Cardinality of Set Union/Examples/Student Subjects/Mathematics and Chemistry

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Example of Use of Cardinality of Set Union

In a particular group of $75$ students, all studied at least one of the subjects mathematics, physics and chemistry.

All candidates attempted at least one of the questions.

$40$ students studied mathematics.
$60$ students studied physics.
$25$ students studied chemistry.

Also:

exactly $5$ students studied all $3$ subjects.


It follows that:

no more than $20$ students studied both mathematics and chemistry.


Proof

Let:

$S_1$ denote the set of students who studied mathematics.
$S_2$ denote the set of students who studied physics.
$S_3$ denote the set of students who studied chemistry.

Knowledge of the total number of students gives us:

$S_1 \cup S_2 \cup S_3 = 75$


Let $N$ denote the number of students $N$ who studied both mathematics and chemistry:

$N = S_1 \cap S_3$


From the question:

\(\ds \card {S_1}\) \(=\) \(\ds 40\)
\(\ds \card {S_2}\) \(=\) \(\ds 60\)
\(\ds \card {S_3}\) \(=\) \(\ds 25\)


First we calculate how many students took just mathematics or chemistry, but who did not take physics.

We have

\(\ds \card {\paren {S_1 \cup S_3} \setminus S_2}\) \(=\) \(\ds \card {\paren {S_1 \cup S_2 \cup S_3} \setminus S_2}\)
\(\ds \) \(=\) \(\ds \card {S_1 \cup S_2 \cup S_3} - \card {S_2}\)
\(\ds \) \(=\) \(\ds 75 - 60\)
\(\ds \) \(=\) \(\ds 15\)


So only $15$ students did not take physics.

Thus no more than $15$ students can have taken both mathematics and chemistry, without taking physics.

However, we are also told that $5$ students took all $3$ courses.

So, in addition to the maximum $15$ who took mathematics and chemistry, without taking physics, this takes the total to a maximum of $20$ students all told who took both mathematics and chemistry.

$\blacksquare$


Sources