They are asking to convert Complete Undirected graph into Directed graph without cycle by choosing direction for the edges.

See this $K_{3}$ graph-

(Image ref)

By this time you must have got Problem statement meaning. Your resultant graph should be acyclic.

Lets say you have a complete graph $G$ which has $n$ vertices, $v_1, v_2,\dots v_n$. To convert it into the resultant graph we have to assign direction to each edge. Now see, our resultant graph is acyclic therefore it must have topological order.

(I have not drawn all edges except $V_1$ edges.)

here every rearrangement of vertices in topological sort leads to one particular combination to choose the direction of edges.

Hence - $n!$ is answer.

Just to illustrate more, see one of the permutation out of $n!$

These two permutation shows that undirected edge between $V_1$ and $V_2$, was first chosen as $V_1 \rightarrow V_2$ and then $V_2 \rightarrow V_1$

Don't think about the labeling of vertices, If I do unlabelling of all $n!$ permutations then all structures are same. But it doesn't matter If I am arriving at the same structure, What matters is, In how many ways I can reach to that.

See this-

All these structures are Isomorphic...But still, there are $3!$ ways to reach such structure.

let there are 3 vertices,

Howmany ways we can arrage them ? it is 3! = 6

i) a,b,c

ii) a,c,b

iii) b,a,c

iv) b,c,a

v) c,a,b

vi) c,b,a

let take each order, and check howmany ways you make it as directed edges ( remember there are 3 edges in K_{3} ) with it

ex :- b,a,c ===> we make b ---> a ----> c, note that remaining edge must be b ---> c, there is no other possibility for it,due to if you make it c ---> b then it is cyclic.

So, there is only one way to assign edges, to each order !

Now, you got doubt that " **There may be many assignment of direction of edges than permutations ? **"

Answer for this question is, Should be NO. ( think about it, let there is an edge u --> v, then u should be before v )

in how many ways can you choose a direction for the edges so that there are no directed cycles? ==> No.of permutations of vertices = n !