Number of edges of the graph $G$

Question

Let $G$ be a graph of which the vertices are the permutations of $\{1,2,3,4,5,6,7,8,9,9,9\}$ with the property that two vertices $(\epsilon_1, \epsilon_2, \ldots, \epsilon_{11})$, $(\epsilon_1', \epsilon_2', \ldots, \epsilon_{11}')$ are connected with an edge if and only if the one is resulted from the other by exchanging the positions of two different integers.

Could you give me an example of what exactly this property means, because I haven't really understood that.

Also how can we calculate the number of edges of the graph $G$ ?

Answer 1

No, The answer isn't $11!$ Frankly, that sounds like just a guess. Pick a vertex. How many other vertices is it adjacent to? Is the answer the same for every vertex? How many vertices are there? Now recall that twice the number of edges is the sum of the vertex degrees.

You should really try to work out the answers to all these questions by yourself. You can't expect to answer math questions instantly, at least not while you're learning the subject. It's one thing to ask what the question means when you can't understand it, because staring at it is unlikely to bring enlightenment, but once you know what the problem is, I really recommend that you work on it for at least half an hour before you decide that you are stuck. You might want to work it out first for the case where there are no repeated numbers, or for a smaller example. What is the answer if the original numbers are $1,2,3?$ What if they are $1,2,3,3$?

EDIT

I'll show you a simple example. Suppose the numbers are $1,2,3,4$. Then we have $24$ vertices. How many neighbors does a vertex have? How may ways are there to swap two vertices? ${4\choose2}=6$ (Why?) The sum of the vertex degrees is $24\cdot6=144$ and this is twice the number of edges, so there are $72$ edges.

Can you modify this for your problem?

EDIT

There are ${11!\over3!}$ vertices. Each vertex is adjacent to ${8\choose2}+8\cdot3=52$ other vertices, since we can choose two of the first $8$ numbers or one of the first $8$ and one of the $3$ nines. That gives ${11!\cdot26\over3!}$ edges.