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How many perfect matching are there in a complete graph of 6 vertices?

1. 15
2. 24
3. 30
4. 60

Perfect matching is a set of edges such that each vertex appears only once and all vertices appear at least once (EXACTLY one appearance). So for $n$ vertices perfect matching will have $n/2$ edges and there won't be any perfect matching if n is odd.

For $n = 6$, we can choose the first edge in ${}^6C_2=15$ ways, second in ${}^4C_2= 6$ ways and third in ${}^2C_2 = 1$ way. So, total number of ways $= 15\times 6=90$. But perfect matching being a set, order of elements is not important. i.e., the $3!$ permutations of the $3$ edges are same only. So, total number of perfect matching $= 90/3!= 90/6 = 15$.

Alternatively we can also say there are 3 identical buckets to be filled from 6 vertices such that 2 should go to each of them. Now the first vertex can combine with any of the other 5 vertices and go to bucket 1- 5 ways. Now only 4 vertices remain and 2 buckets. We can take one vertex and it can choose a companion in 3 ways and go to second bucket- 3 ways. Now only a single bucket and 2 vertices remain- so just 1 way to fill the last one. So total ways=5*3=15.
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Sir,

How is it different from the following statement from at MIT Lecture:

"When a k regular graph can be colored with exactly k colors, a remarkable thing happens – every vertex appears in every mono-chromatic sub-graph, i.e. the coloring decomposes the graph into k perfect matchings"

Considering K6 as a 5 regular graph, there will be 5 different perfect matchings (as the above statement says).

So my confusion arises here !!

Are these 5 perfect matching a part of those 15 perfect matchings

OR

Am I getting some wrong ideas about the above quoted statement.

@Arjun how u get this

So for n vertices perfect matching will have 3 edges and there won't be any perfect matching if n is odd.

bold part i got .

Sorry. That's n/2, not 3. Corrected now.
@Arjun Sir

Why did you divide the result by 3! in the first method and not in the second method?

Note: To understand the solution please go through the definitions of perfect matching

The complete graph kn have a perfect matching only when n is even.. So let n=2m.

Let the vertices be V1 , V2 ,......,V2m.

v1 can be joined to any other 2m-1 vertices

v2 can be joined to any other 2m-3 vertices

Similarly go till V2m which will have only one vertex to be joined with..

No of Perfect matches= (2m-1)(2m-3)(2m-5).....(3)(1)

In the above question 2m=6

So No. of perfect matches=5*3*1=15

edited by

"v2 can be joined to any other 2m-3 vertices"

Let n = 4 (2m = 2x2)
Then as per the above stmnt. v2 will have only one option but actually v2 will have 2 options.. is something wrong in my understanding?

Let v1,v2,v3,v4 be the vertices... So for v1 we need to take any of the other 3 vertices.. Now we have only two vertices left which can be joined in only one way.. Hope you understand.. :)

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.

edited
how 15 ?
2n = 6,

n = 3

Updated.

For complete graph of 6 vertices

Number of perfect matching= n(n-1)/2

= 6*5/2=15
Your formulae is wrong. It just hit here correctly by luck . Try for n =8

Akash Kanase sir is right