a) Given four distinct points $A, B, C$ and $D$, all on the same line, as indicated in the figure below, determine the number of distinct segments that can be formed with vertices at these points.

![](https://cdn.mathpix.com/cropped/2024_05_01_80f7f1185898b39e7876g-07.jpg?height=54&width=506&top_left_y=1983&top_left_x=872)

b) With 10 distinct points on a segment, what would be the new answer?

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Solution

a) There are 6 line segments with vertices at these 4 points: $A B, A C, A D, B C, B D$ and $C D$. Note that the answer would not be different if the points were not collinear.
b) Let's call the 10 points: $A, B, C, D, E, F, G, H, I$ and $J$. Observing the solution from part a), we can organize our count by initially listing all segments with vertex $A$ and then all those that have vertex $B$ and have not yet been counted in our list. After counting for vertex $B$, we repeat this procedure for the other vertices.

With 10 points, there are 9 segments in which one of its vertices is point $A$, as it is sufficient to connect $A$ to each of the other 9 points. With vertex $B$, there are also 9 segments, but one of them has already been counted, as it also has vertex $A$. Thus, we only need to add 8 more new segments. With vertex $C$, there are $9-1-1=7$ new segments because we have already counted the segments $A C$ and $B C$. For each new vertex, the number of new segments is reduced by one unit. Thus, the total number of segments is

$$
9+8+7+6+5+4+3+2+1=45
$$