Amount of lengths between k points

[Einstein's Cat]

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?

 

[Mark44]

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.

 

[Einstein's Cat]

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

 

[mfb]

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

 

[Einstein's Cat]

however k=4 gave 6 lengths and k=5 gave 10, this defies such a pattern surely?

All is well

 

[Mark44]

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.