Calculating number of items in a summed series - non-repeating connections between points

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I would like to calculate the number of possible connections in a set of points, and although I can express the idea in a mathematical formula, I don't know how to "reduce" it to a working calculation.

Let's say I have points A to F that I want to interconnect. I can illustrate the task like so:

A  (B C D E F)
B  (C D E F)
C  (D E F)
D  (E F)
E  (F)
F  --

We start with A, connecting to B C D E F, then move on to B and connect to C D E F, omitting A as we already have the A-B connection from the previous step. And so on until F, where we create no new connections, given that all points are already connected to F.

Mathematically I can state this summing as:

(N-1)+(N-2)+(N-3) ... + (N-(N-1))

That's as far as I can get. I'm not a mathematician and don't know how to go from there to a solution.

I also suspect that what I'm doing here is a known problem in math. Does it have a name that I could look up?

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Okay, I'll make the comment an answer. The derivation of AP sum is here.
Another way you can count pairs of elements of a set of cardinality $n$ is ${n\choose 2}$ (MathJax ${n\choose 2}$) -- is number of ways to choose $2$ items out of $n$ distinct items irrecpective of order.
The formula of number of $k-$combinations is $${n\choose k}=\frac{n!}{k!(n-k)!}$$ and for $k=2$ this gives the same formula $${n\choose k}=\frac{n(n-1)}{2}.$$