Discrete Mathematics Two useful principles

I have 50 pairs of socks of which 30 are black and 35 are cotton. How many pairs of socks are black and cotton? If I call the set of black socks A and the set of cotton socks B, we are looking at |A ∩ B|. It is not hard to see that: |A ∪ B| = |A| + |B| − |A ∩ B| because every element that belongs to both A and B is counted twice by |A|+|B|, so we subtract the size of the intersection once. Therefore, there are 30+3550=15 socks that are black and cotton. Note that this assumes that I don’t have any red socks for instance. Otherwise, the best claim we can make is that we have at least 15 socks and that black and cotton (because |A∪B| would be less or equal to 50 and not exactly 50). The formula above represents the simplest form of the inclusion-exclusion principle. We first include the contribution of each set, then we exclude the contribution of their intersection. Here’s another example: at a party of 12 people, each person knows at least m others. How large must m be to guarantee that we can find three people that know each other? Observe that when m = 6 there is no such guarantee because it is possible to have two groups of size 6 where knowledge is across groups only. What if m > 6? Here’s a proof that this is sufficient. Pick any person p1 and let S1 be the set of people known to p1. Now |S1| > 6, so pick a person in S1, say p2, and let S2 be the set of people known to p2. Note that |S2| > 6 also. If we can prove that S1 ∩ S2 is not empty, then we find a person known to p1 and p2 and we are done. Now |S1| > 6, |S2| > 6, and |S1 ∪ S2| ≤ 12. Therefore, |S1 ∩ S2| = |S1| + |S2| − |S1 ∪ S2| > 6 + 6 − 12 = 0