At a party there are \(n\) people, with \(n\ge2\). Each person writes down how many handshakes they gave during the party. Is it possible that the \(n\) numbers noted are all different?
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The squeeze party
The scene is social, but the heart of the problem is structural. It is enough to look carefully at what the extremes really mean for the impossibility to suddenly appear.
Hints
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- The numbers posibles of apretones go of 0 a n-1.
- If they were all different, all those values would have to appear exactly.
- See what happens if you try to make 0 and n-1 coexist.
Solution
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Answer: No, it is impossible. If the $n$ counts were all different, then they would necessarily have to be:
$$ 0,1,2,\dots,n-1. $$ But that would force them to coexist: - a person with 0 grips; - a person with $n-1$ grips. The one from $n-1$ had to greet everyone, including the one from 0. Contradiction. Therefore, in any party with $n\ge2$, there are always at least two people who end up with the same number of squeezes. $$
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