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.
In graph theory terms, you would like to partition a complete graph into matchings ($1$-regular graphs). There is a well-known algorithm for doing that. Your students would be the vertices of this graph, and a pair of students working together would be an edge of this graph.
First, label your $n$ students (where $n$ is even) as follows: $\infty,0,1,2,\dots,n-2$. Think of the finite (non-$\infty$) labels as remainders modulo $n-1$. Now, for each $i=0,1,2,\dots,n-2$, pair your students as follows: $(\infty,i)$ and $(i+k,i-k)$ for $k=1,2,\dots,\frac{n}{2}-1$. In other words, except for the special pair with $\infty$, the sum of the two labels in a pair should be $2i$ modulo $n-1$.
For example, let $n=8$, then $n-1=7$ (so all labels are remainders modulo $7$), and your vertex labels are $\infty,0,1,2,3,4,5,6$. The matchings you need are as follows:
$$
\begin{matrix}
i=0: & (\infty,0), (1,6), (2,5), (3,4)\\
i=1: & (\infty,1), (2,0), (3,6), (4,5)\\
i=2: & (\infty,2), (3,1), (4,0), (5,6)\\
i=3: & (\infty,3), (4,2), (5,1), (6,0)\\
i=4: & (\infty,4), (5,3), (6,2), (0,1)\\
i=5: & (\infty,5), (6,4), (0,3), (1,2)\\
i=6: & (\infty,6), (0,5), (1,4), (2,3)
\end{matrix}
$$
To see that this as a picture, imagine the labels $0,1,2,\dots,n-2$ as spaced evenly around a circle while $\infty$ is placed in the center. Then $(\infty,i)$ is a "minute hand" pointing at $i$, while the rest of the pairs are all edges perpendicular to it.
Best Answer
I think there are $5$ ways to do it, not $3$.
Identify each student with one of the $16$ points of the affine plane over the field with $4$ elements. To divide the students into $4$ groups, we can group them according to a pencil of $4$ parallel lines. Then the question becomes, "How many parallel directions are there in the plane?"
Of course, this is just the number of lines through the origin. We can join the origin to any of the other $15$ points, but each line is counted $3$ times, once for every point on it other than the origin, so there are $5$ lines.
In general, this is a problem in combinatorial designs, and there are a lot of open questions, I believe.
EDIT
Here is a python script that computes the answer. I don't know how to draw a comprehensible picture of the plane over $\mathbb{F}_4$. I think of a $4\times4$ array of dots for the points. The problem is that there are $20$ lines, and I don't see how to draw them without making a mess.
Here is my script:
This produced the output
Where the last line is a spurious error message due to sloppy programming.
EDIT
In response to user's comments.
I thought of a way to present this that might make it a little clearer. In each group, the $16$ numbers represent the points of the plane, and points of the same color lie on the same line. In the first two groupings, this looks pretty dull; the lines are rows or columns. In the last three groupings, the lines don't look like what we normally think of as lines, but they satisfy the same kind of linear equations as the lines you're used to.
This isn't the place to try to explain that. You can search the Web for "finite geometries", but you'll have to learn a little bit about finite fields before it makes sense.