In a certain city, there are 10 cinemas. On the same day, 7 boys from a class go to the cinema. Assuming equal probabilities, what is the probability that no two boys end up in the same cinema?
I. solution: If we consider every possible case with the 7 boys' lineup fixed, each of the 10 cinemas is characterized by a 7th class repetitive variation. Thus, the number of possible cases is $l=V_{10}^{i, 7}=10^{7}$.

Among these, only the cases where there is no repetition are favorable, so the number of favorable cases is $k=V_{10}^{7}=$ $10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4$

Therefore, the sought probability is

$$
\begin{aligned}
v=\frac{k}{l}=\frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4}{10^{7}} & =\frac{604800}{10^{7}} \approx 0.060 \\
& \text { Gyula Parlagh (Kecskemét, Katona József g. I. o. t.) }
\end{aligned}
$$

II. solution: The probability that the first boy goes to any cinema is 1, that the second boy goes to a cinema different from the first is $\frac{9}{10}$, for the third boy, only 8 cinemas remain, and so on.

Thus, according to the multiplication rule, the sought probability is

$$
v=\frac{9}{10} \cdot \frac{4}{5} \cdot \frac{7}{10} \cdot \frac{3}{5} \cdot \frac{1}{2} \cdot \frac{2}{5}=\frac{1512}{25000}=0.06048
$$

Miklós Csapody (Bp., VIII., Piarist g. I. o. t.)