cdot1}\frac{3}{1}=21\cdot3=63$.
b)Hence here $C(7,3)C(3,0)=\frac{7\cdot6\cdot5}{3\cdot2\cdot1}\cdot1=35\cdot1=35$.
c)Hence here $C(7,0)C(3,3)=1C(3,0)=1\cdot1=1$.
