You can use the so-called "binomial encoding", described in one of my answers on math.stackexchange. You have to consider, though, whether it will be worthwhile. The naive encoding takes $n \log_2 k$ bits, and the best encoding saves about $O(\log n)$ bits (the best encoding uses $\log_2 \frac{n!}{f!^k}$ bits). Plug in your actual numbers to see what kind of saving you can expect.
Edit: I can't find my supposed post on math.stackexchange, so here is a summary of binomial coding (or rather multinomial coding), as it applies in this case. For a vector $x$ of length $k$, define
$$ w(x) = \frac{(\sum_i x_i)!}{\prod_i x_i!}. $$
Note that for $x \neq 0$,
$$ w(x) = \sum_{i\colon x_i > 0} w(x-e_i), $$
where $e_i$ is the $i$th basis vector. For example, if $k = 2$ and $x_1,x_2>0$ then
$$ w(x_1,x_2) = w(x_1-1,x_2) + w(x_1,x_2-1). $$
This is just Pascal's identity
$$ \binom{x_1+x_2}{x_1} = \binom{x_1+x_2}{x_1-1} + \binom{x_1+x_2}{x_1}. $$
For every vector $x$ and every $i$ such that $x_i > 0$, define
$$ w(x,i) = \sum_{j < i\colon x_j > 0} w(x-e_j). $$
Given a sequence $\sigma$ of letters in $\{1,\ldots,k\}$, let $H(\sigma)$ be its histogram, a vector of length $k$. For a non-empty word $\sigma \neq \epsilon$, let $\sigma_1$ be the first symbol, and $\sigma_{>1}$ be the rest, so that $\sigma = \sigma_1 \sigma_{>1}$. We are now ready to define the encoding:
$$ C(\sigma) = \begin{cases} w(H(\sigma),\sigma_1) + C(\sigma_{>1}) & \sigma \neq \epsilon \\ 0 & \sigma = \epsilon. \end{cases} $$
$C(\sigma)$ is an integer between $0$ and $\frac{n!}{f!^k}-1$.
Here is an example, with $k=f=2$. Let us encode $\sigma = 0101$:
$$
\begin{align*}
C(0101) &= w((2,2),0) + w((1,2),1) + w((1,1),0) + w((0,1),1) \\ &=
0 + w(0,2) + 0 + 0 = 1.
\end{align*}
$$
You can check that the inverse images of $0,1,2,3,4,5$ are
$$
0011,0101,0110,1001,1010,1100.
$$
These are just all the solution in increasing lexicography order.
How do we decode? By reversing the encoding procedure. We know the initial histogram $H_1$. Given an input $c_1$, we find the unique symbol $\sigma_1$ such that $w(H_1,\sigma_1) \leq c_1 < w(H_1,\sigma_1+1)$. We then put $H_2 = H_1 - e_{\sigma_1}$, $c_2 = c_1 - w(H_1,\sigma_1)$, and continue. To test your understanding, try to decode $0101$ back from its encoding $1$.
