Let $E_{2p,m}= \Big\{(x_1, x_2, \ldots, x_{2p}): x_d \in \{0,1,2, \ldots, m\} \mbox{ and } \sum_{d=1}^{2p} \epsilon_dx_d=0 \Big\}$, where $\epsilon_d \in \{-1, +1\}$. Then what is the cardinality of the set $E_{2p,m}$.
Hints: Generating function technique (comparing the coefficient).
By direct counting, we have the following: $\# E_{4,1}=8$ and $\# E_{4,2}=33$, and set $E_{4,1}$ is
$$\{(0,0,0,0), (0,0,1,1), (0,1,0,1), (1,0,0,1), (0,1,1,0), (1,0,1,0), (1,1,0,0), (1,1,1,1) \}.$$
$\textbf{Some clarification regarding $\epsilon$ set:}$ Note that $E_{2p,m}$ is the set of sequences $(x_1,\ldots, x_{2p})$ such that there exists a sequence $(\epsilon_1, \epsilon_2, \ldots, \epsilon_{2p}) \in \{-1,+1\}^{2p}$ such that $\sum_{d=1}^{2p} \epsilon_d x_d=0$.
Observe that the $\epsilon$-set depends on the x-sequence. But if for two different (or more than two) $\epsilon$ sets, we are getting same $x$-sequence, then we will count that only one time.
For example: In the set $ E_{4,1}$, for the element $(0,0,1,1)$, we have many $\epsilon$ sets; $(1,1,1,-1)$, $(1,1,-1,1)$, $(-1,-1,-1,1)$, $\ldots$ such that the $\sum_{d=1}^{4} \epsilon_d x_d=0$. But we will count $(0,0,1,1)$ only one time in $\# E_{4,1}$.