B Amount of lengths between k points

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The discussion centers on calculating the total number of lengths between k distinct points on a straight line. For k=2, there is one length, while for k=3, there are three lengths, and for k=4, six lengths, following the combination formula for selecting pairs. The pattern emerges that the number of lengths is given by the binomial coefficient, specifically, the formula for combinations of k points taken 2 at a time, represented as C(k, 2). This leads to the conclusion that the total lengths increase as more points are added, following the mathematical principle of combinations. Understanding these combinations is crucial for determining the relationships between multiple points on a line.
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Say that there lies k points upon a straight line.

My question is this: what is the total amount of lengths between these points?

To elaborate, say k is 2; there are two points and so there is only one length between the two points. How many lengths would there be with k > 2?

What is the proof for a solution to this?
 
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Einstein's Cat said:
Say that there lies k points upon a straight line.

My question is this: what is the total amount of lengths between these points?

To elaborate, say k is 2; there are two points and so there is only one length between the two points. How many lengths would there be with k > 2?

What is the proof for a solution to this?
What do you get if there are three points? Four? Five?

BTW, it's probably reasonable to assume that the points are all distinct.
 
Mark44 said:
What do you get if there are three points? Four? Five?

BTW, it's probably reasonable to assume that the points are all distinct.
I can represent each length between points like this for k= 3 (each pair of integers represents the length between the points that are labelled such)

1, 2
1,3
2,3

for k= 4:

1,2
1,3
1,4
2,3
2,4
3,4

for k= 5

1,2
1,3
1,4
1,5
2,3
2,4
2,5
3,4
3,5
4,5
 
Going from k=2 with 1 distance to k=3, you got 2 additional distances (those to the additional point). Going to k=4, you got 3 additional distances. Do you see a pattern?

Edit: Typo
 
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Einstein's Cat said:
however k=4 gave 6 lengths and k=5 gave 10, this defies such a pattern surely?
All is well
 
Einstein's Cat said:
I can represent each length between points like this for k= 3 (each pair of integers represents the length between the points that are labelled such)

1, 2
1,3
2,3

for k= 4:

1,2
1,3
1,4
2,3
2,4
3,4

for k= 5

1,2
1,3
1,4
1,5
2,3
2,4
2,5
3,4
3,5
4,5
You need to be more specific about where the points are located. In your example for k = 3, the points appear to be at 1, 2, and 3 units, respectively. For three points, there are ##\binom 3 2## ways (the number of combinations of 3 things taken 2 at a time) of choosing two points among the three, where ##\binom 3 2 = \frac{3!}{2! 1!}## . For four points, there are ##\binom 4 2 = \frac{4!}{2! 2!}## = 6 ways. For each pairing of points, you need to calculate the distance between the two points.
 
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