For a graph with odd number of vertices we cannot have a perfect matching.
If there are $2n$ vertices in a complete graph we can surely find a perfect matching for it.
to find perfect matching we need to group vertices of the graph into disjoint sets of 2 vertices. doing so we are able to pull out an edge from the graph which is between those 2 vertices of a set. also, the set is disjoint so pulling out an edge will select only distinct vertices each time, till all vertices are covered.
this will give rise to n sets each of 2 vertices.
Now, the problem is reduced to a combinatorics problem of distributing $2n$ vertices into $n$ sets, each of 2 vertices. also, these $n$ sets are indistinguishable as they contain same number of elements so, counting all ways by which we can do that is given by:$$\frac{(2n)!}{(2!)^n\ n!}$$
here, $2n=6$ so, answer = 15.