What's wrong with this problem (Inclusion-Exclusion principle)
Let $A$, $B$, and $C$ be the set of pupils that have access to a PC running Windows, Apple, and Linux, respectively.

We know that
\begin{align*}
  |A \cup B \cup C| &amp;= 120 \\
  |A| &amp;= 80 \\
  |B| &amp;= 40 \\
  |C| &amp;= 10
\end{align*}

As @Tom van der Zanden pointed already pointed out in the comments, the wording of "... running the Apple or Windows operating system" is unclear. Thus I will cover all three interpretations of that wording.



Interpretation 1: "... running the Apple or Windows operating system" means the PC has installed both the Apple and Windows OS.

This means
\begin{align*}
  |A \cap B| &amp;= 30 \\
  |A \cap C| &amp;= 7 \\
  |B \cap C| &amp;= 5 
\end{align*}
Using inclusion-exclusion, we come to the result that
\begin{align*}
  |A \cap B \cap C| &amp;= |A \cup B \cup C| - |A| - |B| - |C| + |A \cap B| + |A \cap C| + |B \cap C|\\
  &amp;= 120 - 80 - 40 - 10 + 30 + 7 + 5 \\
  &amp;= 32
\end{align*}
So your calculation is correct and there is a contraction. That means that either


the inclusion-exclusion princliple does not hold in general, or
the exercise description is inconsitent, i.e. some of the cardinalities are wrong.


Since the inclusion-exclusion principle is proven to hold for any arbitrary set of sets, the latter case must be true.



Interpretation 2: "... running the Apple or Windows operating system" means that the PC runs either the Apple or the Windows OS.

We can calculate the cardinalities of the intersections $A \cap B$, $A \cap C$, and $B \cap C$. However, we also run into a contradiction by using this interpretation:

\begin{align*}
  &amp; |(A \cup C) \setminus (A \cap C)| = |A \cup C| - |A \cap C| = |A| + |C| - 2|A \cap C| \\
\implies &amp; |A \cap C| = \frac{|(A \cup C) \setminus (A \cap C)| - |A| - |C|}{-2} = \frac{7 - 80 - 10}{-2} = 41.5
\end{align*}
The above is a contradiction, since the cardinality of a finite set must be a natural number.



Interpretation 3: "... running the Apple or Windows operating system" is a regular logic or.

We have $|A| = 80 &gt; 30 = |A \cup B|$. This is also a contradiction, since the cardinality of a set cannot be greater than the cardinality of its superset.



For either interpretation we show that there is a contradiction, therefore there must be a problem in the exercise description