Question: There are 12 points on a circle, one of which is painted red, and another is painted blue, with the remaining 10 points unpainted. Convex polygons formed by these points are classified as follows: polygons whose vertices include both the red and blue points are called bicolored polygons; polygons that include only the red (or blue) point are called red (or blue) polygons; polygons that do not include the red or blue points are called colorless polygons.

How many more bicolored polygons are there compared to colorless polygons, among all convex polygons (with sides ranging from triangles to dodecagons) formed by these 12 points?
(Ninth Zu Chongzhi Cup Mathematics Invitational Competition)
Simplified: From the problem, we know that a bicolored polygon must include two points of different colors, while a colorless polygon must not include any of these points. Since all 12 points are on the circumference of the circle, the number of bicolored polygons is $\sum_{i=1}^{10} C_{10}^{i}$, and the number of colorless polygons is $\sum_{i=3}^{10} C_{10}^{i}$. Also, $\sum_{i=1}^{10} C_{10}^{i}-\sum_{i=3}^{10} C_{10}^{i}=55$, so the number of bicolored polygons is 55 more than the number of colorless polygons.

Generalization 1 For $n$ points on the circumference of a circle $(n \geqslant 3)$, then ?.

Generalization 2 If there are $n$ points on the circumference of a circle $(n \geqslant 3)$, and 3 of these points are painted red, blue, and yellow, respectively, the number of colorless, monochromatic, bichromatic, and trichromatic polygons are $2^{n-3}-\frac{n^{2}-5 n+8}{2}$, $3\left(2^{n-3}-n+2\right)$, $3\left(2^{n-3}-1\right)$, and $2^{n-3}$, respectively.

Generalization 3 If there are $n$ points on the circumference of a circle $(n \geqslant 3)$, and $k$ of these points $(0 \leqslant k \leqslant n)$ are painted with $k$ different colors, the number of $m$-colored polygons $(0 \leqslant m \leqslant k)$ is
$$
\left\{\begin{array}{l}
C_{k}^{m} \sum_{i=3-m}^{n-k} C_{n-k}^{i}, 0 \leqslant m \leqslant 2 ; \\
C_{k}^{m} 2^{n-k}, 3 \leqslant m \leqslant n .
\end{array}\right.
$$

Proof omitted.