Energy distribution

The number of permutations for a given choice of occupation numbers is $Z(n_1,n_2,n_3) = \frac{N!}{n_1! \cdot n_2! \cdot n_3!}$ Therefore, the numbers of microstates are $\begin{aligned} Z(2,7,0) &= 36 \\ Z(3,5,1) = Z(5,1,3) &= 504 \\ Z(4,3,2) &= 1260 \end{aligned}$ The most probable distribution $(n_1,n_2,n_3) = (4,3,2)$ thus accounts for 55% of all possible microstates.

If we assume a Boltzmann distribution, the occupation numbers results $\begin{aligned} & \quad n_i = N \frac{e^{-E_i/k T}}{\sum_{i = 1}^3 e^{-E_i/k T}} = 9 \frac{e^{-i/3}}{\sum_{i = 1}^3 e^{-i/3}} \\ \Rightarrow & \quad (n_1,n_2,n_3) \approx (4.04, 2.89, 2.07) \end{aligned}$ which comes close to the distribution $(n_1,n_2,n_3) = (4, 3, 2)$ . The Boltzmann distribution correspond to the most probable energy distribution for a given internal energy $U$ . For a large number $N$ of particles almost all microstates come close to the Boltzmann distribution.