For 2nd perfect matching no of options = 3 (two vertexes are already used)

For third perfect matching no of options = 1( 4 vertices are already used)

so total = 5 * 3 * 1 = 15

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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 $=\frac{90}{3!}= \frac{90}{6} = 15$.

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$=5\times 3=15.$

Correct Answer: $A$

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 $=\frac{90}{3!}= \frac{90}{6} = 15$.

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$=5\times 3=15.$

Correct Answer: $A$

Imagine a world with $6$ people.

Assume a graph, where each person is a vertex and there's an edge between two vertices, if the two persons representing the vertices like each other.

So, the question represents a scene where everyone loves everyone else, and there's no conflicts of gender whatsoever (because the graph is complete, and not bipartite).

We, the gods, as a wedding planner, have a job to unite those who like each other. If we do our job perfectly, there's no lonely person in the end, and every person is married to exactly one person. We call it a "perfect matching".

If everyone likes everyone else, then it's very easy for us.

We pick first candidate to marry, there's $6$ choices for that.

Now, we pick second candidate to marry the first one we picked, there's $5$ choices for that.

So, total possible matches are $6 \times 5 = 30$

But, who we picked first, and who we picked second doesn't matter, because in the end, they're going to be the same couple.

So, total possible matches will be $ = \frac{30}{2} = 15$

Assume a graph, where each person is a vertex and there's an edge between two vertices, if the two persons representing the vertices like each other.

So, the question represents a scene where everyone loves everyone else, and there's no conflicts of gender whatsoever (because the graph is complete, and not bipartite).

We, the gods, as a wedding planner, have a job to unite those who like each other. If we do our job perfectly, there's no lonely person in the end, and every person is married to exactly one person. We call it a "perfect matching".

If everyone likes everyone else, then it's very easy for us.

We pick first candidate to marry, there's $6$ choices for that.

Now, we pick second candidate to marry the first one we picked, there's $5$ choices for that.

So, total possible matches are $6 \times 5 = 30$

But, who we picked first, and who we picked second doesn't matter, because in the end, they're going to be the same couple.

So, total possible matches will be $ = \frac{30}{2} = 15$

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@coder_k The next paragraph in the mit-notes gives the following example:

“As an example, consider the graph K4. The edges of K4 can be properly colored with 3 colors. This is easy to see if we draw K4 as an equilateral triangle with a vertex in the center connected to each of the corners oriented so that it’s base is horizontal Take the horizontal along with the vertical edge at the top and color them R, then take the congruent pair of edges which are rotated 120 degrees and color them G, and then color the final pair (rotated another 120 degrees) blue. The cube is another example of a 3-regular graph whose edges can be properly colored with 3 colors.”

If you use the logic given in the answers here, you’ll also arrive at the answer for K4 as 3. (1st node, say A, has 3 choices B,C,D. If it chooses B, now C(or D) has only one choice left. therefore 3x1 = 3 matchings possible, which is the same conclusion as in the above example.)

Now, for your confusion, read your highlighted quote carefully – “* When a k regular graph can be colored with exactly k colors….” – *is the part you overlook.

K4, which is a **3-regular** graph with 4 vertices, can be **edge colored** with **exactly 3 colors**. But it is **not necessary that every k-regular graph be edge colored by exactly k colors**.

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47 votes

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

The complete graph $k_{n}$ have a perfect matching only when n is even. So let $n=2m.$

Let the vertices be $V_{1} , V_{2},\ldots ,V_{2m}$._{ }

$v_{1}$ can be joined to any other $\left(2m-1\right)$ vertices.

$v_{2}$ can be joined to any other $\left(2m-3\right)$ vertices.

Similarly, go till $V_{2m}$ which will have only one vertex to be joined with.

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

In the above question $2m=6.$

So, No. of perfect matches = $5\times 3\times 1=15.$

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22 votes

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**.

2 votes

No. of perfect matchings in a complete graph $K_{2n}$ =

Please note it is for $K_{2n}$ and not $K_{n}$.

Here, 2n = 6.

So, $\frac{6*5*4*3*2}{3*2*8}=15$

As for Cyclic graphs $C_n$

Two matchings if n is even, none otherwise.

https://math.stackexchange.com/questions/1981111/number-of-distinct-perfect-matchings-in-a-cycle