= Solution
For the <one-dimensional freely jointed chain>, counting right- and left-pointing links gives
$$
N_{\rightarrow}+N_{\leftarrow}=N,
\qquad
a(N_{\rightarrow}-N_{\leftarrow})=L.
$$
Therefore
$$
\boxed{N_{\rightarrow}=\frac12\left(N+\frac La\right),
\qquad
N_{\leftarrow}=\frac12\left(N-\frac La\right).}
$$
Choosing which links point right gives the <state degeneracy>
$$
\boxed{\Omega(L,N)=\binom N{N_{\rightarrow}}
=\frac{N!}{N_{\rightarrow}!N_{\leftarrow}!}.}
$$
Put $x=L/(Na)$. Applying the <Stirling formula> to $S=k_B\log\Omega$ and discarding terms subleading in $N$ yields
$$
\boxed{S(L,N)=k_BN\left[
\log2-\frac{1+x}{2}\log(1+x)
-\frac{1-x}{2}\log(1-x)
\right],
\qquad x=\frac{L}{Na}.}
$$
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