# Seemingly negative set cardinality in textbook

So in a textbook, the question states:

• There are 90 students and each of them must study at least one of Biology, Physics, or Chemistry. There are 36 students who study Biology, 42 who study Physics, and 40 who study Chemistry. Moreover, 9 study Biology and Physics, 8 study Biology and Chemistry, and 7 study Physics and Chemistry. How many students study all 3 subjects?

Initially, I tried to solve it with the inclusion-exclusion principle. As such, I rearranged the equation $n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C)$ into $n(A\cap B\cap C)=n(A\cup B\cup C)-n(A)-n(B)-n(C)+n(A\cap B)+n(A\cap C)+n(B\cap C)$, and substituted the values in, which is $n(A\cap B\cap C)=90-36-42-40+9+8+7$. However, this means that $n(A\cap B\cap C)=-4$, and hence the set has negative cardinality unless I have done something wrong. The answer is the back of the book is $4$, which leads me to believe that there is a mistake in the question.

I also checked with a venn diagram, which also supports the answer being $-4$.

Are there any issues with my working, or is the question incorrect?

I will draw a Venn diagram and split it into seven areas:

Assume that the "areas" $1;2;3;4;5;6;7$ contains $a_{1};a_{2};a_{3};a_{4};a_{5};a_{6};a_{7}$ students respectively.

I will interpret each part of the problem:

There are $90$ students and each of them must study at least one of Biology, Physics, or Chemistry.

This implies $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7}=90$

There are $36$ students who study Biology, $42$ who study Physics, and $40$ who study Chemistry.

This implies $a_{1}+a_2+a_4+a_5=36$; $a_{2}+a_3+a_5+a_6=42$; $a_{4}+a_5+a_6+a_7=40$.

Moreover, $9$ study Biology and Physics, $8$ study Biology and Chemistry, and $7$ study Physics and Chemistry.

We will have $a_2+a_5=9$; $a_4+a_5=8$; $a_5+a_6=7$.

We have this set of equations: ${\begin{cases}a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7}=90\\a_{1}+a_2+a_4+a_5=36\\a_{2}+a_3+a_5+a_6=42\\a_{4}+a_5+a_6+a_7=40\\a_2+a_5=9\\a_4+a_5=8\\a_5+a_6=7\end{cases}}$

Take the first equation and substract it with all other equations, we will have:

${\begin{cases}a_3+a_6+a_7=54\\a_1+a_4+a_7=48\\a_1+a_2+a_3=50\\a_1+a_3+a_4+a_6+a_7=81\\a_1+a_2+a_3+a_6+a_7=82\\a_1+a_2+a_3+a_4+a_7=83\end{cases}}$

• $6$th equation substract by $3$rd equation: $a_4+a_7=33$, combine with $2$nd equation we will have $a_1=15$.

• $5$th equation substract by $3$rd equation: $a_6+a_7=32$, combine with $1$st equation we will have $a_3=22$, combine with $a_1=15$ and the $3$rd equation, we will have $a_2=13$.

Combine with $a_2+a_5=9$ (first set of equaions), we have $a_5=-4$, so you are doing nothing wrong and the problem is incorrect.