Unique Partitions of Combinations Real world problem.  We have $32$ possible options.  We offer $10$ choices in a configuration without repetition and without order. Simple combinatorics problem that yields $64,512,240$ possible configuration ($32$ choose $10$).  
The rub is that we want to break each configuration into two groups of $5$.  Then we want to produce only unique groups of $5$ that we can combine to produce the total number of configurations. 
What is the formula and associated math for determining the number of these unique groups of $5$ elements? And, more interestingly, the math for the general cases of similar problems and associated proofs.
Simple example using $6$ options choosing $4$ at a time for $15$ configurations.
$(1,2,3,4), (1,2,3,5), (1,2,3,6), (1,2,4,5), (1,2,4,6), (1,2,5,6), (1,3,4,5)$, etc.
If we split these combinations of 4 elements into groups of $2$ there are $11$ unique groups of $2$ elements.  $(1,2), (1,3), (1,4), ... (5,6)$ etc.
Forgive any vagueness in my choice of language as it's been about $25$ years since my formal math education.
 A: Working your $6 \choose 4$ example is enlightening.  There are ${6\choose 2}=15$ pairs.  The four that are missing from your list are $(1,6),(1,5),(2,6),(2,5)$.  They are missing because $1$ and $2$, if present, must be part of the first pair.  $5$ and $6$, if present, must be part of the second pair.  
By the same logic, in your $32\choose 10$ case, choices $1$ through $5$ must be part of the first half and choices $28$ through $32$ must be part of the last half.  No group that contains one of each of these sets need be included.  The bad news is that won't save you much compared to the ${32 \choose 5}=201\ 376$ groups without restriction.  We can catalog the cases we can skip by the number of items from each end.
Cases with  one of the first five, three of the middle, and one of the last five, are $5\cdot {22\choose 3}\cdot 5=38\ 500$.
Cases of two of one end, two in the middle, and one of the other end are $2\cdot {5\choose 2}\cdot {22 \choose 2}\cdot 5=23100$
Cases of two of each end are ${5 \choose 2}^2\cdot 22=2200$
Cases of three of one end and one of the other are $2{5 \choose 3}\cdot 22\cdot 5=2200$
Cases of three of one end and two of the other are $2{5 \choose 3}{5\choose 2}=200$
The total left is $135\ 176$, or about $2/3$ of all the five element subsets.  I am surprised it is so small.
