Number of vertices of the graph = No of 4 element subsets of 10 element set

= ^{10}C_{4}

Now it is given :

Two vertices are adjacent iff they are disjoint subsets.

Now for disjointedness , it is necessary that the vertex should be adjacent iff the adjacent vertex contains no number in the subset represented by it which is in the given vertex representing the number ..

e.g. say we have a vertex for {1,2,3,4} , then there will not be an edge from it to {1,4,5,6} as 1 is common..

So in short for adjacency (to have an edge) we have only such vertices whose labels will be selected from rest of the 6 vertices..

So say we have taken the vertex {1,2,3,4} we can have a vertex representing subset which have 4 numbers ranging from 5 to 10 thus only 6 options and hence which can be done in ^{6}C_{4} = 15 ways in order to have an edge in between and hence incident on vertex labelled {1,2,3,4}

This is nothing but a degree of a single vertex ..So degree of vertex = 15

This will be true for all vertices

So applying handshaking theorem , we have :

2 e = 15 + 15 ...............^{10}C_{4} times [As number of vertices = ^{10}C_{4}]

==> 2 e = (15 * 10 * 9 * 8 * 7) / 24

==> e = 15 * 5 * 3 * 7

**==> e = 1575**

**Thus the given graph contains 1575 edges.. **

*For directed graph however it will be 3150 edges as say we have edge from vertex 1 to vertex 2 and vertex 2 to vertex 1 ..So both are considered to be different edges in case of directed graph and one and the same edge in case of undirected graph..By default we take undirected graph..*