I have the following double product $$ \prod_{a=0}^{3}\prod_{b=0}^{3-i} \Big((p-b)u + (q-a)v\Big) $$ and it does not matter what these variable really are. I want to analytically expand it so I first choose $a=0$ since $a$ can run from $0$ to $3$. Then I get $$ \prod_{b=0}^{3-0} \Big((p-b)u + (q-0)v\Big) = \Big((p-0)u + qv\Big)\Big((p-1)u + qv\Big)\Big((p-2)u + qv\Big)\Big((p-3)u + qv\Big) $$ Is this correct way to expand this product? Then for the next term, when I choose $a=1$, I have $$ \prod_{b=0}^{2} \Big( (p-b)u+(q-1)v \Big)=\Big( (p-0)u+(q-1)v \Big)\Big( (p-1)u+(q-1)v \Big)\Big( (p-2)u+(q-1)v \Big) $$ Then for the next term I choose $a=2$ and the product becomes $$ \prod_{b=0}^{1} \Big( (p-b)u+(q-2)v \Big)= \Big( (p-0)u+(q-2)v \Big)\Big( (p-1)u+(q-2)v \Big) $$ and finally I can choose $a=3$ to get $$ \prod_{b=0}^{0} \Big( (p-b)u+(q-3)v \Big)$$ and I think that this product is 1-term but I am not sure. It makes no sense to me that $b$ runs from $0$ to $0$. I think it means set $j=0$ and that's it.
So I would just like your help in order to confirm this expansion is correct and I would like to know if there exists a more efficient way to make such expansions.