Why does reducing the pressure in an equilibrium reaction favour the side with more moles of gas? I was trying to explain it in terms of reaction rates and such; more than just 'Le Chatelier's principle states ...', but I can't figure out the underlying reason as to why. I can explain why for increasing the pressure, temperature changes and such, but not for this specific one. If there is more space for the particles to move without collided, this reduces the frequency of successful collisions, reducing the reaction rate of both forward and reverse reactions. But, in $$\ce{A + B&lt;=&gt;3C}$$ 
there is initially more moles of C than A and B combined. So, isn't it reasonable to think C will collide more frequently than A and B, meaning the reaction rate of the reverse is quicker, rather than the reaction rate of the forward being quicker?
For your reaction, $\Delta n_g = 3-(1+1)=1$ (assuming all A,B,C are gases). $K_p = \frac{[C]^3}{[A][B]} (\frac{P_{total}}{\Sigma n})^{\Delta n_g}$, where $\Sigma n$ is the total number of moles of gases in the reaction mixture(even of those species which are not taking part in the reaction). Here, it is a constant.

When pressure is decreased($P_{total}$), $ (\frac{P_{total}}{\Sigma n})^{\Delta n_g}$ decreases as $\Delta n_g &gt;0$.

In order to keep $K_p$ constant (it is always constant at a given temperature), $[C]$ has to increase or $[A],[B]$ has to decrease. Therefore the reaction proceeds/shifts more in the forward direction, which has more number of gaseous moles.

Similarly, increase in pressure favours the side with lesser number of gaseous moles.