# Finding number of configurations possible for $3$ particles using Combinatorics

I'm studying Thermodynamics and Statistical Entropy more specifically. There is a formula formulated by Ludwig Boltzmann for the Entropy $$S$$ (degree of randomness) of a system.

$$S=k\ln(\text{W})$$

where $$k$$ is a constant and $$\text{W}$$ is the weight of the configuration, i.e. the number of ways that molecules in the system can be arranged provided they correspond to the same total energy.

I tried to figure out number of configurations possible for $$3$$ molecules with total energy $$3\epsilon$$ that could occupy equally spaced (equally probable) levels of energies 0, $$\epsilon$$ and $$2\epsilon$$.

I'm doubtful of using the formula for number of positive integral solutions to an equation to get the answer for this problem because that would not count the order of the appearance of the solutions. Another that won't work is because we sort of have a weight attached to each sample.

How could I possibly employ Permutations and Combinations' techniques to come up with the solution?

If we lob off the $$\epsilon$$ and just look at the integer coefficient, we let the number of ways $$n$$ molecules with maximum permissible energy $$k$$ can be chosen. We can set up recurrence:
$$f(n, k) = \sum_{i=0}^k f(n-1, k-i)$$
This follows from being able to choose the energy of the next molecule as a free choice from $$\{0, \dots, k\}$$ and the rest of the molecules with the remaining energy.
As base cases we have $$f(n, 0) = 1$$, as there is only one way the remaining $$n$$ molecules can be chosen if their total sum of energy must be $$0$$, and $$f(1, k) = 1$$, if we have only 1 molecule left to assign an energy to we must assign the remaining $$k$$ energy to it.
Your question is now the same as computing $$f(3, 3)$$. Can you take it from here?