In a competition, five students participated: $A, B, C, D$, and $E$. Someone had predicted that the ranking order of the participants would be $A B C D E$, but as it turned out later, they did not guess the position of a single participant correctly, not even the order in which two consecutive participants finished. Another person predicted the order $D A E C B$. This turned out to be much better, as in this prediction, the positions of exactly two participants matched the actual results, and in two cases, the order in which two consecutive participants finished was also correct. So, what was the actual outcome of the competition?
I. solution. 1. The two pairs whose consecutive positions were correctly guessed by the second tip cannot come from a situation where 3 contestants followed each other in the tip's order. Indeed, in such a case, they could not have finished in the positions where the tip mentions them, otherwise the tip would have correctly guessed 3 positions; in this case, however, at least one of the remaining 2 contestants would appear in the tip in a position where one of the 3 actually ended up, which means the tip would have guessed at most 1 contestant's position correctly. Therefore, the correctly guessed sequences can only be the following:

$$
D A, E C \quad \text { or } \quad D A, C B \quad \text { or } \quad A E, C B .
$$

2. Of the two pairs whose consecutive positions were correctly guessed by the tip, at most one could have also had their position correctly guessed, and one of them must have been guessed correctly, otherwise the tip would have guessed at most the position of the fifth contestant. It cannot be the case that in the first scenario listed under (1), $E$ and $C$, or in the third scenario, $A$ and $E$, have the same positions as in the tip, because then the other pair could only have finished consecutively (in the given order) if they finished in the position indicated by the tip.
3. In the first scenario under (1), the actual order could only have been $D, A, B, E, C$, but then the first tip would also have correctly guessed the consecutive positions of $A, B$. In the third scenario, $A$ would have had to finish in first place, as indicated by the first tip, so this cannot be the actual order. In the second scenario, the first pair could not have finished in the position indicated by the tip, because then $C$ would have had to finish in third place, as indicated by the first tip.

Thus, the correct order could only have been

$$
E, D, A, C, B
$$

and this order also satisfies all the conditions of the problem.

János Balázs (Balassagyarmat, Szántó Kovács J. g. III. o. t.)

II. solution. Let's consider the possibilities that remain after the statements made about the first tip, first for the position of $A$, and then relative to $B$. These are as follows:

$$
\begin{array}{cccccc|cccccc|ccccccc|ccccccc}
1 . & B & A & . & . & . & 4 . & B & . & A & . & . & 6 . & B & . & . & A & . & 8 . & B & . & . & . & A \\
2 . & . & A & . & B & . & 5 . & . & . & A & . & B & 7 . & . & . & B & A & . & 9 . & . & . & B & . & A \\
3 . & . & A & . & . & B & & & & & & & & & & & & & 10 . & . & . & . & B & A
\end{array}
$$

The first attempt cannot be continued, because $C$ cannot remain in third place, but it also cannot go to the fourth or fifth place, because in the first case, neither $D$ nor $E$ can go behind it, and in the second case, neither can go in front of it. From the further attempts, while still keeping the first tip in mind, the following 14 orders result:

| $2 . a$ | $C$ | $A$ | $E$ | $B$ | $D$ | 1 |  | $7 . a$ | $C$ | $E$ | $B$ | $A$ | $D$ | 0 |  |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | ---: | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| $3 . a$ | $D$ | $A$ | $E$ | $C$ | $B$ | 5 |  | $7 . b$ | $E$ | $C$ | $B$ | $A$ | $D$ | 0 |  |
| $3 . b$ | $E$ | $A$ | $D$ | $C$ | $B$ | 3 |  | $7 . c$ | $E$ | $D$ | $B$ | $A$ | $C$ | 0 |  |
| $4 . a$ | $B$ | $D$ | $A$ | $E$ | $C$ | 0 |  | $8 . a$ | $B$ | $E$ | $D$ | $C$ | $A$ | 1 |  |
| $5 . a$ | $D$ | $C$ | $A$ | $E$ | $B$ | 2 | 1 | $9 . a$ | $D$ | $C$ | $B$ | $E$ | $A$ | 1 |  |
| $5 . b$ | $E$ | $D$ | $A$ | $C$ | $B$ | 2 | 2 | $10 . a$ | $C$ | $E$ | $D$ | $B$ | $A$ | 0 |  |
| $6 . a$ | $B$ | $E$ | $D$ | $A$ | $C$ | 0 |  | $10 . b$ | $D$ | $C$ | $E$ | $B$ | $A$ | 2 | 0 |

In these, we examine the fulfillment of the statements obtained from the second tip. We have noted next to each order how many positions match the $D A E C B$ order, and where we found this to be 2, we counted the consecutive pairs that match the $D A E C B$ tip. This number is only 2 in $5 b$, so the sought order is $E, D, A, C, B$.