Please refer to this link. It is Erdos and Renyi's first paper on Random Graphs (1959). I am trying to work through it.
I'm struggling with equations (16), (17) and (21).
(16) They split the sum in (13) for $M<s\leq \frac n 2$ and $\frac n 2 < s < n- \frac {2 N_c} n$. In the first case they use (14) and, since the terms are positive, they enlarge the sum to $M<s<\infty$. In the second range they use (15) to estimate the sum, then they substitute $s'=n-s$, and then they extend to $\frac {2N_c}n <s'<\infty$.
The elementary estimates (14) and (15) are not redundant, because they show that the same quantity (the LHS) is estimated differently in case $s\leq \frac n 2$ or $s\geq \frac n 2$. Of course (15) can be proved from (14) by symmetry and vice-versa. This explains why, if you substitute $s'=n-s$ before applying (14) and (15), then you need only (14).
(17) They replace $M$ with $\log \log n$ because they claimed at the beginning of the proof: "let $M$ be a large enough number that we will choose later". That's the moment when they choose it. (16) implies (17).
In more detail, they choose this value $M=\log\log n$, not too big and not too small, because it works conveniently for both $E_{\log \log n}$ and $\bar E_{\log \log n}$. Indeed their true objective is to prove that $P(\bar A,n,N_c)\to 0$, and they split the estimate in two parts. First they show that $P(\bar A\bar E_{\log \log n},n,N_c)\leq P(\bar E_{\log \log n},n,N_c)\to 0$, and then they show more directly that $P(\bar A E_{\log \log n},n,N_c)\to 0$.
(21) The formula contains $(-1)^k$ because it is a direct application of the inclusion-exclusion principle.
They want to count only the graphs without isolated vertices. So first they count ($k=0$) all the possible graphs. Then ($k=1$) they subtract the number of graphs with at least one chosen isolated vertex. Then they add up again the graphs with at least $k=2$ chosen (in $\binom n k$ ways) isolated vertices...
