Cardinality of Set Union/Examples/Examination Candidates

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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:

$N = S_1 \cup S_2 \cup S_3$


From Cardinality of Set Union: 3 Sets:

\(\ds \card {S_1 \cup S_2 \cup S_3}\) \(=\) \(\ds \card {S_1} + \card {S_2} + \card {S_3}\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \card {S_1 \cap S_2} - \card {S_1 \cap S_3} - \card {S_2 \cap S_3}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \card {S_1 \cap S_2 \cap S_3}\)


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:

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

$\blacksquare$


Sources

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