Six, (Full marks 12 points) On a circle, there are
12 points, one of which is painted red, and another is painted blue, with the remaining 10 points unpainted. Among the convex polygons formed by these points, those whose vertices include both the red and blue points are called bicolored polygons; those that include only the red (blue) point are called red (blue) polygons, and those that include neither the red nor the blue point are called colorless polygons.

Try to determine, among all the convex polygons (which can range from triangles to 12-gons) formed by these 12 points, 1. whether the number of bicolored polygons is greater than the number of colorless polygons, and by how many.
For any bicolored $n(n \geqslant 5)$-gon, naturally, removing the red and blue vertices results in a colorless $(n-2)$-gon. Different bicolored $n$-gons, after removing the red and blue vertices, yield different colorless $(n-2)$-gons; conversely, for any colorless polygon, adding red and blue vertices can always result in a bicolored polygon. Therefore, the number of colorless polygons (from triangles to decagons) is equal to the number of bicolored polygons (from pentagons to decagons).

Thus, the number of bicolored polygons is greater.
The additional number is exactly the number of bicolored triangles and bicolored quadrilaterals. It is easy to calculate that there are 10 bicolored triangles; bicolored quadrilaterals have $\frac{1}{2} \times 10 \times 9=45$ (units), meaning bicolored polygons are 55 more.