I have a universe (or total number of people polled who are distributed amongst these sets) of $151$ persons.
These sets correspond to which TV shows they watch (i.e., each set represents one TV show): $$n(A) = 68; n(B) = 61; n(C) = 52.$$
Also provided are the cardinalities of their paired intersections: $$n(A\cap B) = 16; n(A\cap C) = 25; n(B\cap C) = 19.$$
Given this information how do I go about determining the cardinality of the intersection of all three sets? Step by step, formulaic, with basic explanations where appropriate, please. External references are appreciated but I need to see a solution in context. Thanks.
Certainly, $n(A\cap B\cap C)\le n(A\cap B)=16$. Just draw a Venn diagram and observe that indeed every value $0\ldots 16$ is possible. (The principle of inclusion and exclusion would only give a much weaker bound in this specific example!) To be explicit, with $0\le k\le 16$, we can let $$\begin{align} n(A\cap B\cap C)&=k\\ n(A\cap B\cap \bar C)&=16-k\\ n(A\cap \bar B\cap C)&=25-k\\ n(A\cap \bar B\cap \bar C)&=27+k\\ n(\bar A\cap B\cap C)&=19-k\\ n(\bar A\cap B\cap \bar C)&=26+k\\ n(\bar A\cap \bar B\cap C)&=8+k\\ n(\bar A\cap \bar B\cap \bar C)&=30-k\\ \end{align} $$ For example, you will verify that this makes $n(C)=n(\bar A\cap \bar B\cap C)+n(\bar A\cap B\cap C)+n( A\cap \bar B\cap C)+n( A\cap B\cap C) = (8+k)+(19-k)+(25-k)+k = 52 $ as required. Especially, we find out that at least $14$ people dont watch any of the shows (for $30-k\ge 14$).