edited by
10,620 views

2 Answers

Best answer
17 votes
17 votes
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 $K_{2n}=\dfrac{(2n!)}{( 2^n  *  n! )}$
edited by
15 votes
15 votes
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).

Related questions

1 votes
1 votes
2 answers
1
0 votes
0 votes
1 answer
2
atul_21 asked Dec 21, 2017
510 views
I am not convinced by this. Please explain or please tell me the source from where I can clear this out.
0 votes
0 votes
0 answers
3
Jaspreet Kaur Bains asked Dec 19, 2017
1,689 views
Consider complete graphs K5 and K6 . Let X5 and X6 are number of perfect matching of K5 and K6 respectively. Then X5 + X6 = ________.