Counting quadruples of subsets is difficult even when n is small. Let's instead saw we had, for some m, just one ordered quadruple (A,B,C,D) with all A,B,C,D ⊆{1,2,...,m}. Can we think of this in a way that's easier to count? Well, imagine if we looked at each element of {1,2,...n}, looked at out quadruple (A,B,C,D), and then "tagged" the element with an "a" if it's in set A, a "b" as well if it's in set B, so on so forth. So if 2 is in sets A and C, we tag it with "ac".
Thinking of it this way, if we have two different quadruples (A,B,C,D) and (A',B',C',D') and perform this number tagging for them, they'll get two different number taggings. And we can always produce a number tagging from an ordered quadruple.
If we had an arbitrary number tagging, could we always get an ordered quadruple with the stated restrictions out of it? No. If a number was tagged "abc", that would mean that number is in both A and B and therefore A ∩B, but only in C and not D and therefore not in C∩D, breaking the assurance that A∩B ⊆C∩D.
Taking cases on a single number, though, is doable. The possible tags that work with our restriction are "abcd", "ac","acd","ad","bc","bcd","bd", "c","d","cd", and "" (no tags). This is ten possible combinations of tags. Let's call this set of ten tags T.
This is all the intuition needed for a more formal proof that there is a bijection between quadruples (A,B,C,D) with the stated restrictions and A,B,C,D ⊆{1...m} and mappings from {1...m} to T. The latter is easy to count, it's just |T|m=10m = n(m). The problem does not state A,B,C,D are subsets of {1...m} but that's a necessary condition to determine n in the first place, as we need to consider a finite number of sets.
