Calculating number of microstates to find entropy

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UnderLaplacian
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In the Boltzmann entropy formula , the number of microstates is calculated according to Maxwell-Boltzmann statistics , i.e. , W = n!/Πki! , Σki = n . Why cannot we use some other method , such as Bose-Einstein or Fermi-Dirac statistics ?
 
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I am fairly sure that the formula you cite for the number of microstates is a general combinatoric formula. What do you think would make it different if the particles were distinguishable, fermionic, or bosonic?
 
Say , for example , we consider the problem of placing 2 balls in 2 bins . If we treat the balls as identical , we have 3 ways , if not , we have 4 ways . Please point out if I am making some mistake in my interpretation .
 
You are correct. You cannot use the formula $$W = \frac{n!}{k_{1}! ... k_{r}!}$$ to calculate the number of possible states that a system of n identical particles can be distributed among r different energy levels. However, if you change your interpretation, you can use this formula to calculate the number of ways that a virtual collection of N distinguishable systems each of n identical particles can be distributed among the combined energy levels (possible energy levels of the imaginary collection). Then, the Boltzmann entropy formula applies even to systems in which the individual particles are indistinguishable. This is how the core formulas of statistical mechanics are justified in the quantum domain. Does that answer the question?
 
Twigg said:
However, if you change your interpretation, you can use this formula to calculate the number of ways that a virtual collection of N distinguishable systems each of n identical particles can be distributed among the combined energy levels (possible energy levels of the imaginary collection).
Could you please explain your above statement in some more detail ? I did not really get what you meant .