Have to prove in a combinatorial and with pascal's pyramid I have to prove that:
$$\sum_{k=0}^{(\lfloor \frac{n}{2} \rfloor)} { n \choose 2k} = 2^{n-1}$$
in two ways: A combinatorial way, and with the help of the Pascal Pyramid.
For the combinatorial way I thought about possibilities of choosing a set with an even number of elements, but the $$n \choose 0$$ ruins it all!
Any help will be appreciated

Your combinatorial argument is right, both sides of the equation count the number of subsets of $\{1,2,\ldots,n\}$ of even cardinality.  Note that $\binom{n}{0}$ counts the empty subset.  You should also probably add an argument as to why $2^{n-1}$ is the number of subsets of even cardinality.
As for Pascal's Triangle (Pyramid?), I don't want to spoil it for you, but here's an illustration that indicates the relevant points:

(There is a slight difference between the $n$ even and $n$ odd cases.)  Hopefully this provides sufficient inspiration.  The relevant identity is $$\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}.$$
