Specific subset of verticies are degree 1 in random graph probability On a random graph $G_{n,p}$, I want to know the probability that some subset of verticies is degree one. Take it to be the verticies $A = \{1,2,\dots,k\}$. 
My approach for  this is to notice that if they are all degree one, then there must be an even number of verticies in $A$ that all pair with one another, and the rest of the verticies that don't pair within another vertex in $A$ must connect to exactly one vertex in $[n] - A$. 
I them sum over the probability of even numbers $2j\leq k$ such that the $2j$ verticies all pair with one another and nothing else, and the $k-2j$ verticies all connect to exactly one vertex in $[n]-A$ and nothing else. 
I want to know if I'm missing anything, or if this is a valid approach. 
 A: This is valid, especially if you're trying to compute the exact probability. (You should also, if you're not doing so already, include a factor for the number of ways to choose $j$ edges between $k$ vertices.)
If the random graph is sparse (which it ought to be: for $p \gg \frac{\log n}{n}$, we don't get any degree-one vertices with high probability) and $A$ is not too large, then looking at edges inside $A$ is more work than you need to do, when all you want is an asymptotic probability.
The probability that there are any edges inside $A$ is at most $\binom{|A|}{2}p$. When this is a $o(1)$ term, we can ignore it, and assume $A$ is independent. This makes the rest of the calculation much simpler, because you can just multiply the probabilities together for each vertex in $A$.
If you are later planning to sum over all choices of $A$, you do have to say a bit more than "this is a $o(1)$ term" because there are $O(n^{|A|})$ such choices. However, even then, the case where $A$ is independent dominates all other cases: when $A$ is independent, you get to choose $O(n^{|A|})$ neighbors for the degree $1$ vertices of $A$, and when $A$ contains $j>0$ edges, you only get to choose $O(n^{|A|-2j})$ neighbors.
