How many elements are in $A_1 \cup A_2$ if there are 12 elements in $A_1$, 18 elements in $A_2$ and
a) $A_1 \cap A_2 = \phi$
$n(A_1 \cap A_2) = 0$
$n(A_1 \cup A_2) = n(A_1) + n(A_2) - n(A_1 \cap A_2) = 12 + 18 - 0 = 30$
b) $|A_1 \cap A_2| = 1$
$n(A_1 \cap A_2) = 1$
$n(A_1 \cup A_2) = n(A_1) + n(A_2) - n(A_1 \cap A_2) = 12 + 18 - 1 = 29$
c) $|A_1 \cap A_2| = 6$
$n(A_1 \cap A_2) = 6$
$n(A_1 \cup A_2) = n(A_1) + n(A_2) - n(A_1 \cap A_2) = 12 + 18 - 6 = 24$
d) $A_1 \subseteq A_2$
then $A_1 \cap A_2 = A_1$, now $n(A_1 \cap A_2) = n(A_1)$
So that $n(A_1 \cup A_2) = n(A_1) + n(A_2) - n(A_1 \cap A_2) = 12 + 18 - 12 = 18$
There are 345 students at a college who have taken a course in calculus, 212 who taken a course in discrete mathematics, and 188 who have taken courses in both calculus and discrete mathematics. How many students have taken a course in either calculus or discrete mathematics?