## Task 4 - 181224

Thomas presents Jürgen with the following problem:

(1) In my class, exactly 15 students are involved in extracurricular sports, and only the sports of football, swimming, gymnastics, or athletics are practiced.

(2) Each of the mentioned sports is practiced by at least one student.

(3) No student practices more than two of these sports.

(4) Every student who practices swimming or athletics also practices a second sport.

(5) Exactly 3 students practice both football and swimming, exactly 2 students practice both swimming and athletics; no student practices both football and gymnastics.

(6) The number of football players is greater than the number of swimmers, which in turn is greater than the number of gymnasts, and this is greater than the number of athletes.

(7) The number of football players is equal to the sum of the number of gymnasts and the number of athletes.

In (6) and (7), "football player," "swimmer," etc., each refer to a student who practices the respective sport (alone or in addition to a second sport).

Give the number of football players, swimmers, gymnasts, and athletes in my class!

After some consideration, Jürgen says that this problem is not uniquely solvable. Determine all solutions to this problem.
For a simpler way of speaking, we will use the abbreviations F, S, T, L for the four sports Football, Swimming, Gymnastics, and Track and Field, and also for any combinations thereof. Furthermore, let $f, s, t, l$ be the numbers of students who have chosen one of the sports F, S, T, L (possibly in combination with another).

For simplicity, we imagine the 15 students numbered such that students numbered $1-5$ have all chosen S in combination with SF or SL, which exhausts these combination possibilities for S according to the given conditions. For the remaining 10 students, only the 5 possibilities F, T, FL, ST, TL remain.

In particular, from these 10 students, either F or T (but not both!) must be chosen, which leads to the important relationship

$$
(f-3)+t=10
$$

between $f$ and $t$. Substituting this into $t+l=f$, which should hold according to (7), yields the equation

$$
2 t+l=13 \quad(*)
$$

which is the "crux" of this problem. Since according to (5) $l \geq 2$ holds, $l \geq 5$ would immediately lead to the contradiction $t \leq 4 < l$ according to (6), and $l$ must be odd according to (*), the only remaining possibility is

$$
l=3, t=5, f=t+l=8
$$

Of the remaining 10 students, L (each in combination with F or T) must be chosen exactly once, and S (each in combination with T) must be chosen once or twice, which, taking (**) into account, satisfies all conditions of the problem. In total, we have

$$
f=8, s \in\{6,7\}, t=5, l=3
$$

with the only ambiguity concerning $s$.