[ Pairing and grouping; bijections ] [ Examples and counterexamples. Constructions ]  Tolya laid out 101 coins of 1, 2, and 3 kopecks in a row. It turned out that between any two 1-kopeck coins there is at least one coin, between any two 2-kopeck coins there are at least two coins, and between any two 3-kopeck coins there are at least three coins. How many 3-kopeck coins could Tolya have?
Consider four coins lying in a row. According to the condition, there cannot be more than one three-kopeck coin, more than two two-kopeck coins, and more than two one-kopeck coins.

If there are two two-kopeck coins in the set, then between them lie two coins, one of which is a three-kopeck coin: one-kopeck coins cannot lie next to each other. Thus, among any four coins lying in a row, there is exactly one three-kopeck coin.

Since \(101 = 4 \cdot 25 + 1\), the number of three-kopeck coins is either 25 or 26.

Examples: for 25 coins: \(1, 3, 1, 2, 1, 3, 1, 2, \ldots, 1, 3, 1, 2, 1\); for 26 coins: \(3, 1, 2, 1, 3, 1, 2, 1, \ldots, 3, 1, 2, 1, 3\).

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