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 Two rights out of 24 wrongs (Posted on 2005-07-27)
At a party, my wife set 24 place cards on our round table, but everybody sat down randomly, and nobody was in the right place.

My wife commented: "let's turn the table round, until at least two will be in the right place."

Can this always be so, or was it just a fluke?

 See The Solution Submitted by e.g. Rating: 3.3000 (10 votes)

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 Solution | Comment 1 of 11

Imagine turning the table around to each of the 24 configurations. Clearly, my card will be in front of me exactly once, and the same goes for every other guest. So the sum of the number of correct placements for all 24 configurations must be exactly 24 (feel free to use "N" in place of 24 if you want to generalize.)

Now, since in the current configuration there are zero correct placements, the sum of the correct placements in the other 23 configurations must still be 24. Since the largest possible sum that *doesn't* have a member > 1 is only 23, there must be at least one configuration with two or more correct placements.

 Posted by Paul on 2005-07-27 00:54:49

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