# Cardinality of Set Union/Examples/Examination Candidates

## Example of Use of Cardinality of Set Union

In a particular examination, there were $3$ questions.

All candidates attempted at least one of the questions.

$40$ candidates attempted question $1$.
$47$ candidates attempted question $2$.
$31$ candidates attempted question $3$.

Also, it was apparent that:

$9$ candidates attempted at least questions $1$ and $2$.
$15$ candidates attempted at least questions $1$ and $3$.
$11$ candidates attempted at least questions $2$ and $3$.

and:

exactly $6$ candidates attempted all $3$ questions.

It follows that $89$ candidates sat the examination in total.

## Proof

Let:

$S_1$ denote the set of candidates who attempted question $1$.
$S_2$ denote the set of candidates who attempted question $2$.
$S_3$ denote the set of candidates who attempted question $3$.

The number of candidates $N$ who sat the examination in total is therefore:

$N = S_1 \cup S_2 \cup S_3$
 $\ds \card {S_1 \cup S_2 \cup S_3}$ $=$ $\ds \card {S_1} + \card {S_2} + \card {S_3}$ $\ds$  $\, \ds - \,$ $\ds \card {S_1 \cap S_2} - \card {S_1 \cap S_3} - \card {S_2 \cap S_3}$ $\ds$  $\, \ds + \,$ $\ds \card {S_1 \cap S_2 \cap S_3}$

From the question:

 $\ds \card {S_1}$ $=$ $\ds 40$ $\ds \card {S_2}$ $=$ $\ds 47$ $\ds \card {S_3}$ $=$ $\ds 31$ $\ds \card {S_1 \cap S_2}$ $=$ $\ds 9$ $\ds \card {S_1 \cap S_3}$ $=$ $\ds 15$ $\ds \card {S_2 \cap S_3}$ $=$ $\ds 11$ $\ds \card {S_1 \cap S_2 \cap S_3}$ $=$ $\ds 6$

Hence:

 $\ds \card {S_1 \cup S_2 \cup S_3}$ $=$ $\ds 40 + 47 + 31$ $\ds$  $\, \ds - \,$ $\ds 9 - 15 - 11$ $\ds$  $\, \ds + \,$ $\ds 6$ $\ds$ $=$ $\ds 89$

$\blacksquare$