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PLEASE answer it again. I could not find it.
20 couples go to 5 progressive dinners, 4 courses at each dinner. What is the formula so that no 2 couples are together more than 1 time, and no couple serves a course more than 1 time?
20 couples go to 5 progressive dinners, 4 courses and 4 couples at each dinner. What is the formula so that no 2 couples are together more than once, and no couple serves the same course more than once?
thank you bobbym, you have been a great help.
thanki you
bobym, The problem is that you have #1 serving the 1st course of 5 dinners. I am trying to get it so that no couple serves the same course more than once. If some couples are together more than once, than so be it. They just can't serve the same course more than once.
bobbym, No, there are still 20 couples. However, if some of the couples see each other more than once, than so be it. They just can't serve the same course more than once. I really appreciate ypour help.
bobbym, Yes, your suggestion is fine. Any solutions you can give me would be gresatly appreciated.
bobbyum, Last year I was able to work it out for 16 couples, 4 to a group, 4 dinners. This year we added 4 more couples, but still only 4 dinners having 5 groups for each dinner. Yes, days=dinners.
Hi bobbym, I guess I didn't explain myself very well. There are only 4 dinners. 5 groups of 4 each. The numbers must be scrambled so that each course is by a different couple. For example, you have #1 serving the 1st course each dinner.
There are 4 progressive dinners, 5 groups of 4 couples each (20 couples alltogether). Each dinner has 4 courses. Using each couple as a number from 1 to 20, how can it work out so that each couple (number) does only 1 course and is never with any other couple more than once?
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