Andryusha has $100$ stones of different weight and he can distinguish the stones by appearance, but does not know their weight. Every evening, Andryusha can put exactly $10$ stones on the table and at night the brownie will order them in increasing weight. But, if the drum also lives in the house then surely he will in the morning change the places of some $2$ stones.Andryusha knows all about this but does not know if there is a drum in his house. Can he find out?
1. **Initial Setup**: Andryusha has 100 stones of different weights. He can distinguish the stones by appearance but does not know their weights. Every evening, he can put exactly 10 stones on the table, and at night, the brownie will order them in increasing weight. However, if a drum also lives in the house, it will change the places of some 2 stones in the morning. Andryusha needs to determine if there is a drum in his house.

2. **Step 1**: Let Andryusha consider three disjoint 10-tuples \( A, B, \) and \( C \) from the set \( S \) of 100 stones. He should note down all the greater-smaller relations as they occur after a night has passed. This means he will have the ordered sequences of \( A, B, \) and \( C \) as given by the brownie (or potentially altered by the drum).

3. **Step 2**: Consider all 10-tuples of the following type: 2 stones from each of \( A, B, \) and \( C \) (6 stones in total) and a fixed 4-tuple from \( S \setminus \{A, B, C\} \). This ensures that all of the \(\binom{10}{2} \binom{10}{2} \binom{10}{2}\) 6-tuples occur in one such 10-tuple. This means that one of these 10-tuples must contain all 3 bad pairs of \( A, B, \) and \( C \), wlog \((a_1, a_2), (b_1, b_2), (c_1, c_2)\), respectively, with \( a_1 < a_2 \), \( b_1 < b_2 \), \( c_1 < c_2 \) in the actual weight catalog but with \( a_1 > a_2 \), \( b_1 > b_2 \), \( c_1 > c_2 \) in the drum's catalog and the one that Andryusha has noted down.

4. **Step 3**: Since the drum can only change 2 of these 6 stones, one of these pairs remains unchanged. This means that at least one of the pairs \( (a_1, a_2), (b_1, b_2), (c_1, c_2) \) will not be altered by the drum. Andryusha can then compare his notes with the actual order of the stones. If there is a contradiction in any of the pairs, it means there is a drum in the house. If after the above procedure none of the notes he has written are contradicted, then the drum is not in his house.

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