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.