The Handshake Conundrum: Applying the Pigeonhole Principle to a Party Scenario

Problem)
There are n individuals at a party, and some of them have shaken hands.
Prove that there are two individuals who have participated in the same number of handshakes.

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Solution)
To prove that there are two individuals who have participated in the same number of handshakes at a party with n individuals, we can use the pigeonhole principle.

First, let's consider the possible number of handshakes an individual can have. The minimum number of handshakes an individual can have is 0 (if they shake hands with no one), and the maximum number of handshakes is n−1 (if they shake hands with everyone else at the party).

However, it's important to note that not all of these possibilities can occur at the same time. Specifically, if one person has shaken hands with everyone else (n−1 handshakes), no one can have 0 handshakes, because everyone would have shaken hands with that person. Similarly, if one person has 0 handshakes, no one can have n−1 handshakes.

Therefore, the range of possible handshake counts for any individual at the party is from 0 to n−1, but we can't have both extremes (0 and n−1) simultaneously. This gives us n possible handshake counts (from 0 to n−1, inclusive), but only n−1 of them can occur at any given time.

Since there are n individuals at the party and only n−1 distinct handshake counts that can occur, by the pigeonhole principle, at least two individuals must have participated in the same number of handshakes. This is because there are more individuals (pigeons) than there are distinct handshake counts (pigeonholes), ensuring that at least one pigeonhole (handshake count) contains more than one pigeon (individual).

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