getting all possible combinations of a list in a form of sublists

Question

I wonder if someone can help with the following task: What is the way to get all combinations a list can be split into sublists, when order does not matter?

Let's say I have a list of 4 items:

import itertools as it

a = [1, 2, 3, 4]
print(list(it.combinations(a, 2)))

That will give me a list of 6 possible pairs:

[(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)]

How to make (out of it? or any other way) a set of lists that contain original [1, 2, 3, 4] sequence in any order? So for this example it will contain three sublists:

 [(1, 2), (3, 4)]
 [(1, 3), (2, 4)]
 [(1, 4), (2, 3)]

UPDATE: A small clarification: In other words, I need to get all sets of n-tuples such that their members contain all the population of original list, when the order within an n-tuple does not matter. Thus [(1, 2), (3, 4)] is ok, but [(2, 1), (3, 4)] is not needed because it is the same as the first set, if we ignore the order.

UPDATE2: So for the list of length 6, and for chunks of size 2 this fun function should work as follows:

import itertools as it
a = [1, 2, 3, 4, 5, 6,]
r = 2

# fun(a,r):
# OUT:
# [
#    (1, 2), (3, 4), (5, 6)
#    (1, 3), (2, 4), (5, 6),
#    (1, 4), (2, 3), (5, 6),
#    (1, 5), (2, 3), (4, 6),
#    (1, 6), (2, 3), (4, 5),
#  ]

Answer 1

Just zip the combinations, with its reverse and take only the first half of the resulting list

>>> import itertools as it
>>> lst = [1, 2, 3, 4]
>>> r = len(lst)//2
>>> combs = list(it.combinations(lst, r))
>>> list(it.islice(zip(combs, reversed(combs)), len(combs)//2))
[((1, 2), (3, 4)), ((1, 3), (2, 4)), ((1, 4), (2, 3))]