hjnet

Just imagine 12 girls, even if you'd only need to group them in groups of 2 you'd only need 121 combinations (11x11), if you can group them in groups of 3 then you'd need less combinations to ensure that "each girl performs at least once with each of the other girls", not more.

Or just imagine 4 Girls, to make it simplier let's call them A;B;C and D. To group them in groups of 3 so that each girl is ounce with each other you'd have to make the following combinations ABC; DAB and DC?. So you'd only need to make 3 combinations and you still have one place left for an additional girl (so you can save a few combinations if you group them smart) With your calculation you'd need 4 combinations. If you take 5 Girls A;B;C;D;E then you'd need ABC; DAB; EAB and ECD, 4 combinations, not 10 as with your calculation. So I don't know what you've calculated there, but 220 isn't right.

But I have to say that I can't mathematical proof my own number of combinations (34) at this time and it might be possible that I'm wrong too, but I can give you some kind of "circumstantial evidence"

If you'd just have that you'd have to group in groups of random 3 then you'd have with:

3 Girls, 1 combination neccessary
4 Girls, 3 combinations neccessary
5 Girls, 5 combinations neccessary
6 Girls, 8 combinations neccessary
7 Girls, 11 combinations neccessary
8 Girls, 15 combinations neccessary
9 Girls, 19 combinations neccessary
10 Girls, 24 combinations neccessary
11 Girls, 29 combinations neccessary
12 Girls, 35 combinations neccessary

If you can group them in a smart way you can save a few combinations:

3 Girls, 1 combination neccessary
4 Girls, 3 combinations neccessary
5 Girls, 4 combinations neccessary
6 Girls, 7 combinations neccessary
7 Girls, 9 combinations neccessary
8 Girls, 13 combinations neccessary
9 Girls, 16 combinations neccessary
10 Girls, 21 combinations neccessary
11 Girls, 27 combinations neccessary
12 Girls, 33 combinations neccessary