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Is there a way to find no of perfect matchings in a complete graph Kwhere n could be either even or odd..?

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if n is odd then perfect matching 0. because in perfect matching degree of each vertex must be 1, which is not possible if n is odd.

and if n is even then num of perfect matching in $K_{2n}=\dfrac{(2n!)}{( 2^n * n! )}$
by Loyal (8.1k points)
edited by
+2

explain how  K2n=( 2n! ) / ( 2^n  *  n! )

+5
\begin{align} \#\text{Perfect matchings of $K_{2n}$} &= (2n-1)(2n-3)(2n-5)\cdots 5.3.1\\ &=\dfrac{(2n)(2n-1)(2n-2)(2n-3)\cdots 5.4.3.2.1}{(2n)(2n-2)\cdots4.2}\\ &=\dfrac{(2n)!}{(2*n)(2*(n-1))\cdots(2*2)(2*1)}\\ &= \bbox[yellow,5px,border: 2px solid red]{\dfrac{(2n)!}{2^n\times n!}} \end{align}
For Kn

if n is odd , then there is no perfect matching.

n is even then you can count it like ->

(n-1) (n-3) (n-5)...1 (This will end in 1 as n is even).
by Boss (41.5k points)

+1 vote