I am looking at Bruen and Rotschild's article on lower bounds for line blocking sets in the finite projective plane of order $q$. In their proof on page 8, they use the summations \begin{align} \sum_{i=1}^{k} i n_{i}=(k+q)(q+1) \\ \sum_{i=1}^{k}\left(\begin{array}{c}{i} \\ {2}\end{array}\right) n_{i}=\left(\begin{array}{c}{k+q} \\ {2}\end{array}\right) \end{align} which I do seem to understand. $n_i$ denotes here the number of lines containing exactly $i$ points of the line blocking set, and we assume that the blocking set contains $n+k$ points. Here is the explanation the authors give:
(The first sum) counts in two ways the pairs (P; L) where P is a point of S and L a line containing it. (The second sum) counts all ({ P, Q} L) where P and Q are points of S on L.
However I do not understand this explanation. Could someone help?
The first paragraph of the proof explains why each line contains at least $1$ point of $S$, and at most $k$ points of $S$. Hence the left hand side of the first expression sums, for every line, the number of points of $S$ it contains. Of course every point of $S$ is contained in precisely $q+1$ lines, and so $$\sum_{i=1}^{k} i n_{i}=|S|\cdot(q+1)=(k+q)(q+1).$$ The left hand side of the second expression counts the number of ways to first choose any line, and then any pair of points on that line that are contained in $S$. Of course this is the same as the number of ways to choose two points from $S$, as any pair of points uniquely defines a line. Hence $$\sum_{i=1}^k\binom{i}{2}n_i=\binom{|S|}{2}=\binom{k+q}{2}.$$