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My problem: There are $20$ students in a class. $13$ of them study chemistry and $16$ of them study physics. $3$ of them do neither.

My workings:

$13 + 16 = 29$, but there are only $20$ people in the class, that means some people have to do both right? But how do I determine $C \cap P$? So, I said $20 -3 =17$, then $C \cup P$ must have a total of $17$ people. But then I am stuck.

Can someone please visually represent this? Mathematically showing how to find $C \cap P$ is also fine.

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As you correctly observed, $17$ study either chemistry or physics. As $16$ study physics only $1$ student studies only chemistry. Similarly, $4$ students study only physics. This results in $12$ students studying both.

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You have a big set of Students $S$ and you have two subsets $A,B$ of students who study chemistry or physics respectively. You know that $\#(A\cup B)=20-3=17$. Also note that $A\cap B = (A^c\cup B^c)^c$ where $c$ denotes the complement. But you do know how many people do NOT study chemistry or physics.

Do you know how to continue from here?

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If $C$ is the set containing all chemistry students and $P$ is the set containing all physics students, we have:

$$|C| + |P| - |C \cap P| = 20 - 3 \iff 13 + 16 - |C \cap P| = 17$$ $$\iff |C \cap P| = 29 - 17 = 12$$

We thus find $12$ students studying both topics, $13 - 12 = 1$ studying only chemistry and $16 - 12 = 4$ studying only physics. The Venn diagram looks as follows:

enter image description here

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