In a circle, 15 white points and 1 black point were marked. Consider all possible polygons (convex) with their vertices at these points.

Let's separate them into two types:

- Type 1: those that have only white vertices.
- Type 2: those that have the black point as one of the vertices.

Are there more polygons of type 1 or type 2? How many more are there?
Notice that for each type 1 polygon, we can construct a type 2 polygon by adding the black point.

On the other hand, if we have a type 2 polygon and remove the black point, the only way not to generate a polygon is if exactly two white points remain.

Therefore, there are more type 2 polygons than type 1 polygons.

To calculate the difference, we just need to count the number of pairs of white points. For this, observe that each white point can form a pair with each of the other 14 white points. Thus, since there are 15 white points, we will have \(15 \times 14\) ordered pairs. It follows that we have \(15 \times 14 / 2 = 105\) pairs of points.

Observation: It is possible to determine the number of type 1 and type 2 polygons. See the box "Counting Subsets" on page 118.
Suggestion: Construct a type 2 polygon from a type 1 polygon.
![](https://cdn.mathpix.com/cropped/2024_05_01_9f6ce5ffa52bbf7500dfg-3.jpg?height=842&width=330&top_left_y=1124&top_left_x=1502)

## Suggestion:

(a) Suppose

$$
a \leqslant b \leqslant c \leqslant d \leqslant e
$$

What can we say about \(a + b\)? And about \(d + e\)? And about \(a + c\)?

(b) Carlos will not be able to achieve his goal because there are two sets of four numbers that generate the numbers 10, 20, 22, 24, 26, and 36.