Probability graph will have some nodes with full-mesh connectivity. There is a graph with $n$ nodes. This makes for $n \choose 2$ edges. We know for sure $l$ of these $n \choose 2$ edges are present and the rest are absent. What is the probability that for some $k<\min(l,n)$ (in fact we should have ${k \choose 2} \leq l$), there will be a sub-graph with full-mesh connectivity between those $k$ nodes (meaning every one of those $k$ nodes is connected to every other one of them). For example, when $n=5$, $k=3$ and $l=3$, I think there are $5 \choose 3$ ways of forming graphs where $3$ nodes form a full-mesh and $10 \choose 3$ ways of having three edges connected in general (${5 \choose 2 } =10$). So this probability should become: $\frac{5 \choose 3}{10 \choose 3}$. I'm particularly interested in the cases where $n=7$ and $k=4$.

My attempt: first, we choose $k$ nodes nodes that are going to form a full mesh. Number of ways of doing this is - $n \choose k$. Now, we have remaining $u = l - {k \choose 2}$ of the edges which are present. Overall, there are $v = {n \choose 2}-{k \choose 2}$ edges (present or absent) remaining. So the probability should become:
$$\frac{{v \choose u}{n \choose k}}{{n \choose 2} \choose l}$$
I feel I might be double-counting some cases. Is there anyway to verify this formula independently or even a simple pair of second eyes will help.
 A: Your approach is a good one but you double count cases where there are two different sets of $k$ vertices with a full mesh.  I think it will be a difficult inclusion/exclusion calculation because you might just have a full mesh on $k+1$ vertices, you might have a full mesh on $k+1$ with one or a few edges missing, or you might have two disjoint sets of $k$ vertices.
A: As Ross notes, it's a complicated  computation in general.
Let's try your example with $n=7$, $k=4$, $l=10$.  For each $4$-element subset $S$ of $[1..n]$ let 
$A(S)$ be the event that $S$ is a clique (which is the actual graph-theory term for what you call a "full mesh").  For each of the ${7 \choose 4}$ sets of $4$ vertices, 
$\mathbb P(A(S)) = {15 \choose 4}/{21 \choose 10} = \frac{5}{1292}$.  Given that $S$ is a clique, that has $6$ edges, leaving just $4$ more edges.  In particular, you can't have two edge-disjoint $4$-cliques, or even two that share a single edge (i.e. two vertices).  You could have two $4$-cliques sharing three vertices (say $\{a,b,c,d\}$ and $\{b,c,d,e\}$).  Either $a$ and $e$ are also joined, so this is a $5$-clique (which has not just two but $5$ $4$-cliques), or they are not.  The first case accounts for all $10$ edges, while in the second case there is one other edge which could be any of $11$ possibilities.  Thus the probability of $5$ $4$-cliques is ${7 \choose 5}/{21 \choose 10} = 1/16796$.  The probability of $2$ $4$-cliques is ${7 \choose 3}{4 \choose 2}\cdot 11/{21 \choose 10} = 55/8398$.  The expected number of $4$-cliques is ${7 \choose 4} {15 \choose 4}/{21 \choose 10} = 175/1292$, so the probability of $1$ $4$-clique must be
$$\frac{175}{1292} - 2 \cdot \frac{55}{8398} - 5 \cdot \frac{1}{16796} = \frac{1025}{8398}$$
