Electrons on multiple levels of energy I was asked to find the number of posibilities to find $6$ electrons in $8$ levels of energy, without restrictions, I was intuiting that this problem is totally equivalent to put $n$ balls in $k$ boxes, so in order to calculate I use 
\begin{equation}
\binom{n+k-1}{k-1}
\end{equation}
where $n$ is the number of balls (electrons) and $k$ is the number of boxes (energy levels). So with this, I found that exists 3432 possibilities to put the electrons in those levels. There is no restriction about empty levels.
Is this even right?

Your strategy is correct, but your answer has an extra factor of $2$.  
Since there are six electrons to be distributed to eight levels of energy, we want the number of solutions of the equation 
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 = 6$$
in the nonnegative integers, where $x_i$, $1 \leq i \leq 8$, is the number of electrons in the $i$th energy level.  A particular solution corresponds to the placement of seven addition signs in a row of six ones.  For instance,
$$1 1 1 + 1 1 + 1 + + + + +$$
which corresponds to the solution $x_1 = 3$, $x_2 = 2$, $x_3 = 1$, $x_4 = x_5 = x_6 = x_7 = x_8 = 0$.  The number of such solutions is the number of ways we can place seven addition signs in a row of six ones, which is 
$$\binom{6 + 8 - 1}{8 - 1} = \binom{13}{7} = 1716$$
