How to find the number of perfect matchings in complete graphs?

It's just the number of ways of partitioning the six vertices into three sets of two vertices each, right? So that's 15; vertex 1 can go with any of the 5 others, then choose one of the 4 remaining, it can go with any of three others, then there are no more choices to make. $5\times3=15$.


If you just want to get the number of perfect matching then use the formula $\dfrac{(2n)!}{2^n\cdot n!}$ where $2n =$ number of vertices in the complete graph.

Detailed Explaination:- You must understand that we have to make n different sets of two vertices each.

First take a vertex . Now we have (2n-1) ways to select another vertex to make the pair. Now to make another pair we take a vertex and now we have (2n-3) ways to select another vertex. This is because we have already used 2 vertices in first pair and one vertex is currently in use to make 2nd pair. Similarly for 3rd pair we will have (2n-5) ways . When we are making nth pair we will have just one way.

Multiplying all we get (2n-1)(2n-3) ...........1 Now multiply and divide it by even terms as follows ((2n)(2n-1)(2n-2)(2n-3)..................1)/((2n)(2n-2)(2n-4).....*2)

now the numerator will become (2n)! and take 2 common from each term in denominator . You will get 2^n * (n*(n-1)(n-2).......1) Hence the denominator will become 2^n * n! hence we get the formula as (2n)!/(2^n*n!). Hope this will help.

Tags:

Graph Theory