Number of labeled graphs satisfying a degree sequence Say we have two sequences of integers $d^\text{in}$ and $d^\text{out}$ representing the in- and out-degree sequences of a directed graph. How many (possibly isomorphic) graphs are there that satisfy this sequence? 
One paper seems to think this is a simple matter, and uses the formula
$$
\frac{m!}{\prod_i d^\text{in}_i!\prod_i d^\text{out}_i!}
$$
My knowledge of combinatorics isn't good enough to prove the correctness of this, but I can sort of see how it works, tying up the outgoing links to ingoing links and dividing out the results that produce the same graph.
Another paper, published later, gives only an approximate formula using a far more complex method.
Is the first method incorrect? If so why?
Additionally, if I want to get a decent estimate for this problem (directed or undirected graphs) without too much fuss, is there a simple method available?
 A: Partial answer:  That first formula does not work for counting directed graphs with prescribed degrees (I assume multiple edges and loops are allowed).  One example where it fails is if $d^{out} = (2, 2, 0)$ and $d^{in} = (1, 1, 2)$.  In this case, the formula gives $3$ but there are $4$ such graphs.  
To see the problem, consider trying to count the situation where $u$ has edges going to $v$ and $w$.  The formula does do well to divide out by $d^{out}(u)!$, since the $m!$ total would count a vertex $u$ connecting to $v$ and $w$ different from that vertex connecting with $w$ and $v$.   Since we want to count these as the same, we divide by $d^{out}(u)! = 2$.  Similarly with dividing out by $d^{in}(v)$.  The problem is if you had two edges going from $u$ to $v$.  The $m!$ counts two different possibilities, but then we divide by $d^{out}(u)! = 2$ and $d^{in}(v)! = 2$, thus dividing by $4$.  This is essentially an over-count of how much we should divide.  
I don't know how you'd find a better answer besides the second paper you linked to.
