A game of darts (sort of)

Let $T$ be the final score.

Since we can choose how the darts are arranged and there is an infinitely sized plane, assume that no three darts of the same colour are collinear and no three lines are concurrent.

If you chose any 4 of the darts of the same colour you would be able to form one convex quadrilateral. If you join the 2 diagonals of this convex quadrilateral you get 1 intersection. Therefore, for every convex quadrilateral, you get 1 intersection. Therefore, we have to find the number of ways that you can make a convex quadrilateral. To make a quadrilateral you need to choose four points from all the points available. However, to ensure no concave quadrilaterals are made, arrange all the darts of the same colour in a circle so that no three lines are concurrent.

Therefore, the number of intersections with $n$ darts is $\begin{pmatrix} n \\ 4 \end{pmatrix}$ .

Therefore, the number of intersections between 7 darts is $\begin{pmatrix} 7 \\ 4 \end{pmatrix} = 35$ and the number of intersections between $x$ darts is $\begin{pmatrix} x \\ 4 \end{pmatrix}$ . Thus the sum of these intersections and the number of green darts is $35+\begin{pmatrix} x \\ 4 \end{pmatrix}$ .

Therefore, $T$ (the number of green intersections) $= \begin{pmatrix} 35+\begin{pmatrix} x \\ 4 \end{pmatrix} \\ 4 \end{pmatrix}$ .

Now we can plug in values where $x \ge 4$ because that is the number of darts you need for 1 intersection.

When $x = 4$ , $T = 58905$ .

When $x = 5$ , $T = 91300$ .

When $x = 6$ , $T = 230300$ .

...

When $x = 10$ , $T = 146475945$ .

When $x = 11$ , $T = 727441715$ .

When $x = 12$ , $T = 3250609780$ .

$727441715 < 1000000000 < 3250609780$

$f\left( 11 \right) <1,000,000,000<f\left( 12 \right)$

Therefore the maximum amount of blue darts he can have is $\boxed { 11 }$ .