68. A large group of tourists set off on an international trip. Out of them, 28 people speak English, 13 speak French, 10 speak German, 8 speak both English and French, 6 speak both English and German, 5 speak both French and German, 2 speak all three languages, and 41 people do not speak any of the three languages. How many tourists are in the group
Let's denote the set of tourists in the group who speak English, French, or German, respectively, as $A$, $B$, and $C$. According to the problem,

\[
\begin{gathered}
n(A)=28, \quad n(B)=13, \quad n(C)=10, \quad n(A \cap B)=8 \\
n(A \cap C)=6, \quad n(B \cap C)=5, \quad n(A \cap B \cap C)=2
\end{gathered}
\]

First, we need to find the number of tourists who speak at least one of the three foreign languages, i.e., $n(A \cup B \cup C)$. For this, we will use Venn diagrams (Fig. 27).

Let's calculate the sum $n(A) + n(B) + n(C)$. Since each of the numbers $n(A \cap B)$, $n(A \cap C)$, and $n(B \cap C)$ is included twice in this sum, we need to subtract the sum $n(A \cap B) + n(A \cap C) + n(B \cap C)$ from it.

Now, let's determine how many times the number $n(A \cap B \cap C)$ is included in the expression

\[
n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C)
\]

It is included three times with a positive sign (in each of the terms $n(A)$, $n(B)$, and $n(C)$) and three times with a negative sign (in each of the terms $n(A \cap B)$, $n(A \cap C)$, and $n(B \cap C)$). Therefore, to avoid losing the tourists who belong to the set $A \cap B \cap C$, we need to add the number $n(A \cap B \cap C)$. We get:

\[
\begin{gathered}
n(A \cup B \cup C) = n(A) + n(B) + n(C) - \\
- n(A \cap B) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C).
\end{gathered}
\]

Then we have:

\[
n(A \cup B \cup C) = 28 + 13 + 10 - 8 - 6 - 5 + 2 = 34.
\]

Thus, the total number of tourists in the group is $34 + 41 = 75$.

Answer: $75$.

Formula (2) is valid for any three finite sets $A$, $B$, and $C$.

Now, using formulas (1) and (2), let's try to guess how, in general, for any finite sets $A_{1}, A_{2}, A_{3}, \ldots, A_{k}$, the value of $n\left(A_{1} \cup A_{2} \cup \ldots \cup A_{k}\right)$ is calculated.

The following formula is valid:

\[
\begin{aligned}
& n\left(A_{1} \cup A_{2} \cup \ldots \cup A_{k}\right) = n\left(A_{1}\right) + n\left(A_{2}\right) + \ldots + n\left(A_{k}\right) - n\left(A_{1} \cap A_{2}\right) - \\
& \quad - n\left(A_{1} \cap A_{3}\right) - \ldots - n\left(A_{k-1} \cap A_{k}\right) + n\left(A_{1} \cap A_{2} \cap A_{3}\right) + \\
& \quad + n\left(A_{1} \cap A_{2} \cap A_{4}\right) + \ldots + (-1)^{k-1} \cdot n\left(A_{1} \cap A_{2} \cap \ldots \cap A_{k}\right).
\end{aligned}
\]

This formula is called the principle of inclusion and exclusion.

In particular, when $k$ is even, the last term in the right-hand side of formula (3) has a negative sign (as in formula (1)), and when $k$ is odd, it has a positive sign (as in formula (2)).