**Consider a tournament with n players where each player plays against every other player. Suppose each player wins at least once. Show that at least 2 of the players have the same number of wins.**

The **Answer to the Question**

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**Here's the Solution to this Question**

The number of wins for a player is at least $1$ and at the most $(n-1).$

These $(n-1)$ numbers correspond to the pigeonholes to accomodate $n$ playesr (pigeons).

Thus, by the Pigeonhole Principle there must be at least two players occupying one pigeonhole (number of wins).

Hence, it is proved.