# Most likely distribution derivation

Given a large number N of distinguishable particles distributed among M boxes, we know that the total number of possible microstates is $$M^N$$ and that the number of microstates with a distribution among the boxes given by the configuration $$[n_1, n_2, ..., n_M]$$ is given by $$\Omega= \frac{N!}{\Pi_{j=1}^M n_j!}$$ I need to show that the most likely distribution sees the particles equally distributed among the M boxes. I know I need to use the Lagrangian variation of parameters, as the number of particles $$N$$ is constant. By doing so I got that $$\ln(n_j)+\alpha =0$$ where $$\alpha$$ is my Lagrangian parameter, for all $$j$$. I just do not know how to get $$n_j=\frac{N}{M}$$ from here... Is my reasoning wrong?

You're essentially done. You know that $$n_j$$ is constant in $$j$$; from the constraint that $$\sum_j n_j=N$$ you get $$n_j=N/M$$. That's the beauty of Lagrange multipliers!
• Oh that's easier than I thought! Just to confirm, I get the condition that $n_j$ is constant in $j$ through the $ln(n_j)+\alpha =0$ equation, right? Apr 2 at 20:39
• Yes, exactly $\log n_j=-\alpha$ for all j Apr 2 at 21:24