jmc

We have $$\begin{array}{rl}\left(\genfrac{}{}{0ex}{}{4}{0}\right)& +\left(\genfrac{}{}{0ex}{}{4}{1}\right)+\left(\genfrac{}{}{0ex}{}{4}{2}\right)+\left(\genfrac{}{}{0ex}{}{4}{3}\right)+\left(\genfrac{}{}{0ex}{}{4}{4}\right)\\ & =\frac{4!}{0!4!}+\frac{4!}{1!3!}+\frac{4!}{2!2!}+\frac{4!}{3!1!}+\frac{4!}{4!0!}\\ & =\frac{4\cdot 3\cdot 2\cdot 1}{4\cdot 3\cdot 2\cdot 1}+\frac{4\cdot 3\cdot 2\cdot 1}{1\cdot 3\cdot 2\cdot 1}+\frac{4\cdot 3\cdot 2\cdot 1}{2\cdot 1\cdot 2\cdot 1}\\ & +\frac{4\cdot 3\cdot 2\cdot 1}{3\cdot 2\cdot 1\cdot 1}+\frac{4\cdot 3\cdot 2\cdot 1}{4\cdot 3\cdot 2\cdot 1}\\ & =1+4+\frac{4\cdot 3}{2}+4+1\\ & =1+4+6+4+1\\ & =\boxed{16}.\end{array}$$Notice that this is $2^4$. There is an explanation for this: $$\left(\genfrac{}{}{0ex}{}{4}{0}\right)+\left(\genfrac{}{}{0ex}{}{4}{1}\right)+\left(\genfrac{}{}{0ex}{}{4}{2}\right)+\left(\genfrac{}{}{0ex}{}{4}{3}\right)+\left(\genfrac{}{}{0ex}{}{4}{4}\right)$$is the sum of the number of ways to choose $0$, $1$, $2$, $3$, or $4$ out of $4$ objects. This accounts for every way of choosing any number of the $4$ objects. There are $2\times 2\times 2\times 2=2^4$ ways of doing so, because there are two options for each object: choose that object or don't choose it.