63rd Czech and Slovak Mathematical Olympiad

We postpone the question of permutations of the five rounds and the three courts to the end of our solution. Denoting first the teams by numbers $1,2,3,4,5,6$ (in a fixed way), we rearrange the five rounds of any satisfactory draw by means of the following numbering: Let $1$ and $2$ be the rounds with matches of the pairs of teams $(1,2)$ and $(1,3)$, respectively. If a pair $(3,a)$ plays in the round $1$ and if a pair $(2,b)$ plays in the round $2$, then $a,b$ are two distinct numbers from $\{4,5,6\}$ (otherwise the third pairs in the rounds $1$ and $2$ are identical). Let $3,4$ and $5$ denote the rounds with pairs $(1,a)$, $(1,b)$ and $(1,c)$ respectively, where $c\in \{4,5,6\}\setminus \{a,b\}$. Up to this moment, we have fixed an uncompleted draw 1: (1, 2), (3, a), 2: (1, 3), (2, b), 3: (1, a), 4: (1, b), 5: (1, c), which can be extended to a the complete draw in the only one way: 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 $(4,5,6)$, the total number of the complete draws (written as above) is $3!=6$. Taking in account the number $5!$ of the possible permutations of the five rounds and the number $3!$ of possible permutations of the three courts, we conclude that the requested number of all draws is equal to $$6\cdot 5!\cdot 6^5=5!\cdot 6^6=2^9\cdot 3^7\cdot 5=5598720.$$

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Alternative solution.

Let us denote the six teams by numbers $1,2,3,4,5,6$ and construct first an "unordered" draw in which the rounds will be "numbered" by the opponents of the team $1$ — see the following table in which the other opponents of the team $2$ are denoted as $a,b,c,d$: 1: (1, 2), 2: (1, 3), (2, a), 3: (1, 4), (2, b), 4: (1, 5), (2, c), 5: (1, 6), (2, d). Note that $(a,b,c,d)$ is a permutation of the quadruple $(3,4,5,6)$ and that the following two restrictions are evident: $$▹3\ne a,4\ne b,5\ne c\text{ and }6\ne d$$ The two-element sets $\{3,a\}$, $\{4,b\}$, $\{5,c\}$, $\{6,d\}$ are pairwise distinct. It is clear that under these two conditions, the third pairs for the rounds $2$-$5$ are uniquely determined, as well as the remaining two pairs for the round $1$. Consequently, we have to calculate the number of permutations $(a,b,c,d)$ of the quadruple $(3,4,5,6)$ which satisfy the two above stated conditions. Using the inclusion-exclusion principle we conclude that the first condition is fulfilled by exactly nine permutations: $$4!-\left(4\cdot 3!-\left(\genfrac{}{}{0ex}{}{4}{2}\right)\cdot 2!+4-1\right)=9.$$ Moreover, exactly three of them do not satisfy the second condition, namely the permutations $(4,3,6,5)$, $(5,6,3,4)$ and $(6,5,4,3)$. Thus the total number of the satisfactory permutations equals $9-3=6$. We have proved that there are six "unordered" draws in the above specified sense. Combining this result with the idea of permuting the rounds and the courts, we conclude that the requested number of the draws is equal to $$6\cdot 6^5\cdot 5!=5!\cdot 6^6=2^9\cdot 3^7\cdot 5=5,598,720.$$