Compressing a bit string when I know how many 1s and 0s there are
The operation of translating between combinatorial objects and their indices in some enumeration is known as ranking/unranking (ranking is to convert an object to a number, unranking is the opposite). In your case, you are interesting in ranking/unranking subsets.
Suppose that you are given a subset $S$ of $\{1,\ldots,n\}$ of size $k$, and want to convert it to an integer in $0,\ldots,\binom{n}{k}-1$. If $k = 0$ or $k = n$ then you can just output $0$. Otherwise, we want to put the $\binom{n-1}{k}$ subsets not containing $n$ before the $\binom{n-1}{k-1}$ subsets containing $n$. Therefore:

If $S$ doesn't contain $n$ then you just output the rank of $S$ as a subset of $\{1,\ldots,n-1\}$ of size $k$.
If $S$ does contain $n$, you compute the rank of $S \setminus \{n\}$ as a subset of $\{1,\ldots,n-1\}$ of size $k-1$, and add to it $\binom{n-1}{k}$.

The decoding procedure is very similar. If $k = 0$ you output $\emptyset$, if $k = n$ you output $\{1,\ldots,n\}$, and otherwise:

If the index is less than $\binom{n-1}{k}$, decode it as encoding a subset of $\{1,\ldots,n-1\}$ of size $k$.
Otherwise, subtract $\binom{n-1}{k}$ from the index, decode it as encoding a subset of $\{1,\ldots,n-1\}$ of size $k-1$, and add the element $n$ to the resulting subset.

You can precompute all relevant binomial coefficients to speed things up. In your case, the index is more than 64-bit long, so you will have to work with 128-bit arithmetic somehow, but the resulting algorithm should be reasonably fast.