The number of microstates of distributing five quanta of energy among four distinguishable particles is :
56
This is a counting problem of the "stars and bars" type. The quanta of energy are indistinguishable from one another — one quantum is the same as any other — while the particles are distinguishable, so it matters which particle receives how many.
The standard result for distributing \(q\) identical quanta among \(N\) distinguishable particles is
\(W = \binom{q + N - 1}{q} = \frac{(q+N-1)!}{q!\,(N-1)!}\).
The reasoning behind it: lay out the \(q\) quanta in a row and insert \(N-1\) dividers to split them among the \(N\) particles. Each distinct arrangement of quanta and dividers is one microstate, and there are \(q + N - 1\) positions of which we choose \(q\) for the quanta.
Substitute \(q = 5\) and \(N = 4\):
\(W = \binom{5 + 4 - 1}{5} = \binom{8}{5} = \frac{8!}{5!\,3!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56\).
Note that \(\binom{8}{5} = \binom{8}{3}\), so the same answer follows from choosing the positions of the three dividers instead — a useful check.
The value 126 is \(\binom{9}{4}\), obtained by miscounting the number of dividers, and 35 is \(\binom{7}{3}\). The quantity \(W\) computed this way is exactly the statistical weight that enters the Boltzmann entropy \(S = k\ln W\).
Hence the number of microstates is 56.
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