4. (4 points) When preparing for the exam, three schoolchildren solved 100 problems. Each schoolchild solved 60 problems, and every problem was solved by someone. A problem is considered difficult if only one schoolchild solved it. An easy problem is one that was solved by all three schoolchildren. Which type of problems is more - easy or difficult? By how much?
Solution. Let

$x_{i}$ - the number of problems solved only by the $i$-th student,

$y_{i, j}$ - the number of problems solved only by the $i$-th and $j$-th students,

$z$ - the number of problems solved by all students (the number of easy problems)

Then the number of difficult problems is $x_{1}+x_{2}+x_{3}$.

By the condition, we have a system of four linear equations in 7 unknowns

$$
\left\{\begin{array}{c}
x_{1}+x_{2}+x_{3}+y_{1,2}+y_{1,3}+y_{2,3}+z=100 \\
x_{1}+y_{1,2}+y_{1,3}+z=60 \\
x_{2}+y_{1,2}+y_{2,3}+z=60 \\
x_{3}+y_{1,3}+y_{2,3}+z=60
\end{array}\right.
$$

From this system, we can find that $x_{1}+x_{2}+x_{3}-z=20$. The last equality gives the answer to the problem.

Answer: there are 20 more difficult problems than easy ones.

Another solution involves applying the inclusion-exclusion principle for counting the size of the union of three sets

$$
|A \cup B \cup C|=|A|+|B|+|C|-|A \cap B|-|A \cap C|-|C \cap B|+|A \cap B \cap C|
$$

Here $A$ is the set of problems solved by the 1st student, $|A|=60$

$B$ is the set of problems solved by the 2nd student, $|B|=60$

$C$ is the set of problems solved by the 3rd student, $|C|=60$

$A \cap B \cap C$ is the set of problems solved by all students (the set of easy problems),

$A \cup B \cup C$ is the set of all problems, $|A \cup B \cup C|=100$.

Then the set of difficult problems is described by the equality

$$
(A \cup B \cup C) \backslash((A \cap B) \cup((A \cap C) \backslash(A \cap B \cap C)) \cup((B \cap C) \backslash(A \cap B \cap C)))
$$

The size of this set is

$$
\begin{aligned}
& |A \cup B \cup C|-|A \cap B|-(|A \cap C|-|A \cap B \cap C|)-(|B \cap C|-|A \cap B \cap C|)= \\
& =|A \cup B \cup C|-|A \cap B|-|A \cap C|-|B \cap C|+2|A \cap B \cap C|
\end{aligned}
$$

Taking into account the inclusion-exclusion principle, we get that the number of difficult problems is

$$
2|A \cup B \cup C|-|A|-|C|-|B|+|A \cap B \cap C|=200-180+|A \cap B \cap C|=20+|A \cap B \cap C|
$$

Thus, there are 20 more difficult problems than easy ones.