Question

There are 40 students in a chemistry class and 60 students in a physics class. Find the number of students which are either in physics class or chemistry class in the following case:

The two classes meet at the same hour

Answer 1

Let A be the set of students in chemistry class and B be the set of students in physics class. It is given that $n\left(A\right)=40$ and $n\left(B\right)=60$.
If two classes meet at the same hour, then there will not be a common student sitting in both the classes.
Therefore, $n(A\cap B)=0$
$\therefore n(A\cup B)=n\left(A\right)+n\left(B\right)-n(A\cap B)$
$\Rightarrow n(A\cup B)=40+60-0=100$