You talking about mosfets not bjt's, remove the resistor and it all goes bang
Woah, woah woah. The BJT does *not* go bang if you remove the resistor, not even close. See attached sim. The top circuit would work perfectly if the LEDs were simple 2V red LEDs, but from context I suspect these are actually LED arrays designed to run at 12V which is why it's not working. I'm going to continue under the former assumption though.
I'm not advocating this design, but let's actually model what we would expect to happen with the top circuit. Firstly, the LED/resistor pairs will never see more than 5V, so the fact that they're being powered off 12V is weird, they may was well be powered off 5V as well. If you want to fix this, the LEDs should be connected to 12V and switched from below (still with the resistors down to ground for the reason explained below). So from +12V, it goes LED, then BJT, then resistor.
That point aside, the bases of all the transistors are going to have the same voltage, because they're shorted together -- let's say 4V for the sake of argument (probably much lower if the resistor is too large, but that's not my point). Since the VBE of a transistor will stick to a narrow range around 0.7V more-or-less regardless of base current, we expect that the emitters of all the transistors will all be around 3.3V.
If the LED resistors are appropriately chosen, this is a perfect outcome, they'll all shine equally brightly regardless of transistor hFE and so on. You do not need the 10k resistor at all, the BJTs will
not blow up because the emitter voltages will rise.
In short, get rid of the base resistors entirely (just short them out) and that's actually the best thing to do, not a recipe for a bang. Assuming those LEDs are just single, naked LEDs as opposed to 12V modules.