Actually I am a little bit confused about this. I have a question:
Out of 30 students in a hostel, 15 study History, 8 study Economics, and 6 study Geography. It is known that 3 students study all these subjects. Show that 7 or more students study none of these subjects.
Here, they gave $n(H) = 15$, $n(E)=8$, $n(G)=6$, $n(H \cap E \cap G)=3$. (For a set $A$, $n(A)$ denotes how many members it has.)
But I know we have a formula that $$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C)$$
But they asked to find $n((H \cup E \cup G)')$?
$|H\cup E \cup G|+|H\cup E \cup G|'=30$
By Inclusion–exclusion principle wiki $|H\cup E \cup G|=|H|+|E|+|G|-|H\cap G|-|G \cap E|-|E\cap H|+|H\cap E \cap G|$
$$|H\cap G|+|G \cap E|+|E\cap H|=|H|+|E|+|G|+|H\cap E \cap G|-|H\cup E \cup G|$$ $$|H\cap G|+|G \cap E|+|E\cap H|=23-|H\cup E \cup G|$$
But $|H\cap G|+|G \cap E|+|E\cap H|\ge 0$, because are non-negative integers.
Then $|H\cup E \cup G|\le 23$ this implies $|H\cup E \cup G|'\ge 7$.