Sum of products of binomial coefficients

It is easy to see that the given sum is going to be the coefficient of the $x^{2016}$ in the expansion of the product $(1+x)^{505}(1+x)^{505}(1+x)^{505}(1+x)^{505}.$ Additionally, $(1+x)^{2020}$ has the same expanded form as this product. Therefore the given sum is equal to $\binom{2020}{2016}=\boxed{691\:677\:274\:345}.$

There are $\displaystyle 4$ types of objects each having $\displaystyle 505$ entities.

We need to select certain number of entities from all the types such that the total number of entities selected is $\displaystyle 2016$ . This is the same as selecting $\displaystyle 2016$ entities out of these $\displaystyle 4 \times 505$ entities.

$\displaystyle \Rightarrow { 2020 \choose 2016} = \boxed{691,677,274,345}$

If the sum of upper indices and lowrer indices are constant the the summation converges to the Sum of the upper indices and Sum of lower indices as the Upper and lower indices.