Example 1-43 Three identical particles are distributed in two identical potential wells, with a total energy of $3 \mathrm{E}_{0}$. Suppose particles at the energy level $k \mathrm{E}_{0}$ obey the Fermi-Dirac distribution, with $2\left(1+k^{2}\right)$ states. How many different configurations are there?

Particles in two different potential wells are independent of each other, although the Fermi-Dirac distribution requires that particles at the same energy level cannot be in the same state, this applies only to particles within the same well, and there is no such restriction between the two wells. The two potential wells are identical, meaning they are indistinguishable.
When 3 identical particles are distributed in two potential wells, there are only two scenarios: (1) all 3 particles are in the same potential well; (2) one potential well has one particle, and the other two particles are in the other potential well. If the 3 particles must be distributed in two potential wells, i.e., one potential well cannot be empty, then the first scenario is excluded. We will discuss the following cases separately:

(1) All 3 particles are in the same potential well, and the distribution states are $(0,0,3),(0,1,2),(1,1,1)$, 3 in total. The problem is the same as Example 1-42, and will not be discussed in detail again. The number of different distribution images is
$$
N_{1}=20+2 \times 4 \times 10+\binom{4}{3}=104
$$
(2) The 3 particles are distributed in two potential wells, and the distribution states with a total energy level of $3 \mathrm{E}_{0}$ are as follows:
$$
(0 ; 0,3),(0 ; 1,2),(1 ; 0,2),(1 ; 1,1),(2 ; 1,0),(3 ; 0,0)
$$

Here, $(a ; b, c)$ indicates that one potential well has one particle with an energy level of $a$; the other two particles are in another potential well with energy levels $b$ and $c$; and $a+b+c=3$.
According to the addition rule and multiplication rule, we get
$$
\begin{aligned}
N_{2} & =2 \times 2 \times 20+2 \times 4 \times 10+4 \times 2 \times 10+4 \times\binom{ 4}{2}+10 \times 2 \times 4 \\
& =80+80+80+80+24+20=364
\end{aligned}
$$

If the 3 particles are distributed in two potential wells, excluding the case where one potential well is empty, there are 364 distribution images.
If we consider scenario (1), the total number of images is
$$
N=104+364=468
$$