Example 1-42 4 identical particles, with a total energy of $4 \mathrm{E}_{0}$, where $\mathrm{E}_{0}$ is a constant. Each particle's energy level can be $k \mathrm{E}_{0}$, $k=0,1,2,3,4$.
(1) If the particles obey the Bose-Einstein distribution, i.e., a particle at the $k \mathrm{E}_{0}$ energy level can have $k^{2}+1$ states, and particles at the same energy level can be in the same state. How many different distribution patterns are there?
(2) If the particles obey the Fermi-Dirac distribution, i.e., a particle at the $k \mathrm{E}_{0}$ energy level has $2\left(1+k^{2}\right)$ states, and no two particles at the same energy level can be in the same state, how many different distribution patterns are there?
For a total energy of $4 \mathrm{E}_{0}$ among 4 particles, there are the following 5 possible distributions:
$$
(0,0,0,4),(0,0,1,3),(0,0,2,2)(0,1,1,2),(1,1,1,1)
$$

where $\left(n_{1}, n_{2}, n_{3}, n_{4}\right)$ represents the energy levels of the 4 particles as $n_{1} \mathrm{E}_{0}, n_{2} \mathrm{E}_{0}, n_{3} \mathrm{E}_{0}, n_{4} \mathrm{E}_{0}, n_{1}+n_{2}+n_{3}+n_{4}=4$. For example, $(0,0,2,2)$ indicates that two particles are at level 0 and two particles are at level $2 \mathrm{E}_{0}$.
(1) First, consider the Bose-Einstein distribution. The number of states for each energy level is:
\begin{tabular}{c|ccccc} 
Energy level $k \mathrm{E}_{0}$ & 0 & $\mathrm{E}_{0}$ & $2 \mathrm{E}_{0}$ & $3 \mathrm{E}_{0}$ & $4 \mathrm{E}_{0}$ \\
\hline Number of states & 1 & 2 & 5 & 10 & 17
\end{tabular}
(1) For the $(0,0,0,4)$ distribution, there are
$$
1 \times 17=17
$$

images. The 3 particles at level 0 can be considered as 3 indistinguishable balls placed in distinguishable boxes, with each box capable of holding more than one ball.
(2) For the $(0,0,1,3)$ distribution, there are
$$
1 \times 2 \times 10=20
$$

images.
(3) For the $(0,0,2,2)$ distribution, it needs to be proven that there are 5 states for $k=2$, and two particles at the same energy level can be in the same state. Viewing the states of the energy levels as labeled boxes, the problem is equivalent to placing 2 indistinguishable balls into 5 boxes, allowing more than one ball in the same box, hence there are $\binom{5+2-1}{2}=\binom{6}{2}=15$ images.
(4) For the $(0,1,1,2)$ distribution, there are
$$
\binom{2+2-1}{2}\binom{5}{1}=15
$$

images.
(5) For the $(1,1,1,1)$ distribution, there are
$$
\binom{2+4-1}{4}=\binom{5}{4}=5
$$

images.
According to the addition rule, the total number of images is
$$
N=17+20+15+15+5=72
$$
(2) If the particles follow the Fermi-Dirac distribution, i.e., the number of states at level $k \mathrm{E}_{0}$ is $2\left(1+k^{2}\right)$, and no two particles at the same energy level can be in the same state.
\begin{tabular}{c|ccccc} 
Energy level $k \mathrm{E}_{0}$ & 0 & $\mathrm{E}_{0}$ & $2 \mathrm{E}_{0}$ & $3 \mathrm{E}_{0}$ & $4 \mathrm{E}_{0}$ \\
\hline Number of states & 2 & 4 & 10 & 20 & 34
\end{tabular}

Discuss each case as follows:
(1) Level 0 has only two states, and 3 particles cannot be placed in the same state, so this distribution is impossible.
(2) For the $(0,0,1,3)$ distribution, there are
$$
4 \times 20=80
$$

different images.
(3) For the $(0,1,1,2)$ distribution, there are
$$
2 \times\binom{4}{2} \times 10=120
$$

different images.
(4) For the $(0,0,2,2)$ distribution, there are
$$
\binom{10}{2}=45
$$

different images.
(5) For the $(1,1,1,1)$ distribution, there are
$$
\binom{4}{4}=1
$$

different images.
According to the addition rule, there are
$$
80+120+45+1=246
$$

different images.