Combinatorics: Steiner Triple System

[Robben]

Homework Statement

A Steiner Triple System, denoted by $STS(v),$ is a pair $(S,T)$ consisting of a set $S$ with $v$ elements, and a set $T$ consisting of triples of $S$ such that every pair of elements of $S$ appear together in a unique triple of $T$.

Homework Equations

None

The Attempt at a Solution

My book goes on to say that the number of triples of a $STS(n)$ disjoint from a given triple is $(n-3)(n-7)/6$ but I am not sure how they got that result?

I know that there are $n(n-1)/6$ triples altogether where each point of a triple lies in $(n-1)/2$ triples but I am not sure how they got that $(n-3)(n-7)/6.$

 

[certainly]

Try subtracting the number of non-disjoint sets from the total.

 

[Robben]

Try subtracting the number of non-disjoint sets from the total.

Can you elaborate please? What does it mean when a STS is disjoint from a given triple?

 

[certainly]

say the first triple is (a,b,c), for a triple to be disjoint to this triple it must not contain any of the elements a, b, c i.e. the union of 2 disjoint triples will be the null set.
[EDIT:- so you are to find all triples in the STS that do not contain any of the elements a,b or c.]

 

[Robben]

say the first triple is (a,b,c), for a triple to be disjoint to this triple it must not contain any of the elements a, b, c i.e. the union of 2 disjoint triples will be the null set.
[EDIT:- so you are to find all triples in the STS that do not contain any of the elements a,b or c.]

Oh, I see thank you.

 

[certainly]

Were you able to prove the desired result ?

 

[Robben]

Using your suggestion I got that $n(n−1)/6−3(n−1)/2$ but I am still not sure how they got $(n−3)(n−7)/6?$

 

[certainly]

Let

be a set of

elements together with a set

of 3-subset (triples) of

such that every 2-subset of

occurs in exactly one triple of [PLAIN]http://mathworld.wolfram.com/images/equations/SteinerTripleSystem/Inline6.gif. Then http://mathworld.wolfram.com/images/equations/SteinerTripleSystem/Inline7.gif is called a Steiner triple system.
Let's use this definition henceforth. It is not only much simpler, but also a lot more clear.
You are forgetting to subtract the original set and you are also forgetting that more than one 2-subsets were covered in the original triple. And since every 2-subset has a unique triple you need to take those into account.