# Too many school assignments

This year we have to make our school assignments in pairs.
With each classmate must be made exactly one of those assignments.
Exactly 30% of the assignments will be made by a pair of girls.

How many assignments do I have to make?

Say we have $$b$$ boys and $$g$$ girls, then we have $${g\choose 2} = {3\over 10}{b+g\choose 2}$$ so we have this equation $$3b^2 +3b(2g-1)= 7g(g-1)$$

Now for each $$g$$ find $$b$$. If we put $$x = 2(b+g)-1$$ and $$y= 2g-1$$ we get $$\boxed{10y^2-7=3x^2}$$

Here are some values • Looks like a variant of Pell's equation, so I would try to build infinite solutions with the same algorithm used in Pell's equation. I don't know the exact theory for this variant, though. Apr 24 at 18:36
• This doesn’t directly answer the original question. Also it doesn’t work for y=x=1 since that would mean g=1 and b=0, which would mean there were zero assignments. (I guess technically 30% of zero is zero, but it still seems to go against the spirit of the question.) Apr 24 at 18:57
• @Greedoid much better, though strictly speaking the question was “how many assignments do I have?” Apr 25 at 15:20
• I agree with Nick (both times). An addition that the most reasonable class size is 16 (at least I think so), and esp. that the (probably intended) solution is thus 15 would really make it an answer to my question. Still it seems complete enough to already accept it. Apr 25 at 20:28

I'm not sure if this is the only possible solution to the puzzle, but I found that if:

You have 2 boys and 3 girls in the class, then 10 total assignments will be made where 3 are made by solely girls. This means that you have to make four assignments in total.

Again, there may be more solutions based on the rules outlined in the puzzle.

• How could I have missed that one.. I was looking for the other answer and maybe a proof that there is no solution with very large classes. I guess 4 would already be too many for many students. Apr 24 at 12:11