Find the probability that a system has energy $\epsilon$ at $T = \infty$ 
A system of $8$ identical distinguishable particles is in equilibrium in a heat bath at temperature $T$
  Each particle has $5$ states with equally spaced energy levels ${\epsilon}_{j}=j{\epsilon}$ for $j=0,1,2,3,4$.
What is the entropy of a single particle at $T=\infty$? 
The possible answers are:
$(\mathrm{a}) \, 3.69k_B$
$(\mathrm{b}) \, 1.61k_B$ 
$(\mathrm{c}) \, 1.30k_B$
$(\mathrm{d}) \, k_B$ 
$(\mathrm{e}) \, 0$

I know that the entropy can be calculated via the Gibbs' entropy: $$S_G=-k_B\sum_{j}p_j\ln(p_j)\tag{1}$$ which is a generalization of the Boltzmann entropy: $$S_B=-N\,k_B\sum_{j}p_j\ln(p_j)=k_B\ln\Omega\tag{2}$$
In the answer it states that  

$\color{red}{\fbox{$\text{At}\,T=\infty \,\text{all probabilities are equal}$}}$. 

which I don't understand. 
Using $(2)$ with $N=5$ and $p_j=\left(\dfrac{1}{5}\right) \, \forall\, j$
$$S_B=-N\,k_B\sum_{j}p_j\ln(p_j)=-5k_B\frac{1}{5}\ln\left(\frac{1}{5}\right)\approx 1.609437912k_B\implies (\mathrm{b})$$ 
Which is the correct answer.

Now here is the real question that I have, there is a reason why I put the former question & answer first which will soon become clear:

A system of $8$ identical distinguishable particles is in equilibrium in a heat bath at temperature $T$
  Each particle has $5$ states with equally spaced energy levels ${\epsilon}_{j}=j{\epsilon}$ for $j=0,1,2,3,4$ 
  $\color{blue}{\text{(everything is the same as the previous question)}}$.
What is the probability that the system has energy $\epsilon$ at $T = \infty$? 
The possible answers are:
$(\mathrm{a}) \, 0.2$
$(\mathrm{b}) \, 3.03\times 10^{-2}$ 
$(\mathrm{c}) \, 2.05\times 10^{-5}$
$(\mathrm{d}) \, 2.56\times 10^{-6}$ 
$(\mathrm{e}) \, 0$


My Attempt:
Given that 
$$p_j=\frac{\exp{\left(\frac{-{\Large\epsilon}_j}{k_B\, T}\right)}}{Z}\tag{3}$$
where the partition function $Z$ is given by
$$Z=\sum\limits_j\exp{\left(\frac{-{\epsilon}_j}{k_B\, T}\right)}\tag{4}$$ 
But since $T=\infty$ I won't be able to make any use of $(3)$ or $(4)$.
So I would say that the answer is $(\mathrm{a})$ as I made use of quote from the previous question marked in $\color{red}{\mathrm{red}}$. If all probabilities are equally likely then $\dfrac15$ seems correct to me.
End of attempt.

Now the answer to that question states that 

The probability is $p_0\cdot p_1^{N-1}$ times the number of permutations of one particle in the $j=0$ and $(N-1)$ in the $j=1$ states, which is $N$.

I have no idea what the quote is talking about but the correct numerical answer is $2.0480000000000007\times 10^{-5}$, which is $(\mathrm{c})$.
Could anyone please provide me with some hints or show how the author was able to reach the answer of $(\mathrm{c})$? 
Thank you.

EDIT:
I've just realized the correct answer is $$\frac{8}{5^8}=\frac{8}{390625}\approx 2.0480000000000007\times 10^{-5}$$ which I just found by using $$\frac{\text{Number of microstates of the system that are assigned energy } \Large\epsilon}{\text{Number of all possible microstates}}$$ thanks to the comment below by @JánLalinský. 
Now I would just like to understand the other way of calculating it which was outlined in the final quotation.
Regards.

 A: Tip: Aren't you confusing the probabilty of a particle in a $\varepsilon$ state with the probability of the system ($8$ particles) to be in a $\varepsilon$ state? I mean, you CAN use your (3) and (4) expressions in order to find the probability of a particle of being in an $\varepsilon$ state (as $T=\infty$ stands for the limit $T\rightarrow \infty$):  $$p_j=\frac{e^{\frac{-\varepsilon }{k_BT}}}{1+e^{\frac{-\varepsilon}{k_B T}}+e^{\frac{-2\varepsilon}{k_B T}}+e^{\frac{-3\varepsilon}{k_B T}}+e^{\frac{-4\varepsilon}{k_B T}}},$$
which for $T\rightarrow\infty$ goes to $1/5$, which is not what they are asking for but a correct result. (In fact it can be intuitive to think that if $T\rightarrow \infty$ there is so much thermal energy so the particles have no preference of being in any energy state, so they are equiprobable, imagine a giant trying to be in every step of a human stairs)
- So now in order to find the correct answer you should consider how many configurations (microstates) are possible that give a sum of $\varepsilon$ for the system. (i.e one particle in the $\varepsilon _1$ and all the other on the $\varepsilon _0.)$ 
A: Entropy of equally probable states can be directly calculated from Boltzmann's law,
$S=k_B\ln W=k_B\ln \frac{N}{N_i}=k_B\ln 5=1.609k_B$
where $N$ is number of states and $N_i$ is number of favourable outcomes for $i^{th}$ state.
The meaning of equal probable states is that if $N$ is number of states or arrangement then,
$P=\sum_{i=0}^{N}P_i=NP_i=1\Rightarrow P_i=\frac{1}{N}\tag*{}$
At $T\rightarrow\infty$, all states are equal probable because,
$P_i=\dfrac{e^{\frac{-\epsilon_i}{k_BT}}}{\sum_i^N e{^\frac{-\epsilon_i}{k_BT}}}=\dfrac{1}{\sum_i^N i=N}\tag*{}$
Now the probability of any state is equiprobable at $T\rightarrow\infty$, so probabity of particle in $i^{th}$ state is $p=\frac{1}{5}$ for given problem, then probability of particle is not in $i^{th}$ is $1-p=q$. Number of trails is number of particles. Therefore probability of particle not in rest of states is,
$(1-q)^8=(1-\frac{4}{5})^8=(\frac{1}{5})^8\tag*{}$
But as there are 8 particles or trial and particles are distinguishable then probability isincrease by multiply of 1 out of 8 particles permutation. Particles are indistinguishable then combination. In this case both are same and equal to 8. Thus probability of 1 particle in any given state is,
$8\cdot (\frac{1}{5})^8=2.048\times 10^{-5}\tag*{}$
In my view answer should be,
$8\cdot (\frac{1}{5})(\frac{4}{5})^7=0.33554432\tag*{}$
Reason is obvious, increasing number of particles, increasese number of chances to having given state if number of states remain fixed.
