Is the following reaction allowed and why? $$ \nu_e \to e^- + \mu^+ + \nu_{\mu} $$
I would say it is allowed since individual lepton number and charge are conserved.
It is by lepton number and charge, but you can't get energy/momentum to balance. In the $\nu_e$ rest frame there isn't enough energy to make the products. If there is a nucleus around, you can imagine the $\nu_e$ emitting a virtual $W^+$ making the $e^-$, the $W^+$ scattering electromagnetically off a nucleus to deal with the momentum, then decaying into $\mu^++\nu_\mu$
Also : Lepton number conservation law, respectively Barion number conservation law, are laws which explain if any reaction can occur or not. Every particle has its own leptonic (barionic) number, and for the reaction to occur, the sum of those numbers in the right side of the equation MUST be equal to the sum of leptonic numbers in the left side of the equation. So if you have any doubt in any of these reactions, just look at the leptonic respectively barionic number of the particle.