>Total no. of edges = C(2n,2), Choose 1 edge in C(2n,2) ways.
>Remaining no. of edges C(2n-2,2). (Since 2 vertices already selected.
>So, no. of ways of choosing 2 edges = C(2n,2)*C(2n-2,2).
>Total ways of choosing edges -> C(2n,2)*C(2n-2,2)*..*C(2,2)
> => ((2n)!)/(2^n) *(Note: Missing n! in denominator)*

>Choose a vertex. Choose another vertex (part of same edge): 2n-1 ways.*(Why no. of ways of selecting first vertex not considered??)*
>Choose next vertex. Choose the pair vertex (part of same edge): 2n-3 ways.
>So end result - (2n-1)*(2n-3)*..*(3)*(1) (Product of all odd numbers less than 2n)
>Simplification gives -> ((2n)!)/((2^n)*n!)