From the Levi-Civita symbol wikipedia page, in four-dimensions the object $\epsilon_{abcd}\epsilon^{\alpha\beta\rho\sigma}$ yields: $$ \epsilon_{\alpha\beta\rho\sigma} \epsilon^{abcd} \ = \ \det\left( \left[ \begin{matrix} \delta^{\mu}_{\ a} & \delta^{\mu}_{\ b} & \delta^{\mu}_{\ c} & \delta^{\mu}_{\ d} \\ \delta^{\nu}_{\ a} & \delta^{\nu}_{\ b} & \delta^{\nu}_{\ c} & \delta^{\nu}_{\ d} \\ \delta^{\rho}_{\ a} & \delta^{\rho}_{\ b} & \delta^{\rho}_{\ c} & \delta^{\rho}_{\ d} \\ \delta^{\sigma}_{\ a} & \delta^{\sigma}_{\ b} & \delta^{\sigma}_{\ c} & \delta^{\sigma}_{\ d} \end{matrix} \right] \right) $$
So if I want to compute the contraction $\epsilon_{abcd}\epsilon^{ab\rho\sigma}$ I know that I'd have something like: $$ \epsilon_{ab\rho\sigma} \epsilon^{abcd} \ = \ \sum_{a=1}^{4} \sum_{b=1}^{4} \det\left( \left[ \begin{matrix} \delta^{a}_{\ a} & \delta^{a}_{\ b} & \delta^{a}_{\ c} & \delta^{a}_{\ d} \\ \delta^{b}_{\ a} & \delta^{b}_{\ b} & \delta^{b}_{\ c} & \delta^{b}_{\ d} \\ \delta^{\rho}_{\ a} & \delta^{\rho}_{\ b} & \delta^{\rho}_{\ c} & \delta^{\rho}_{\ d} \\ \delta^{\sigma}_{\ a} & \delta^{\sigma}_{\ b} & \delta^{\sigma}_{\ c} & \delta^{\sigma}_{\ d} \end{matrix} \right] \right) $$
Writing the above as a determinant is a really gross thing - and then I'd rather not take 16 sums when I do the contraction.
Surely, there has to be simple way to see what this object becomes? Using the symmetries of $\epsilon$ somehow? I'm wondering what kind of tricks I can employ in order to avoid doing the above computation in brute force.
From that same Wikipedia page $$\epsilon_{ab\rho\sigma} \epsilon^{abcd} = 2!\delta^{cd}_{\rho\sigma} = 2!\epsilon_{\rho\sigma}\epsilon^{cd}$$