Sorry Anonymous, but what I really was looking for was not three groups of seven or seven groups of three....math hasn't ever been my strongest subject, but I do know that 21 is divisible by 3 and 7. What I need is 9 groups of 2 and 1 group of 3, where the groups are always different and the 'third wheel' is not always the same person. I also wanted each person to have at least one pairing with each other person at some point.
In solving the problem, Dad earned himself a bonus by introducing me to a new word - "combinatorics" (in which he took at least two classes in college) and he also made me feel better by assuring me that this truly is a complex problem.
This is a picture of a couple of the many pages I filled with numbers in trying to solve the problem myself. I kept thinking I had it all figured out and then something would go terribly wrong and the headache would start to creep in...
So, thanks Dad! And thanks to any other minds out there that were trying to help with the problem. If you're still interested, I'll see if I can find a way to post the final results, when I actually make the lists for the kids.
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