how to make a $3$-$(10,4,1)$ design using graphs A $t$-$(v,k,\lambda)$ design is defined this way :
We have a set with $v$ elements (called the points). We also have a collection of distinguished subsets each having $k$ elements, we call each of these subsets a block. Each set of $t$ points appear together in exactly $\lambda$ blocks.
I want to make a $3$-$(10,4,1)$ using graphs. According to this link:
http://www.sciencedirect.com/science/article/pii/0012365X83901176
( page 2 )
I know which graph I should use but I don't know how to convert that graph to a $3$-$(10,4,1)$ design.  
Note: one of my friends solved this problem using a $K_5$ graph. He said that we should see every edge of $K_5$ as a vertex in our new graph. But still we don't know why $K_5$ and why we should use this method and how to build that $3$-$(10,4,1)$ design.
 A: You take $\Gamma = K_{5}$, and consider the edges of $\Gamma$ to be the points of your design.  Then you have some specified subgraphs, given in the picture on page 2 that you refer to, they all have $4$ edges.  The blocks of your design are the subgraphs of $\Gamma$ that are isomorphic to one of these specified subgraphs. (considered as sets of edges).
For example, take any $3$ edges of $\Gamma$.  If they all share a common vertex, then they occur together in exactly one block, corresponding to the first picture in the list for these parameters.  If they form a triangle, then they occur in exactly one block, corresponding to the second picture.  Otherwise, they occur in a block corresponding to the third picture (try to show this).  This argument shows that this construction does in fact give a $3$-$(10,4,1)$ design.
The selection of $K_{5}$ is just related to the construction, in this paper they always start with a complete graph $K_{n}$ for some $n$.  Notice that $K_{5}$ has $(5\cdot 4)/2 = 10$ edges, which is the same as the number of points you want for your design.  Also notice that all the graphs they give in the table for this parameter set have $5$ vertices, they are all supposed to be taken as subgraphs of $K_{5}$.  Because of the way this is constructed, this gives the symmetric group $S_{5}$ as the automorphism group of the graph, and so $S_{5}$ will also act as an automorphism group of the design.
