In a remote school of 13 teachers, 8 teach Government (G), 7 teach Geography (P) and 7 teach History (H). 2 teach Government and Geography only, 2 teach Government and History only and 3 teach Geography and History only. Represent the information in a Venn diagram.

here's how to solve this problem involving sets and Venn diagrams:

First, let's define the sets and the given information:
Let $N$ be the total number of teachers in the school. $N = 13$.
Let $G$ be the set of teachers who teach Government. $n(G) = 8$.
Let $P$ be the set of teachers who teach Geography. $n(P) = 7$.
Let $H$ be the set of teachers who teach History. $n(H) = 7$.

We are given the following "only" intersections:
$n(G \cap P \text{ only}) = 2$ (teachers who teach Government and Geography but not History)
$n(G \cap H \text{ only}) = 2$ (teachers who teach Government and History but not Geography)
$n(P \cap H \text{ only}) = 3$ (teachers who teach Geography and History but not Government)

Step 1: Find the number of teachers who teach all three subjects, $n(G \cap P \cap H)$.
Let $x = n(G \cap P \cap H)$.
The number of teachers who teach only Government is $n(G \text{ only}) = n(G) - [n(G \cap P \text{ only}) + n(G \cap H \text{ only}) + x]$.
$n(G \text{ only}) = 8 - (2 + 2 + x) = 8 - (4 + x) = 4 - x$.

The number of teachers who teach only Geography is $n(P \text{ only}) = n(P) - [n(G \cap P \text{ only}) + n(P \cap H \text{ only}) + x]$.
$n(P \text{ only}) = 7 - (2 + 3 + x) = 7 - (5 + x) = 2 - x$.

The number of teachers who teach only History is $n(H \text{ only}) = n(H) - [n(G \cap H \text{ only}) + n(P \cap H \text{ only}) + x]$.
$n(H \text{ only}) = 7 - (2 + 3 + x) = 7 - (5 + x) = 2 - x$.

The sum of all disjoint regions must equal the total number of teachers, assuming all teachers teach at least one subject.

$$n(G \cup P \cup H) = n(G \text{ only}) + n(P \text{ only}) + n(H \text{ only}) + n(G \cap P \text{ only}) + n(G \cap H \text{ only}) + n(P \cap H \text{ only}) + n(G \cap P \cap H)$$ $$13 = (4 - x) + (2 - x) + (2 - x) + 2 + 2 + 3 + x$$ $$13 = 4 - x + 2 - x + 2 - x + 2 + 2 + 3 + x$$ $$$$