Minimum cases of n choose k with respect of n choose q

I have a list

people = ['P1', 'P2', 'P3', 'P4', 'P5', 'P6', 'P7']allComb4 = list(itertools.combinations(people,4)) # n choose k#[('P1', 'P2', 'P3', 'P4'), ('P1', 'P2', 'P3', 'P5'), ('P1', 'P2', 'P3', 'P6'), ('P1', 'P2', 'P3', 'P7'), ('P1', 'P2', 'P4', 'P5'), ('P1', 'P2', 'P4', 'P6'), ('P1', 'P2', 'P4', 'P7'), ('P1', 'P2', 'P5', 'P6'), ('P1', 'P2', 'P5', 'P7'), ('P1', 'P2', 'P6', 'P7'), ('P1', 'P3', 'P4', 'P5'), ('P1', 'P3', 'P4', 'P6'), ('P1', 'P3', 'P4', 'P7'), ('P1', 'P3', 'P5', 'P6'), ('P1', 'P3', 'P5', 'P7'), ('P1', 'P3', 'P6', 'P7'), ('P1', 'P4', 'P5', 'P6'), ('P1', 'P4', 'P5', 'P7'), ('P1', 'P4', 'P6', 'P7'), ('P1', 'P5', 'P6', 'P7'), ('P2', 'P3', 'P4', 'P5'), ('P2', 'P3', 'P4', 'P6'), ('P2', 'P3', 'P4', 'P7'), ('P2', 'P3', 'P5', 'P6'), ('P2', 'P3', 'P5', 'P7'), ('P2', 'P3', 'P6', 'P7'), ('P2', 'P4', 'P5', 'P6'), ('P2', 'P4', 'P5', 'P7'), ('P2', 'P4', 'P6', 'P7'), ('P2', 'P5', 'P6', 'P7'), ('P3', 'P4', 'P5', 'P6'), ('P3', 'P4', 'P5', 'P7'), ('P3', 'P4', 'P6', 'P7'), ('P3', 'P5', 'P6', 'P7'), ('P4', 'P5', 'P6', 'P7')]allComb2 = list(itertools.combinations(people,2)) # n choose q# [('P1', 'P2'), ('P1', 'P3'), ('P1', 'P4'), ('P1', 'P5'), ('P1', 'P6'), ('P1', 'P7'), ('P2', 'P3'), ('P2', 'P4'), ('P2', 'P5'), ('P2', 'P6'), ('P2', 'P7'), ('P3', 'P4'), ('P3', 'P5'), ('P3', 'P6'), ('P3', 'P7'), ('P4', 'P5'), ('P4', 'P6'), ('P4', 'P7'), ('P5', 'P6'), ('P5', 'P7'), ('P6', 'P7')]

I need to find in allComb4 minimum number of elements with respect of allComb2. Desired result like bellow.

output = [['P1', 'P2', 'P5', 'P6'], ['P1', 'P3', 'P4', 'P7'], ['P2', 'P3', 'P5', 'P7'], ['P2', 'P4', 'P5', 'P6'], ['P3', 'P4', 'P6', 'P7']]

That means, any pair I pick up from allComb2 I will find that pair elements in one element of output. How can I do that?

LE: Always q < k