4. A volleyball tournament involved six teams, each playing against each other exactly once. In each of the five rounds, three matches took place simultaneously on three courts 1, 2, and 3. How many possible schedules were there for such a tournament? By a schedule, we mean a $5 \times 3$ table, where at position $(i, j)$, where $i \in\{1,2,3,4,5\}$ and $j \in\{1,2,3\}$, is listed a pair of teams (without specifying the order) that played against each other in the $i$-th round on court number $j$. Instead of the decimal representation of the resulting number, you can provide its prime factorization.

The regional round of category A takes place

## on Tuesday, January 14, 2014

so that it starts in the morning and the participants have 4 hours of clean time to solve the problems. Allowed aids are writing and drawing supplies and school MF tables. Calculators, laptops, and any other electronic aids are not allowed. Each problem can earn the participant 6 points; the score limit $\mathrm{k}$ for determining successful solvers and successful participants will be set centrally after evaluating the statistical results of the scores from all regions. These details will be communicated to the students before the start of the competition.
4. We will evaluate the permutations of courts $\mathrm{v}$ in each round and the permutations of the rounds themselves at the end; first, we will firmly label the teams with numbers $1,2,3,4,5,6$ and according to this, we will uniquely rearrange any schedule of five rounds. We will declare the rounds 1 and 2 to be the rounds in sequence with pairs $(1,2)$ and $(1,3)$; if at the same time in round 1 there is a pair $(3, a)$ and in round 2 a pair $(2, b)$, then $a, b$ are two different numbers from $\{4,5,6\}$, otherwise, we would have the same pair left for the third match in each of the two rounds. We will declare rounds 3, 4, and 5 to be the rounds in sequence with pairs $(1, a), (1, b)$, and $(1, c)$, where $c \in \{4,5,6\} \backslash \{a, b\}$. We thus have a uniquely determined order of all rounds with an incomplete schedule:

$$
\begin{array}{lll}
1: & (1,2), & (3, a), \\
2: & (1,3), & (2, b), \\
3: & (1, a), & \\
4: & (1, b), & \\
5: & (1, c), &
\end{array}
$$

which can clearly be completed in a unique way to a full schedule:

| $1:$ | $(1,2)$, | $(3, a)$, | $(b, c)$, |
| :--- | :--- | :--- | :--- |
| $2:$ | $(1,3)$, | $(2, b)$, | $(a, c)$, |
| $3:$ | $(1, a)$, | $(2, c)$, | $(3, b)$, |
| $4:$ | $(1, b)$, | $(2, a)$, | $(3, c)$, |
| $5:$ | $(1, c)$, | $(2,3)$, | $(a, b)$. |

Since $(a, b, c)$ is any permutation of the triplet $(4,5,6)$, the number of such written schedules is exactly $3! = 6$. In each such schedule, the rounds can be ordered in $5!$ ways, and in each of the five rounds, the pairs can be assigned to courts in $3! = 6$ ways. The sought number is thus equal to the number

$$
6 \cdot 5! \cdot 6^5 = 5! \cdot 6^6 = 2^9 \cdot 3^7 \cdot 5 = 5598720.
$$

Another solution. Let us label the teams with numbers $1,2,3,4,5$, and 6 and first construct an unordered schedule of the tournament such that the individual rounds are "numbered" by the opponents of team number 1 and examine how many ways we can complete the matches of team number 2, choosing the opponents of two only from the four-element set $\{3,4,5,6\}$:

$$
\begin{array}{lll}
1: & (1,2), & \\
2: & (1,3), & (2, a), \\
3: & (1,4), & (2, b), \\
4: & (1,5), & (2, c), \\
5: & (1,6), & (2, d).
\end{array}
$$

For the pairs $(2, a), (2, b), (2, c), (2, d)$, we have the following constraints:

$\triangleright$ Team 2 cannot have the same opponent as team 1 in any round.

$\triangleright$ Both teams 1 and 2 cannot have the same opponents alternately in two different rounds, because then we would have the same pair left for the third match in both rounds.

When these two conditions are met, the remaining pair in each round, in which 1 and 2 do not play against each other, is clearly uniquely determined. After determining them, we are left with exactly two pairs of opponents that must play against each other in the round in which 1 and 2 play against each other.

It remains to determine the number of permutations of the four-element set $\{3,4,5,6\}$ of opponents of team 2 that satisfy the two conditions.

The number of permutations $(a, b, c, d)$ satisfying the first condition, i.e., the inequalities $a \neq 3$, $b \neq 4$, $c \neq 5$, and $d \neq 6$, can be found using the principle of inclusion and exclusion: the number of suitable permutations is thus

$$
4! - \left(4 \cdot 3! - \left(\begin{array}{l}
4 \\
2
\end{array}\right) \cdot 2! + 4 - 1\right) = 9
$$

Among them, there are exactly three permutations $(a, b, c, d)$ that do not satisfy the second condition: these are clearly the permutations $(4,3,6,5), (5,6,3,4)$, and $(6,5,4,3)$ and no others.

Thus, there are a total of 6 possible schedules, in each round we have, however, $3! = 6$ ways to assign the pairs of opponents to individual courts, and a total of $5!$ possible orderings of the individual rounds, in total therefore

$$
6 \cdot 6^5 \cdot 5! = 5! \cdot 6^6 = 2^9 \cdot 3^7 \cdot 5 = 5598720 \text{ possibilities.}
$$

Note. Instead of using the principle of inclusion and exclusion, we can find and list all permutations $(a, b, c, d)$ that satisfy both conditions using a suitable systematic procedure. These are the permutations $(4,5,6,3), (4,6,3,5), (5,3,6,4), (5,6,4,3), (6,3,4,5)$, $(6,5,3,4)$.

On the other hand, the number of permutations satisfying the first but not the second condition can be reached by the following reasoning: we have six ways to choose the alternating pair of opponents, and such pairs are clearly two in each of the counted permutations.

For a complete solution, award 6 points. Award 2 to 3 points if the solver describes a meaningful way to create all possible schedules for the tournament, even if they ultimately fail to derive the correct number of possibilities.


[^0]:    1 The last conclusion holds because the function $x \mapsto x^3$ is increasing, and thus injective, in the real domain.