There are 5 paths leading to the top of a mountain, and two people go up and come down. In how many ways can this happen, if we do not distinguish between the two people and

a) each path can be used by at most one person and only once,

b) each path can be used twice, but only in different directions,

c) there are no restrictions,

d), e) and f) are the same as a), b), and c), but we distinguish between the two people.
When two people are not distinguished personally, for example, choosing the 2nd and 5th path is the same as choosing the 5th and 2nd path. Therefore, in cases a), b), and c), we are dealing with combinations.

a) The ascent of the two people is characterized by a combination of two elements from five. (No repetition, because of the constraint that a path can be used by at most one person.) Therefore, the total number of possible ascents is \( C_{5}^{2} \).

For the descent, according to the condition, only 3 paths are available. Therefore, the total number of possible descents is \( C_{3}^{2} \). Since each ascent can be paired with \( C_{3}^{2} \) descents, the total number of possible cases is

\[
C_{5}^{2} \cdot C_{3}^{2} = \binom{5}{2} \binom{3}{2} = \frac{5 \cdot 4}{1 \cdot 2} \cdot 3 = 10 \cdot 3 = 30
\]

b) The number of possible ascents remains \( \binom{5}{2} \). However, in this case, the same constraint applies to the descent as to the ascent, i.e., that (in one direction) two people cannot use the same path. Therefore, the total number of possible descents is also \( \binom{5}{2} \), and thus the total number of possible cases is

\[
\binom{5}{2} \binom{5}{2} = 10 \cdot 10 = 100
\]

c) In this case, two people can use the same path, so we are dealing with combinations with repetition, and the number of possible cases is the same in both directions: \( C_{5}^{i, 2} \). Therefore, the total number of possible cases is

\[
\left(C_{5}^{i, 2}\right)^{2} = \left(\frac{5 \cdot 6}{1 \cdot 2}\right)^{2} = 15^{2} = 225
\]

When we distinguish the two people personally, the 2nd and 5th path is a different case from the 5th and 2nd path. Therefore, in cases d), e), and f), we are dealing with variations. The reasoning is otherwise identical to the considerations followed in cases a), b), and c).

Accordingly,

d) \( V_{5}^{2} \cdot V_{3}^{2} = (5 \cdot 4) \cdot (3 \cdot 2) = 20 \cdot 6 = 120 \).
e) \( V_{5}^{2} \cdot V_{5}^{2} = (5 \cdot 4)^{2} = 20^{2} = 400 \).
f) \( V_{5}^{i, 2} \cdot V_{5}^{i, 2} = \left(5^{2}\right)^{2} = 25^{2} = 625 \).

Imre Szlanka (Aszód, Petőfi g. II. o. t.)