Formula to calculate $(1n)+(2(n-1))+(3(n-2)+(4(n-3)) ...$ How would I go about calculating something like $(1\times4)+(2\times3)+(3\times2)+(4\times1)$? It is the equivalent of $(4+3+2+1)+(3+2+1)+(2+1)+(1)$.

You may prove by induction on $n$ that
$$1\cdot n+ 2\cdot(n-1)+\ldots+(n-1)\cdot 2+n\cdot 1=\sum_{k=1}^{n} k\cdot(n+1-k) =\binom{n+2}{3}$$
holds. In your case, $n=4$, then answer is so given by $\binom{6}{3}=\color{red}{20}$.
