Let $S$ be the set consisting of the four letters $\{\mathtt{\text{‘A’}},\mathtt{\text{‘E’}},\mathtt{\text{‘F’}},\mathtt{\text{‘R’}}\}$.
For $n\ge 0$, let $S^{\ast }(n)$ denote the set of words of length $n$ consisting of letters belonging to $S$.
We designate the words $\mathtt{\text{FREE}},\mathtt{\text{FARE}},\mathtt{\text{AREA}},\mathtt{\text{REEF}}$ as keywords.

Let $f(n)$ be the number of words in $S^{\ast }(n)$ that contains all four keywords exactly once.

This first happens for $n=9$, and indeed there is a unique 9 lettered word that contain each of the keywords once: $\mathtt{\text{FREEFAREA}}$
So, $f(9)=1$.

You are also given that $f(15)=72863$.

Find $f(30)$.