Math CentralQuandaries & Queries


Question from Mark, a teacher:

Eleven guys are going on a fishing trip and want to rotate so everyone fishes with the other guys with the fewest number of repeats.
They want to fish in groups of two with one fishing alone for
three and a half days so will rotate seven times.
Is there a combination that works and what is it?


There is a schedule with no repeats, but you need 11 trips to have each person with each of the ten possible partners (one week is solo). Try this:

1: 1&10, 2&9, 3&8, 4&7, 5&6, 11
2: 2&11, 3&10, 4&9, 5&8. 6&7, 0
3: 3&1, 4&11, 5&10, 6&9, 7&8, 1

To get the schedule for the next trip, add one to every number in the current one, and wrap 11 back around to 1 as if on an odometer. If you continue long enough, this will give you the 11 trip schedule that was mentioned above. Using the first seven trips will give you a schedule that’s the best you can do, so you can stop after you have seven.


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